## ECE1505H Convex Optimization. Lecture 7: Examples of convex and concave functions, local and global minimums. Taught by Prof. Stark Draper

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper, from [1].

### Today

• Local and global optimality
• Compositions of functions
• Examples

### Example:

\label{eqn:convexOptimizationLecture7:20}
\begin{aligned}
F(x) &= x^2 \\
F”(x) &= 2 > 0
\end{aligned}

strictly convex.

### Example:

\label{eqn:convexOptimizationLecture7:40}
\begin{aligned}
F(x) &= x^3 \\
F”(x) &= 6 x.
\end{aligned}

Not always non-negative, so not convex. However $$x^3$$ is convex on $$\textrm{dom} F = \mathbb{R}_{+}$$.

### Example:

\label{eqn:convexOptimizationLecture7:60}
\begin{aligned}
F(x) &= x^\alpha \\
F'(x) &= \alpha x^{\alpha-1} \\
F”(x) &= \alpha(\alpha-1) x^{\alpha-2}.
\end{aligned}

fig. 1. Powers of x.

This is convex on $$\mathbb{R}_{+}$$, if $$\alpha \ge 1$$, or $$\alpha \le 0$$.

### Example:

\label{eqn:convexOptimizationLecture7:80}
\begin{aligned}
F(x) &= \log x \\
F'(x) &= \inv{x} \\
F”(x) &= -\inv{x^2} \le 0
\end{aligned}

This is concave.

### Example:

\label{eqn:convexOptimizationLecture7:100}
\begin{aligned}
F(x) &= x\log x \\
F'(x) &= \log x + x \inv{x} = 1 + \log x \\
F”(x) &= \inv{x}
\end{aligned}

This is strictly convex on
$$\mathbb{R}_{++}$$, where
$$F”(x) \ge 0$$.

### Example:

\label{eqn:convexOptimizationLecture7:120}
\begin{aligned}
F(x) &= e^{\alpha x} \\
F'(x) &= \alpha e^{\alpha x} \\
F”(x) &= \alpha^2 e^{\alpha x} \ge 0
\end{aligned}

fig. 2. Exponential.

Such functions are plotted in fig. 2, and are convex function for all $$\alpha$$.

### Example:

For symmetric $$P \in S^n$$

\label{eqn:convexOptimizationLecture7:140}
\begin{aligned}
F(\Bx) &= \Bx^\T P \Bx + 2 \Bq^\T \Bx + r \\
\spacegrad F &= (P + P^\T) \Bx + 2 \Bq = 2 P \Bx + 2 \Bq \\
\end{aligned}

This is convex(concave) if $$P \ge 0$$ ($$P \le 0$$).

### Example:

\label{eqn:convexOptimizationLecture7:780}
F(x, y) = x^2 + y^2 + 3 x y,

that is neither convex nor concave is plotted in fig 3.

fig 3. Function with saddle point (3d and contours)

This function can be put in matrix form

\label{eqn:convexOptimizationLecture7:160}
F(x, y) = x^2 + y^2 + 3 x y
=
\begin{bmatrix}
x & y
\end{bmatrix}
\begin{bmatrix}
1 & 1.5 \\
1.5 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix},

and has the Hessian

\label{eqn:convexOptimizationLecture7:180}
\begin{aligned}
&=
\begin{bmatrix}
\partial_{xx} F & \partial_{xy} F \\
\partial_{yx} F & \partial_{yy} F \\
\end{bmatrix} \\
&=
\begin{bmatrix}
2 & 3 \\
3 & 2
\end{bmatrix} \\
&= 2 P.
\end{aligned}

From the plot we know that this is not PSD, but this can be confirmed by checking the eigenvalues

\label{eqn:convexOptimizationLecture7:200}
\begin{aligned}
0
&=
\det ( P – \lambda I ) \\
&=
(1 – \lambda)^2 – 1.5^2,
\end{aligned}

which has solutions

\label{eqn:convexOptimizationLecture7:220}
\lambda = 1 \pm \frac{3}{2} = \frac{3}{2}, -\frac{1}{2}.

This is not PSD nor negative semi-definite, because it has one positive and one negative eigenvalues. This is neither convex nor concave.

Along $$y = -x$$,

\label{eqn:convexOptimizationLecture7:240}
\begin{aligned}
F(x,y)
&=
F(x,-x) \\
&=
2 x^2 – 3 x^2 \\
&=
– x^2,
\end{aligned}

so it is concave along this line. Along $$y = x$$

\label{eqn:convexOptimizationLecture7:260}
\begin{aligned}
F(x,y)
&=
F(x,x) \\
&=
2 x^2 + 3 x^2 \\
&=
5 x^2,
\end{aligned}

so it is convex along this line.

### Example:

\label{eqn:convexOptimizationLecture7:280}
F(\Bx) = \sqrt{ x_1 x_2 },

on $$\textrm{dom} F = \setlr{ x_1 \ge 0, x_2 \ge 0 }$$

For the Hessian
\label{eqn:convexOptimizationLecture7:300}
\begin{aligned}
\PD{x_1}{F} &= \frac{1}{2} x_1^{-1/2} x_2^{1/2} \\
\PD{x_2}{F} &= \frac{1}{2} x_2^{-1/2} x_1^{1/2}
\end{aligned}

The Hessian components are

\label{eqn:convexOptimizationLecture7:320}
\begin{aligned}
\PD{x_1}{} \PD{x_1}{F} &= -\frac{1}{4} x_1^{-3/2} x_2^{1/2} \\
\PD{x_1}{} \PD{x_2}{F} &= \frac{1}{4} x_2^{-1/2} x_1^{-1/2} \\
\PD{x_2}{} \PD{x_1}{F} &= \frac{1}{4} x_1^{-1/2} x_2^{-1/2} \\
\PD{x_2}{} \PD{x_2}{F} &= -\frac{1}{4} x_2^{-3/2} x_1^{1/2}
\end{aligned}

or
\label{eqn:convexOptimizationLecture7:340}
=
-\frac{\sqrt{x_1 x_2}}{4}
\begin{bmatrix}
\inv{x_1^2} & -\inv{x_1 x_2} \\
-\inv{x_1 x_2} & \inv{x_2^2}
\end{bmatrix}.

Checking this for PSD against $$\Bv = (v_1, v_2)$$, we have
\label{eqn:convexOptimizationLecture7:360}
\begin{aligned}
\begin{bmatrix}
v_1 & v_2
\end{bmatrix}
\begin{bmatrix}
\inv{x_1^2} & -\inv{x_1 x_2} \\
-\inv{x_1 x_2} & \inv{x_2^2}
\end{bmatrix}
\begin{bmatrix}
v_1 \\ v_2
\end{bmatrix}
&=
\begin{bmatrix}
v_1 & v_2
\end{bmatrix}
\begin{bmatrix}
\inv{x_1^2} v_1 -\inv{x_1 x_2} v_2 \\
-\inv{x_1 x_2} v_1 + \inv{x_2^2} v_2
\end{bmatrix} \\
&=
\lr{ \inv{x_1^2} v_1 -\inv{x_1 x_2} v_2 } v_1 +
\lr{ -\inv{x_1 x_2} v_1 + \inv{x_2^2} v_2 } v_2
\\
&=
\inv{x_1^2} v_1^2
+ \inv{x_2^2} v_2^2
-2 \inv{x_1 x_2} v_1 v_2 \\
&=
\lr{
\frac{v_1}{x_1}
-\frac{v_2}{x_2}
}^2 \\
&\ge 0,
\end{aligned}

so $$\spacegrad^2 F \le 0$$. This is a negative semi-definite function (concave). Observe that this check required checking PSD for all values of $$\Bx$$.

This is an example of a more general result

\label{eqn:convexOptimizationLecture7:380}
F(x) = \lr{ \prod_{i = 1}^n x_i }^{1/n},

which is concave (prove on homework).

### Summary.

If $$F$$ is differentiable in \R{n}, then check the curvature of the function along all lines. i.e. At all locations and in all directions.

If the Hessian is PSD at all $$\Bx \in \textrm{dom} F$$, that is

\label{eqn:convexOptimizationLecture7:400}
\spacegrad^2 F \ge 0 \, \forall \Bx \in \textrm{dom} F,

then the function is convex.

### Example:

Over $$\textrm{dom} F = \mathbb{R}^n$$

\label{eqn:convexOptimizationLecture7:420}
F(\Bx) = \max_{i = 1}^n x_i

i.e.
\label{eqn:convexOptimizationLecture7:440}
\begin{aligned}
F((1,2) &= 2 \\
F((3,-1) &= 3
\end{aligned}

### Example:

\label{eqn:convexOptimizationLecture7:460}
F(\Bx) = \max_{i = 1}^n F_i(\Bx),

where

\label{eqn:convexOptimizationLecture7:480}
F_i(\Bx)
=
… ?

max of a set of convex functions is a convex function.

### Example:

\label{eqn:convexOptimizationLecture7:500}
F(x) =
x_{[1]} +
x_{[2]} +
x_{[3]}

where

$$x_{[k]}$$ is the k-th largest number in the list

Write

\label{eqn:convexOptimizationLecture7:520}
F(x) = \max x_i + x_j + x_k

\label{eqn:convexOptimizationLecture7:540}
(i,j,k) \in \binom{n}{3}

### Example:

For $$\Ba \in \mathbb{R}^n$$ and $$b_i \in \mathbb{R}$$

\label{eqn:convexOptimizationLecture7:560}
\begin{aligned}
F(\Bx)
&= \sum_{i = 1}^n \log( b_i – \Ba^\T \Bx )^{-1} \\
&= -\sum_{i = 1}^n \log( b_i – \Ba^\T \Bx )
\end{aligned}

This $$b_i – \Ba^\T \Bx$$ is an affine function of $$\Bx$$ so it doesn’t affect convexity.

Since $$\log$$ is concave, $$-\log$$ is convex. Convex functions of affine function of $$\Bx$$ is convex function of $$\Bx$$.

### Example:

\label{eqn:convexOptimizationLecture7:580}
F(\Bx) = \sup_{\By \in C} \Norm{ \Bx – \By }

fig. 3. Max length function

Here $$C \subseteq \mathbb{R}^n$$ is not necessarily convex. We are using $$\sup$$ here because the set $$C$$ may be open. This function is the length of the line from $$\Bx$$ to the point in $$C$$ that is furthest from $$\Bx$$.

• $$\Bx – \By$$ is linear in $$\Bx$$
• $$g_\By(\Bx) = \Norm{\Bx – \By}$$ is convex in $$\Bx$$ since norms are convex functions.
• $$F(\Bx) = \sup_{\By \in C} \Norm{ \Bx – \By }$$. Each $$\By$$ index is a convex function. Taking max of those.

### Example:

\label{eqn:convexOptimizationLecture7:600}
F(\Bx) = \inf_{\By \in C} \Norm{ \Bx – \By }.

Min and max of two convex functions are plotted inĀ fig. 4.

fig. 4. Min and max

The max is observed to be convex, whereas the min is not necessarily so.

\label{eqn:convexOptimizationLecture7:800}
F(\Bz) = F(\theta \Bx + (1-\theta) \By) \ge \theta F(\Bx) + (1-\theta)F(\By).

This is not necessarily convex for all sets $$C \subseteq \mathbb{R}^n$$, because the $$\inf$$ of a bunch of convex function is not necessarily convex. However, if $$C$$ is convex, then $$F(\Bx)$$ is convex.

### Consequences of convexity for differentiable functions

• Think about unconstrained functions $$\textrm{dom} F = \mathbb{R}^n$$.
• By first order condition $$F$$ is convex iff the domain is convex and
\label{eqn:convexOptimizationLecture7:620}
F(\Bx) \ge \lr{ \spacegrad F(\Bx)}^\T (\By – \Bx) \, \forall \Bx, \By \in \textrm{dom} F.

If $$F$$ is convex and one can find an $$\Bx^\conj \in \textrm{dom} F$$ such that

\label{eqn:convexOptimizationLecture7:640}

then

\label{eqn:convexOptimizationLecture7:660}
F(\By) \ge F(\Bx^\conj) \, \forall \By \in \textrm{dom} F.

If you can find the point where the gradient is zero (which can’t always be found), then $$\Bx^\conj$$ is a global minimum of $$F$$.

Conversely, if $$\Bx^\conj$$ is a global minimizer of $$F$$, then $$\spacegrad F(\Bx^\conj) = 0$$ must hold. If that were not the case, then you would be able to find a direction to move downhill, contracting the optimality of $$\Bx^\conj$$.

### Local vs Global optimum

fig. 6. Global and local minimums

Definition: Local optimum
$$\Bx^\conj$$ is a local optimum of $$F$$ if $$\exists \epsilon > 0$$ such that $$\forall \Bx$$, $$\Norm{\Bx – \Bx^\conj} < \epsilon$$, we have

\begin{equation*}
F(\Bx^\conj) \le F(\Bx)
\end{equation*}

fig. 5. min length function

Theorem:
Suppose $$F$$ is twice continuously differentiable (not necessarily convex)

• If $$\Bx^\conj$$ is a local optimum then\begin{equation*}
\begin{aligned}
\end{aligned}
\end{equation*}
• If
\begin{equation*}
\begin{aligned}
\end{aligned},
\end{equation*}then $$\Bx^\conj$$ is a local optimum.

Proof:

• Let $$\Bx^\conj$$ be a local optimum. Pick any $$\Bv \in \mathbb{R}^n$$.\label{eqn:convexOptimizationLecture7:720}
\lim_{t \rightarrow 0} \frac{ F(\Bx^\conj + t \Bv) – F(\Bx^\conj)}{t}
= \lr{ \spacegrad F(\Bx^\conj) }^\T \Bv
\ge 0.

Here the fraction is $$\ge 0$$ since $$\Bx^\conj$$ is a local optimum.

Since the choice of $$\Bv$$ is arbitrary, the only case that you can ensure that $$\ge 0, \forall \Bv$$ is

\label{eqn:convexOptimizationLecture7:740}

( or else could pick $$\Bv = -\spacegrad F(\Bx^\conj)$$.

This means that $$\spacegrad F(\Bx^\conj) = 0$$ if $$\Bx^\conj$$ is a local optimum.

Consider the 2nd order derivative

\label{eqn:convexOptimizationLecture7:760}
\begin{aligned}
\lim_{t \rightarrow 0} \frac{ F(\Bx^\conj + t \Bv) – F(\Bx^\conj)}{t^2}
&=
\lim_{t \rightarrow 0} \inv{t^2}
\lr{
F(\Bx^\conj) + t \lr{ \spacegrad F(\Bx^\conj) }^\T \Bv + \inv{2} t^2 \Bv^\T \spacegrad^2 F(\Bx^\conj) \Bv + O(t^3)
– F(\Bx^\conj)
} \\
&=
\inv{2} \Bv^\T \spacegrad^2 F(\Bx^\conj) \Bv \\
&\ge 0.
\end{aligned}

Here the $$\ge$$ condition also comes from the fraction, based on the optimiality of $$\Bx^\conj$$. This is true for all choice of $$\Bv$$, thus $$\spacegrad^2 F(\Bx^\conj)$$.

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.

## ECE1505H Convex Optimization. Lecture 6: First and second order conditions. Taught by Prof.\ Stark Draper

February 1, 2017 ece1505 , , , , , , , ,

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper, from [1].

### Today

• First and second order conditions for convexity of differentiable functions.
• Consequences of convexity: local and global optimality.
• Properties.

### Quasi-convex

$$F_1$$ and $$F_2$$ convex implies $$\max( F_1, F_2)$$ convex.

fig. 1. Min and Max

Note that $$\min(F_1, F_2)$$ is NOT convex.

If $$F : \mathbb{R}^n \rightarrow \mathbb{R}$$ is convex, then $$F( \Bx_0 + t \Bv )$$ is convex in $$t\,\forall t \in \mathbb{R}, \Bx_0 \in \mathbb{R}^n, \Bv \in \mathbb{R}^n$$, provided $$\Bx_0 + t \Bv \in \textrm{dom} F$$.

Idea: Restrict to a line (line segment) in $$\textrm{dom} F$$. Take a cross section or slice through $$F$$ alone the line. If the result is a 1D convex function for all slices, then $$F$$ is convex.

This is nice since it allows for checking for convexity, and is also nice numerically. Attempting to test a given data set for non-convexity with some random lines can help disprove convexity. However, to show that $$F$$ is convex it is required to test all possible slices (which isn’t possible numerically, but is in some circumstances possible analytically).

### Differentiable (convex) functions

Definition: First order condition.

If

\begin{equation*}
F : \mathbb{R}^n \rightarrow \mathbb{R}
\end{equation*}

is differentiable, then $$F$$ is convex iff $$\textrm{dom} F$$ is a convex set and $$\forall \Bx, \Bx_0 \in \textrm{dom} F$$

\begin{equation*}
F(\Bx) \ge F(\Bx_0) + \lr{\spacegrad F(\Bx_0)}^\T (\Bx – \Bx_0).
\end{equation*}

This is the first order Taylor expansion. If $$n = 1$$, this is $$F(x) \ge F(x_0) + F'(x_0) ( x – x_0)$$.

The first order condition says a convex function \underline{always} lies above its first order approximation, as sketched in fig. 3.

fig. 2. First order approximation lies below convex function

When differentiable, the supporting plane is the tangent plane.

Definition: Second order condition

If $$F : \mathbb{R}^n \rightarrow \mathbb{R}$$ is twice differentiable, then $$F$$ is convex iff $$\textrm{dom} F$$ is a convex set and $$\spacegrad^2 F(\Bx) \ge 0 \,\forall \Bx \in \textrm{dom} F$$.

The Hessian is always symmetric, but is not necessarily positive. Recall that the Hessian is the matrix of the second order partials $$(\spacegrad F)_{ij} = \partial^2 F/(\partial x_i \partial x_j)$$.

The scalar case is $$F”(x) \ge 0 \, \forall x \in \textrm{dom} F$$.

An implication is that if $$F$$ is convex, then $$F(x) \ge F(x_0) + F'(x_0) (x – x_0) \,\forall x, x_0 \in \textrm{dom} F$$

Since $$F$$ is convex, $$\textrm{dom} F$$ is convex.

Consider any 2 points $$x, y \in \textrm{dom} F$$, and $$\theta \in [0,1]$$. Define

\label{eqn:convexOptimizationLecture6:60}
z = (1-\theta) x + \theta y \in \textrm{dom} F,

then since $$\textrm{dom} F$$ is convex

\label{eqn:convexOptimizationLecture6:80}
F(z) =
F( (1-\theta) x + \theta y )
\le
(1-\theta) F(x) + \theta F(y )

Reordering

\label{eqn:convexOptimizationLecture6:220}
\theta F(x) \ge
\theta F(x) + F(z) – F(x),

or
\label{eqn:convexOptimizationLecture6:100}
F(y) \ge
F(x) + \frac{F(x + \theta(y-x)) – F(x)}{\theta},

which is, in the limit,

\label{eqn:convexOptimizationLecture6:120}
F(y) \ge
F(x) + F'(x) (y – x),

completing one direction of the proof.

To prove the other direction, showing that

\label{eqn:convexOptimizationLecture6:140}
F(x) \ge F(x_0) + F'(x_0) (x – x_0),

implies that $$F$$ is convex. Take any $$x, y \in \textrm{dom} F$$ and any $$\theta \in [0,1]$$. Define

\label{eqn:convexOptimizationLecture6:160}
z = \theta x + (1 -\theta) y,

which is in $$\textrm{dom} F$$ by assumption. We want to show that

\label{eqn:convexOptimizationLecture6:180}
F(z) \le \theta F(x) + (1-\theta) F(y).

By assumption

1. $$F(x) \ge F(z) + F'(z) (x – z)$$
2. $$F(y) \ge F(z) + F'(z) (y – z)$$

Compute

\label{eqn:convexOptimizationLecture6:200}
\begin{aligned}
\theta F(x) + (1-\theta) F(y)
&\ge
\theta \lr{ F(z) + F'(z) (x – z) }
+ (1-\theta) \lr{ F(z) + F'(z) (y – z) } \\
&=
F(z) + F'(z) \lr{ \theta( x – z) + (1-\theta) (y-z) } \\
&=
F(z) + F'(z) \lr{ \theta x + (1-\theta) y – \theta z – (1 -\theta) z } \\
&=
F(z) + F'(z) \lr{ \theta x + (1-\theta) y – z} \\
&=
F(z) + F'(z) \lr{ z – z} \\
&= F(z).
\end{aligned}

### Proof of the 2nd order case for $$n = 1$$

Want to prove that if

\label{eqn:convexOptimizationLecture6:240}
F : \mathbb{R} \rightarrow \mathbb{R}

is a convex function, then $$F”(x) \ge 0 \,\forall x \in \textrm{dom} F$$.

By the first order conditions $$\forall x \ne y \in \textrm{dom} F$$

\label{eqn:convexOptimizationLecture6:260}
\begin{aligned}
F(y) &\ge F(x) + F'(x) (y – x)
F(x) &\ge F(y) + F'(y) (x – y)
\end{aligned}

Can combine and get

\label{eqn:convexOptimizationLecture6:280}
F'(x) (y-x) \le F(y) – F(x) \le F'(y)(y-x)

Subtract the two derivative terms for

\label{eqn:convexOptimizationLecture6:340}
\frac{(F'(y) – F'(x))(y – x)}{(y – x)^2} \ge 0,

or
\label{eqn:convexOptimizationLecture6:300}
\frac{F'(y) – F'(x)}{y – x} \ge 0.

In the limit as $$y \rightarrow x$$, this is
\label{eqn:convexOptimizationLecture6:320}
\boxed{
F”(x) \ge 0 \,\forall x \in \textrm{dom} F.
}

Now prove the reverse condition:

If $$F”(x) \ge 0 \,\forall x \in \textrm{dom} F \subseteq \mathbb{R}$$, implies that $$F : \mathbb{R} \rightarrow \mathbb{R}$$ is convex.

Note that if $$F”(x) \ge 0$$, then $$F'(x)$$ is non-decreasing in $$x$$.

i.e. If $$x < y$$, where $$x, y \in \textrm{dom} F$$, then

\label{eqn:convexOptimizationLecture6:360}
F'(x) \le F'(y).

Consider any $$x,y \in \textrm{dom} F$$ such that $$x < y$$, where

\label{eqn:convexOptimizationLecture6:380}
F(y) – F(x) = \int_x^y F'(t) dt \ge F'(x) \int_x^y 1 dt = F'(x) (y-x).

This tells us that

\label{eqn:convexOptimizationLecture6:400}
F(y) \ge F(x) + F'(x)(y – x),

which is the first order condition. Similarly consider any $$x,y \in \textrm{dom} F$$ such that $$x < y$$, where

\label{eqn:convexOptimizationLecture6:420}
F(y) – F(x) = \int_x^y F'(t) dt \le F'(y) \int_x^y 1 dt = F'(y) (y-x).

This tells us that

\label{eqn:convexOptimizationLecture6:440}
F(x) \ge F(y) + F'(y)(x – y).

### Vector proof:

$$F$$ is convex iff $$F(\Bx + t \Bv)$$ is convex $$\forall \Bx,\Bv \in \mathbb{R}^n, t \in \mathbb{R}$$, keeping $$\Bx + t \Bv \in \textrm{dom} F$$.

Let
\label{eqn:convexOptimizationLecture6:460}
h(t ; \Bx, \Bv) = F(\Bx + t \Bv)

then $$h(t)$$ satisfies scalar first and second order conditions for all $$\Bx, \Bv$$.

\label{eqn:convexOptimizationLecture6:480}
h(t) = F(\Bx + t \Bv) = F(g(t)),

where $$g(t) = \Bx + t \Bv$$, where

\label{eqn:convexOptimizationLecture6:500}
\begin{aligned}
F &: \mathbb{R}^n \rightarrow \mathbb{R} \\
g &: \mathbb{R} \rightarrow \mathbb{R}^n.
\end{aligned}

This is expressing $$h(t)$$ as a composition of two functions. By the first order condition for scalar functions we know that

\label{eqn:convexOptimizationLecture6:520}
h(t) \ge h(0) + h'(0) t.

Note that

\label{eqn:convexOptimizationLecture6:540}
h(0) = \evalbar{F(\Bx + t \Bv)}{t = 0} = F(\Bx).

Let’s figure out what $$h'(0)$$ is. Recall hat for any $$\tilde{F} : \mathbb{R}^n \rightarrow \mathbb{R}^m$$

\label{eqn:convexOptimizationLecture6:560}
D \tilde{F} \in \mathbb{R}^{m \times n},

and
\label{eqn:convexOptimizationLecture6:580}
{D \tilde{F}(\Bx)}_{ij} = \PD{x_j}{\tilde{F_i}(\Bx)}

This is one function per row, for $$i \in [1,m], j \in [1,n]$$. This gives

\label{eqn:convexOptimizationLecture6:600}
\begin{aligned}
\frac{d}{dt} F(\Bx + \Bv t)
&=
\frac{d}{dt} F( g(t) ) \\
&=
\frac{d}{dt} h(t) \\
&= D h(t) \\
&= D F(g(t)) \cdot D g(t)
\end{aligned}

The first matrix is in $$\mathbb{R}^{1\times n}$$ whereas the second is in $$\mathbb{R}^{n\times 1}$$, since $$F : \mathbb{R}^n \rightarrow \mathbb{R}$$ and $$g : \mathbb{R} \rightarrow \mathbb{R}^n$$. This gives

\label{eqn:convexOptimizationLecture6:620}
\frac{d}{dt} F(\Bx + \Bv t)
= \evalbar{D F(\tilde{\Bx})}{\tilde{\Bx} = g(t)} \cdot D g(t).

That first matrix is

\label{eqn:convexOptimizationLecture6:640}
\begin{aligned}
\evalbar{D F(\tilde{\Bx})}{\tilde{\Bx} = g(t)}
&=
\evalbar{
\lr{\begin{bmatrix}
\PD{\tilde{x}_1}{ F(\tilde{\Bx})} &
\PD{\tilde{x}_2}{ F(\tilde{\Bx})} & \cdots
\PD{\tilde{x}_n}{ F(\tilde{\Bx})}
\end{bmatrix}
}}{ \tilde{\Bx} = g(t) = \Bx + t \Bv } \\
&=
\evalbar{
}{
\tilde{\Bx} = g(t)
} \\
=
\end{aligned}

The second Jacobian is

\label{eqn:convexOptimizationLecture6:660}
D g(t)
=
D
\begin{bmatrix}
g_1(t) \\
g_2(t) \\
\vdots \\
g_n(t) \\
\end{bmatrix}
=
D
\begin{bmatrix}
x_1 + t v_1 \\
x_2 + t v_2 \\
\vdots \\
x_n + t v_n \\
\end{bmatrix}
=
\begin{bmatrix}
v_1 \\
v_1 \\
\vdots \\
v_n \\
\end{bmatrix}
=
\Bv.

so

\label{eqn:convexOptimizationLecture6:680}
h'(t) = D h(t) = \lr{ \spacegrad F(g(t))}^\T \Bv,

and
\label{eqn:convexOptimizationLecture6:700}
h'(0) = \lr{ \spacegrad F(g(0))}^\T \Bv
=

Finally

\label{eqn:convexOptimizationLecture6:720}
\begin{aligned}
F(\Bx + t \Bv)
&\ge h(0) + h'(0) t \\
&= F(\Bx) + \lr{ \spacegrad F(\Bx) }^\T (t \Bv) \\
&= F(\Bx) + \innerprod{ \spacegrad F(\Bx) }{ t \Bv}.
\end{aligned}

Which is true for all $$\Bx, \Bx + t \Bv \in \textrm{dom} F$$. Note that the quantity $$t \Bv$$ is a shift.

### Epigraph

Recall that if $$(\Bx, t) \in \textrm{epi} F$$ then $$t \ge F(\Bx)$$.

\label{eqn:convexOptimizationLecture6:740}
t \ge F(\Bx) \ge F(\Bx_0) + \lr{\spacegrad F(\Bx_0) }^\T (\Bx – \Bx_0),

or

\label{eqn:convexOptimizationLecture6:760}
0 \ge
-(t – F(\Bx_0)) + \lr{\spacegrad F(\Bx_0) }^\T (\Bx – \Bx_0),

In block matrix form

\label{eqn:convexOptimizationLecture6:780}
0 \ge
\begin{bmatrix}
\lr{ \spacegrad F(\Bx_0) }^\T & -1
\end{bmatrix}
\begin{bmatrix}
\Bx – \Bx_0 \\
t – F(\Bx_0)
\end{bmatrix}

With $$\Bw = \begin{bmatrix} \lr{ \spacegrad F(\Bx_0) }^\T & -1 \end{bmatrix}$$, the geometry of the epigraph relation to the half plane is sketched in fig. 3.

fig. 3. Half planes and epigraph.

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.

## ECE1505H Convex Optimization. Lecture 3: Matrix functions, SVD, and types of Sets. Taught by Prof. Stark Draper

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course ECE1505H, Convex Optimization, taught by Prof. Stark Draper.

## Matrix inner product

Given real matrices $$X, Y \in \mathbb{R}^{m\times n}$$, one possible matrix inner product definition is

\label{eqn:convexOptimizationLecture3:20}
\begin{aligned}
\innerprod{X}{Y}
&= \textrm{Tr}( X^\T Y) \\
&= \textrm{Tr} \lr{ \sum_{k = 1}^m X_{ki} Y_{kj} } \\
&= \sum_{k = 1}^m \sum_{j = 1}^n X_{kj} Y_{kj} \\
&= \sum_{i = 1}^m \sum_{j = 1}^n X_{ij} Y_{ij}.
\end{aligned}

This inner product induces a norm on the (matrix) vector space, called the Frobenius norm

\label{eqn:convexOptimizationLecture3:40}
\begin{aligned}
\Norm{X }_F
&= \textrm{Tr}( X^\T X) \\
&= \sqrt{ \innerprod{X}{X} } \\
&=
\sum_{i = 1}^m \sum_{j = 1}^n X_{ij}^2.
\end{aligned}

## Range, nullspace.

Definition: Range: Given $$A \in \mathbb{R}^{m \times n}$$, the range of A is the set:

\begin{equation*}
\mathcal{R}(A) = \setlr{ A \Bx | \Bx \in \mathbb{R}^n }.
\end{equation*}

Definition: Nullspace: Given $$A \in \mathbb{R}^{m \times n}$$, the nullspace of A is the set:

\begin{equation*}
\mathcal{N}(A) = \setlr{ \Bx | A \Bx = 0 }.
\end{equation*}

## SVD.

To understand operation of $$A \in \mathbb{R}^{m \times n}$$, a representation of a linear transformation from \R{n} to \R{m}, decompose $$A$$ using the singular value decomposition (SVD).

Definition: SVD: Given $$A \in \mathbb{R}^{m \times n}$$, an operator on $$\Bx \in \mathbb{R}^n$$, a decomposition of the following form is always possible

\begin{equation*}
\begin{aligned}
A &= U \Sigma V^\T \\
U &\in \mathbb{R}^{m \times r} \\
V &\in \mathbb{R}^{n \times r},
\end{aligned}
\end{equation*}

where $$r$$ is the rank of $$A$$, and both $$U$$ and $$V$$ are orthogonal

\begin{equation*}
\begin{aligned}
U^\T U &= I \in \mathbb{R}^{r \times r} \\
V^\T V &= I \in \mathbb{R}^{r \times r}.
\end{aligned}
\end{equation*}

Here $$\Sigma = \textrm{diag}( \sigma_1, \sigma_2, \cdots, \sigma_r )$$, is a diagonal matrix of “singular” values, where

\begin{equation*}
\sigma_1 \ge \sigma_2 \ge \cdots \ge \sigma_r.
\end{equation*}

For simplicity consider square case $$m = n$$

\label{eqn:convexOptimizationLecture3:100}
A \Bx = \lr{ U \Sigma V^\T } \Bx.

The first product $$V^\T \Bx$$ is a rotation, which can be checked by looking at the length

\label{eqn:convexOptimizationLecture3:120}
\begin{aligned}
\Norm{ V^\T \Bx}_2
&= \sqrt{ \Bx^\T V V^\T \Bx } \\
&= \sqrt{ \Bx^\T \Bx } \\
&= \Norm{ \Bx }_2,
\end{aligned}

which shows that the length of the vector is unchanged after application of the linear transformation represented by $$V^\T$$ so that operation must be a rotation.

Similarly the operation of $$U$$ on $$\Sigma V^\T \Bx$$ also must be a rotation. The operation $$\Sigma = [\sigma_i]_i$$ applies a scaling operation to each component of the vector $$V^\T \Bx$$.

All linear (square) transformations can therefore be thought of as a rotate-scale-rotate operation. Often the $$A$$ of interest will be symmetric $$A = A^\T$$.

## Set of symmetric matrices

Let $$S^n$$ be the set of real, symmetric $$n \times n$$ matrices.

Theorem: Spectral theorem: When $$A \in S^n$$ then it is possible to factor $$A$$ as

\begin{equation*}
A = Q \Lambda Q^\T,
\end{equation*}

where $$Q$$ is an orthogonal matrix, and $$\Lambda = \textrm{diag}( \lambda_1, \lambda_2, \cdots \lambda_n)$$. Here $$\lambda_i \in \mathbb{R} \, \forall i$$ are the (real) eigenvalues of $$A$$.

A real symmetric matrix $$A \in S^n$$ is “positive semi-definite” if

\begin{equation*}
\Bv^\T A \Bv \ge 0 \qquad\forall \Bv \in \mathbb{R}^n, \Bv \ne 0,
\end{equation*}
and is “positive definite” if

\begin{equation*}
\Bv^\T A \Bv > 0 \qquad\forall \Bv \in \mathbb{R}^n, \Bv \ne 0.
\end{equation*}

The set of such matrices is denoted $$S^n_{+}$$, and $$S^n_{++}$$ respectively.

Consider $$A \in S^n_{+}$$ (or $$S^n_{++}$$ )

\label{eqn:convexOptimizationLecture3:200}
A = Q \Lambda Q^\T,

possible since the matrix is symmetric. For such a matrix

\label{eqn:convexOptimizationLecture3:220}
\begin{aligned}
\Bv^\T A \Bv
&=
\Bv^\T Q \Lambda A^\T \Bv \\
&=
\Bw^\T \Lambda \Bw,
\end{aligned}

where $$\Bw = A^\T \Bv$$. Such a product is

\label{eqn:convexOptimizationLecture3:240}
\Bv^\T A \Bv
=
\sum_{i = 1}^n \lambda_i w_i^2.

So, if $$\lambda_i \ge 0$$ ($$\lambda_i > 0$$ ) then $$\sum_{i = 1}^n \lambda_i w_i^2$$ is non-negative (positive) $$\forall \Bw \in \mathbb{R}^n, \Bw \ne 0$$. Since $$\Bw$$ is just a rotated version of $$\Bv$$ this also holds for all $$\Bv$$. A necessary and sufficient condition for $$A \in S^n_{+}$$ ($$S^n_{++}$$ ) is $$\lambda_i \ge 0$$ ($$\lambda_i > 0$$).

## Square root of positive semi-definite matrix

Real symmetric matrix power relationships such as

\label{eqn:convexOptimizationLecture3:260}
A^2
=
Q \Lambda Q^\T
Q \Lambda Q^\T
=
Q \Lambda^2
Q^\T
,

or more generally $$A^k = Q \Lambda^k Q^\T,\, k \in \mathbb{Z}$$, can be further generalized to non-integral powers. In particular, the square root (non-unique) of a square matrix can be written

\label{eqn:convexOptimizationLecture3:280}
A^{1/2} = Q
\begin{bmatrix}
\sqrt{\lambda_1} & & & \\
& \sqrt{\lambda_2} & & \\
& & \ddots & \\
& & & \sqrt{\lambda_n} \\
\end{bmatrix}
Q^\T,

since $$A^{1/2} A^{1/2} = A$$, regardless of the sign picked for the square roots in question.

## Functions of matrices

Consider $$F : S^n \rightarrow \mathbb{R}$$, and define

\label{eqn:convexOptimizationLecture3:300}
F(X) = \log \det X,

Here $$\textrm{dom} F = S^n_{++}$$. The task is to find $$\spacegrad F$$, which can be done by looking at the perturbation $$\log \det ( X + \Delta X )$$

\label{eqn:convexOptimizationLecture3:320}
\begin{aligned}
\log \det ( X + \Delta X )
&=
\log \det ( X^{1/2} (I + X^{-1/2} \Delta X X^{-1/2}) X^{1/2} ) \\
&=
\log \det ( X (I + X^{-1/2} \Delta X X^{-1/2}) ) \\
&=
\log \det X + \log \det (I + X^{-1/2} \Delta X X^{-1/2}).
\end{aligned}

Let $$X^{-1/2} \Delta X X^{-1/2} = M$$ where $$\lambda_i$$ are the eigenvalues of $$M : M \Bv = \lambda_i \Bv$$ when $$\Bv$$ is an eigenvector of $$M$$. In particular

\label{eqn:convexOptimizationLecture3:340}
(I + M) \Bv =
(1 + \lambda_i) \Bv,

where $$1 + \lambda_i$$ are the eigenvalues of the $$I + M$$ matrix. Since the determinant is the product of the eigenvalues, this gives

\label{eqn:convexOptimizationLecture3:360}
\begin{aligned}
\log \det ( X + \Delta X )
&=
\log \det X +
\log \prod_{i = 1}^n (1 + \lambda_i) \\
&=
\log \det X +
\sum_{i = 1}^n \log (1 + \lambda_i).
\end{aligned}

If $$\lambda_i$$ are sufficiently “small”, then $$\log ( 1 + \lambda_i ) \approx \lambda_i$$, giving

\label{eqn:convexOptimizationLecture3:380}
\log \det ( X + \Delta X )
=
\log \det X +
\sum_{i = 1}^n \lambda_i
\approx
\log \det X +
\textrm{Tr}( X^{-1/2} \Delta X X^{-1/2} ).

Since
\label{eqn:convexOptimizationLecture3:400}
\textrm{Tr}( A B ) = \textrm{Tr}( B A ),

this trace operation can be written as

\label{eqn:convexOptimizationLecture3:420}
\log \det ( X + \Delta X )
\approx
\log \det X +
\textrm{Tr}( X^{-1} \Delta X )
=
\log \det X +
\innerprod{ X^{-1}}{\Delta X},

so
\label{eqn:convexOptimizationLecture3:440}

To check this, consider the simplest example with $$X \in \mathbb{R}^{1 \times 1}$$, where we have

\label{eqn:convexOptimizationLecture3:460}
\frac{d}{dX} \lr{ \log \det X } = \frac{d}{dX} \lr{ \log X } = \inv{X} = X^{-1}.

This is a nice example demonstrating how the gradient can be obtained by performing a first order perturbation of the function. The gradient can then be read off from the result.

## Second order perturbations

• To get first order approximation found the part that varied linearly in $$\Delta X$$.
• To get the second order part, perturb $$X^{-1}$$ by $$\Delta X$$ and see how that perturbation varies in $$\Delta X$$.

For $$G(X) = X^{-1}$$, this is

\label{eqn:convexOptimizationLecture3:480}
\begin{aligned}
(X + \Delta X)^{-1}
&=
\lr{ X^{1/2} (I + X^{-1/2} \Delta X X^{-1/2} ) X^{1/2} }^{-1} \\
&=
X^{-1/2} (I + X^{-1/2} \Delta X X^{-1/2} )^{-1} X^{-1/2}
\end{aligned}

To be proven in the homework (for “small” A)

\label{eqn:convexOptimizationLecture3:500}
(I + A)^{-1} \approx I – A.

This gives

\label{eqn:convexOptimizationLecture3:520}
\begin{aligned}
(X + \Delta X)^{-1}
&=
X^{-1/2} (I – X^{-1/2} \Delta X X^{-1/2} ) X^{-1/2} \\
&=
X^{-1} – X^{-1} \Delta X X^{-1},
\end{aligned}

or

\label{eqn:convexOptimizationLecture3:800}
\begin{aligned}
G(X + \Delta X)
&= G(X) + (D G) \Delta X \\
&= G(X) + (\spacegrad G)^\T \Delta X,
\end{aligned}

so
\label{eqn:convexOptimizationLecture3:820}
=
– X^{-1} \Delta X X^{-1}.

The Taylor expansion of $$F$$ to second order is

\label{eqn:convexOptimizationLecture3:840}
F(X + \Delta X)
=
F(X)
+
\textrm{Tr} \lr{ (\spacegrad F)^\T \Delta X}
+
\inv{2}
\lr{ (\Delta X)^\T (\spacegrad^2 F) \Delta X}.

The first trace can be expressed as an inner product

\label{eqn:convexOptimizationLecture3:860}
\begin{aligned}
\textrm{Tr} \lr{ (\spacegrad F)^\T \Delta X}
&=
\innerprod{ \spacegrad F }{\Delta X} \\
&=
\innerprod{ X^{-1} }{\Delta X}.
\end{aligned}

The second trace also has the structure of an inner product

\label{eqn:convexOptimizationLecture3:880}
\begin{aligned}
(\Delta X)^\T (\spacegrad^2 F) \Delta X
&=
\textrm{Tr} \lr{ (\Delta X)^\T (\spacegrad^2 F) \Delta X} \\
&=
\innerprod{ (\spacegrad^2 F)^\T \Delta X }{\Delta X},
\end{aligned}

where a no-op trace could be inserted in the second order term since that quadratic form is already a scalar. This $$(\spacegrad^2 F)^\T \Delta X$$ term has essentially been found implicitly by performing the linear variation of $$\spacegrad F$$ in $$\Delta X$$, showing that we must have

\label{eqn:convexOptimizationLecture3:900}
\textrm{Tr} \lr{ (\Delta X)^\T (\spacegrad^2 F) \Delta X}
=
\innerprod{ – X^{-1} \Delta X X^{-1} }{\Delta X},

so
\label{eqn:convexOptimizationLecture3:560}
F( X + \Delta X) = F(X) +
\innerprod{X^{-1}}{\Delta X}
+\inv{2} \innerprod{-X^{-1} \Delta X X^{-1}}{\Delta X},

or
\label{eqn:convexOptimizationLecture3:580}
\log \det ( X + \Delta X) = \log \det X +
\textrm{Tr}( X^{-1} \Delta X )
– \inv{2} \textrm{Tr}( X^{-1} \Delta X X^{-1} \Delta X ).

## Convex Sets

• Types of sets: Affine, convex, cones
• Examples: Hyperplanes, polyhedra, balls, ellipses, norm balls, cone of PSD matrices.

Definition: Affine set:

A set $$C \subseteq \mathbb{R}^n$$ is affine if $$\forall \Bx_1, \Bx_2 \in C$$ then

\begin{equation*}
\theta \Bx_1 + (1 -\theta) \Bx_2 \in C, \qquad \forall \theta \in \mathbb{R}.
\end{equation*}

The affine sum above can
be rewritten as

\label{eqn:convexOptimizationLecture3:600}
\Bx_2 + \theta (\Bx_1 – \Bx_2).

Since $$\theta$$ is a scaling, this is the line containing $$\Bx_2$$ in the direction between $$\Bx_1$$ and $$\Bx_2$$.

Observe that the solution to a set of linear equations

\label{eqn:convexOptimizationLecture3:620}
C = \setlr{ \Bx | A \Bx = \Bb },

is an affine set. To check, note that

\label{eqn:convexOptimizationLecture3:640}
\begin{aligned}
A (\theta \Bx_1 + (1 – \theta) \Bx_2)
&=
\theta A \Bx_1 + (1 – \theta) A \Bx_2 \\
&=
\theta \Bb + (1 – \theta) \Bb \\
&= \Bb.
\end{aligned}

Definition: Affine combination: An affine combination of points $$\Bx_1, \Bx_2, \cdots \Bx_n$$ is

\begin{equation*}
\sum_{i = 1}^n \theta_i \Bx_i,
\end{equation*}

such that for $$\theta_i \in \mathbb{R}$$

\begin{equation*}
\sum_{i = 1}^n \theta_i = 1.
\end{equation*}

An affine set contains all affine combinations of points in the set. Examples of a couple affine sets are sketched inĀ fig 1.1

For comparison, a couple of non-affine sets are sketched inĀ fig 1.2

Definition: Convex set: A set $$C \subseteq \mathbb{R}^n$$ is convex if $$\forall \Bx_1, \Bx_2 \in C$$ and $$\forall \theta \in \mathbb{R}, \theta \in [0,1]$$, the combination

\label{eqn:convexOptimizationLecture3:700}
\theta \Bx_1 + (1 – \theta) \Bx_2 \in C.

Definition: Convex combination: A convex combination of $$\Bx_1, \Bx_2, \cdots \Bx_n$$ is

\begin{equation*}
\sum_{i = 1}^n \theta_i \Bx_i,
\end{equation*}

such that $$\forall \theta_i \ge 0$$

\begin{equation*}
\sum_{i = 1}^n \theta_i = 1
\end{equation*}

Definition: Convex hull: Convex hull of a set $$C$$ is a set of all convex combinations of points in $$C$$, denoted

\label{eqn:convexOptimizationLecture3:720}
\textrm{conv}(C) = \setlr{ \sum_{i=1}^n \theta_i \Bx_i | \Bx_i \in C, \theta_i \ge 0, \sum_{i=1}^n \theta_i = 1 }.

A non-convex set can be converted into a convex hull by filling in all the combinations of points connecting points in the set, as sketched in fig 1.3.

Definition: Cone: A set $$C$$ is a cone if $$\forall \Bx \in C$$ and $$\forall \theta \ge 0$$ we have $$\theta \Bx \in C$$.

This scales out if $$\theta > 1$$ and scales in if $$\theta < 1$$.

A convex cone is a cone that is also a convex set. A conic combination is

\begin{equation*}
\sum_{i=1}^n \theta_i \Bx_i, \theta_i \ge 0.
\end{equation*}

A convex and non-convex 2D cone is sketched in fig. 1.4

A comparison of properties for different set types is tabulated in table 1.1

## Hyperplanes and half spaces

Definition: Hyperplane: A hyperplane is defined by

\begin{equation*}
\setlr{ \Bx | \Ba^\T \Bx = \Bb, \Ba \ne 0 }.
\end{equation*}

A line and plane are examples of this general construct as sketched in
fig. 1.5

An alternate view is possible should one
find any specific $$\Bx_0$$ such that $$\Ba^\T \Bx_0 = \Bb$$

\label{eqn:convexOptimizationLecture3:740}
\setlr{\Bx | \Ba^\T \Bx = b }
=
\setlr{\Bx | \Ba^\T (\Bx -\Bx_0) = 0 }

This shows that $$\Bx – \Bx_0 = \Ba^\perp$$ is perpendicular to $$\Ba$$, or

\label{eqn:convexOptimizationLecture3:780}
\Bx
=
\Bx_0 + \Ba^\perp.

This is the subspace perpendicular to $$\Ba$$ shifted by $$\Bx_0$$, subject to $$\Ba^\T \Bx_0 = \Bb$$. As a set

\label{eqn:convexOptimizationLecture3:760}
\Ba^\perp = \setlr{ \Bv | \Ba^\T \Bv = 0 }.

## Half space

Definition: Half space: The half space is defined as
\begin{equation*}
\setlr{ \Bx | \Ba^\T \Bx = \Bb }
= \setlr{ \Bx | \Ba^\T (\Bx – \Bx_0) \le 0 }.
\end{equation*}

This can also be expressed as $$\setlr{ \Bx | \innerprod{ \Ba }{\Bx – \Bx_0 } \le 0 }$$.

## Jacobian and Hessian matrices

January 15, 2017 ece1505 , , , , , ,

## Motivation

In class this Friday the Jacobian and Hessian matrices were introduced, but I did not find the treatment terribly clear. Here is an alternate treatment, beginning with the gradient construction from [2], which uses a nice trick to frame the multivariable derivative operation as a single variable Taylor expansion.

## Multivariable Taylor approximation

The Taylor series expansion for a scalar function $$g : {\mathbb{R}} \rightarrow {\mathbb{R}}$$ about the origin is just

\label{eqn:jacobianAndHessian:20}
g(t) = g(0) + t g'(0) + \frac{t^2}{2} g”(0) + \cdots

In particular

\label{eqn:jacobianAndHessian:40}
g(1) = g(0) + g'(0) + \frac{1}{2} g”(0) + \cdots

Now consider $$g(t) = f( \Bx + \Ba t )$$, where $$f : {\mathbb{R}}^n \rightarrow {\mathbb{R}}$$, $$g(0) = f(\Bx)$$, and $$g(1) = f(\Bx + \Ba)$$. The multivariable Taylor expansion now follows directly

\label{eqn:jacobianAndHessian:60}
f( \Bx + \Ba)
= f(\Bx)
+ \evalbar{\frac{df(\Bx + \Ba t)}{dt}}{t = 0} + \frac{1}{2} \evalbar{\frac{d^2f(\Bx + \Ba t)}{dt^2}}{t = 0} + \cdots

The first order term is

\label{eqn:jacobianAndHessian:80}
\begin{aligned}
\evalbar{\frac{df(\Bx + \Ba t)}{dt}}{t = 0}
&=
\sum_{i = 1}^n
\frac{d( x_i + a_i t)}{dt}
\evalbar{\PD{(x_i + a_i t)}{f(\Bx + \Ba t)}}{t = 0} \\
&=
\sum_{i = 1}^n
a_i
\PD{x_i}{f(\Bx)} \\
\end{aligned}

Similarily, for the second order term

\label{eqn:jacobianAndHessian:100}
\begin{aligned}
\evalbar{\frac{d^2 f(\Bx + \Ba t)}{dt^2}}{t = 0}
&=
\evalbar{\lr{
\frac{d}{dt}
\lr{
\sum_{i = 1}^n
a_i
\PD{(x_i + a_i t)}{f(\Bx + \Ba t)}
}
}
}{t = 0} \\
&=
\evalbar{
\lr{
\sum_{j = 1}^n
\frac{d(x_j + a_j t)}{dt}
\sum_{i = 1}^n
a_i
\frac{\partial^2 f(\Bx + \Ba t)}{\partial (x_j + a_j t) \partial (x_i + a_i t) }
}
}{t = 0} \\
&=
\sum_{i,j = 1}^n a_i a_j \frac{\partial^2 f}{\partial x_i \partial x_j} \\
&=
\end{aligned}

The complete Taylor expansion of a scalar function $$f : {\mathbb{R}}^n \rightarrow {\mathbb{R}}$$ is therefore

\label{eqn:jacobianAndHessian:120}
f(\Bx + \Ba)
= f(\Bx) +
\inv{2} \lr{ \Ba \cdot \spacegrad}^2 f + \cdots,

so the Taylor expansion has an exponential structure

\label{eqn:jacobianAndHessian:140}
f(\Bx + \Ba) = \sum_{k = 0}^\infty \inv{k!} \lr{ \Ba \cdot \spacegrad}^k f = e^{\Ba \cdot \spacegrad} f.

Should an approximation of a vector valued function $$\Bf : {\mathbb{R}}^n \rightarrow {\mathbb{R}}^m$$ be desired it is only required to form a matrix of the components

\label{eqn:jacobianAndHessian:160}
\Bf(\Bx + \Ba)
= \Bf(\Bx) +
\inv{2} [\lr{ \Ba \cdot \spacegrad}^2 f_i]_i + \cdots,

where $$[.]_i$$ denotes a column vector over the rows $$i \in [1,m]$$, and $$f_i$$ are the coordinates of $$\Bf$$.

## The Jacobian matrix

In [1] the Jacobian $$D \Bf$$ of a function $$\Bf : {\mathbb{R}}^n \rightarrow {\mathbb{R}}^m$$ is defined in terms of the limit of the $$l_2$$ norm ratio

\label{eqn:jacobianAndHessian:180}
\frac{\Norm{\Bf(\Bz) – \Bf(\Bx) – (D \Bf) (\Bz – \Bx)}_2 }{ \Norm{\Bz – \Bx}_2 },

with the statement that the function $$\Bf$$ has a derivative if this limit exists. Here the Jacobian $$D \Bf \in {\mathbb{R}}^{m \times n}$$ must be matrix valued.

Let $$\Bz = \Bx + \Ba$$, so the first order expansion of \ref{eqn:jacobianAndHessian:160} is

\label{eqn:jacobianAndHessian:200}
\Bf(\Bz)
= \Bf(\Bx) + [\lr{ \Bz – \Bx } \cdot \spacegrad f_i]_i
.

With the (unproven) assumption that this Taylor expansion satisfies the norm limit criteria of \ref{eqn:jacobianAndHessian:180}, it is possible to extract the structure of the Jacobian by comparison

\label{eqn:jacobianAndHessian:220}
\begin{aligned}
(D \Bf)
(\Bz – \Bx)
&=
{\begin{bmatrix}
\lr{ \Bz – \Bx } \cdot \spacegrad f_i
\end{bmatrix}}_i \\
&=
{\begin{bmatrix}
\sum_{j = 1}^n (z_j – x_j) \PD{x_j}{f_i}
\end{bmatrix}}_i \\
&=
{\begin{bmatrix}
\PD{x_j}{f_i}
\end{bmatrix}}_{ij}
(\Bz – \Bx),
\end{aligned}

so
\label{eqn:jacobianAndHessian:240}
\boxed{
(D \Bf)_{ij} = \PD{x_j}{f_i}
}

Written out explictly as a matrix the Jacobian is

\label{eqn:jacobianAndHessian:320}
D \Bf
=
\begin{bmatrix}
\PD{x_1}{f_1} & \PD{x_2}{f_1} & \cdots & \PD{x_n}{f_1} \\
\PD{x_1}{f_2} & \PD{x_2}{f_2} & \cdots & \PD{x_n}{f_2} \\
\vdots & \vdots & & \vdots \\
\PD{x_1}{f_m} & \PD{x_2}{f_m} & \cdots & \PD{x_n}{f_m} \\
\end{bmatrix}
=
\begin{bmatrix}
\vdots \\
\end{bmatrix}.

In particular, when the function is scalar valued
\label{eqn:jacobianAndHessian:261}

With this notation, the first Taylor expansion, in terms of the Jacobian matrix is

\label{eqn:jacobianAndHessian:260}
\boxed{
\Bf(\Bz)
\approx \Bf(\Bx) + (D \Bf) \lr{ \Bz – \Bx }.
}

## The Hessian matrix

For scalar valued functions, the text expresses the second order expansion of a function in terms of the Jacobian and Hessian matrices

\label{eqn:jacobianAndHessian:271}
f(\Bz)
\approx f(\Bx) + (D f) \lr{ \Bz – \Bx }
+ \inv{2} \lr{ \Bz – \Bx }^\T (\spacegrad^2 f) \lr{ \Bz – \Bx }.

Because $$\spacegrad^2$$ is the usual notation for a Laplacian operator, this $$\spacegrad^2 f \in {\mathbb{R}}^{n \times n}$$ notation for the Hessian matrix is not ideal in my opinion. Ignoring that notational objection for this class, the structure of the Hessian matrix can be extracted by comparison with the coordinate expansion

\label{eqn:jacobianAndHessian:300}
=
\sum_{r,s = 1}^n a_r a_s \frac{\partial^2 f}{\partial x_r \partial x_s}

so
\label{eqn:jacobianAndHessian:280}
\boxed{
=
\frac{\partial^2 f_i}{\partial x_i \partial x_j}.
}

In explicit matrix form the Hessian is

\label{eqn:jacobianAndHessian:340}
=
\begin{bmatrix}
\frac{\partial^2 f}{\partial x_1 \partial x_1} & \frac{\partial^2 f}{\partial x_1 \partial x_2} & \cdots &\frac{\partial^2 f}{\partial x_1 \partial x_n} \\
\frac{\partial^2 f}{\partial x_2 \partial x_1} & \frac{\partial^2 f}{\partial x_2 \partial x_2} & \cdots &\frac{\partial^2 f}{\partial x_2 \partial x_n} \\
\vdots & \vdots & & \vdots \\
\frac{\partial^2 f}{\partial x_n \partial x_1} & \frac{\partial^2 f}{\partial x_n \partial x_2} & \cdots &\frac{\partial^2 f}{\partial x_n \partial x_n}
\end{bmatrix}.

Is there a similar nice matrix structure for the Hessian of a function $$f : {\mathbb{R}}^n \rightarrow {\mathbb{R}}^m$$?

# References

[1] Stephen Boyd and Lieven Vandenberghe. Convex optimization. Cambridge university press, 2004.

[2] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.