Matrix Calculus - Linear Algebra

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Question

True or False, the Constrained Extremum Theorem only applies to skew-symmetric matrices.

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Answer

It only applies to symmetric matrices, not skew-symmetric ones. The Constrained Extremum Theorem concerns the maximum and minimum values of the quadratic form when $|\mathbf{x}|=1$ .

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Question

The maximum value of a quadratic form ( is an $n \times n$ symmetric matrix, $|\mathbf{x}|=1$ ) corresponds to which eigenvalue of ?

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Answer

This is the statement of the Constrained Extremum Theorem. Likewise, the minimum value of the quadratic form corresponds to the smallest eigenvalue of .

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Question

What is the the gradient vector of the following function?

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Answer

Recall that

$\nabla_x f(x)=\begin{bmatrix} $\frac{\partial f}{\partial x_1}$\ \ $\frac{\partial f}{\partial x_2}$ \ \ $\frac{\partial f}{\partial x_3}$ \ \vdots \ $\frac{\partial f}{\partial x_n}$ \end{bmatrix}$

All we need to do is calculate 3 partial derivatives, and put them into this form.

$$\frac{\partial f}{\partial x}$=18yz+100xz$

$$\frac{\partial f}{\partial y}$=18xz+30y$

$\frac{\partial f}{\partial $z}$=18xy+50x^2$

Put these into vector form to get

$\nabla_x f(x)=\begin{bmatrix} 18yz+100xz\ 18xz+30y \ $18xy+50x^2$ \end{bmatrix}$

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Question

Find the gradient vector of the following function.

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Answer

To find the gradient vector, we need to find the partial derivatives in respect to x and y.

$\frac{\partial f}{\partial $x}$=2xy^2$+\frac{10}{x}$

$\frac{\partial f}{\partial $y}$=2yx^2$+\frac{10}{y}$

Then our final answer looks like

$\begin{bmatrix} $2xy^2$+\frac{10}{x}$\ \ $2yx^2$+\frac{10}{y}$ \end{bmatrix}$

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Question

$\begin{align}&$\text{Determine the gradient, }$\nabla$\text{, of the function}$\&f(x,y)=5xcos(y)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=5xcos(y)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[cos(u)]=-sin(u)du\&d[a^u$$]=a^uduln$(a)\&\nabla f=\begin{bmatrix}5cos(y)\-5xsin(y)\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Determine the gradient, }$\nabla$\text{, of the function}$\&f(x,y)=15\cdot $4^{(2x)}$ln(3y) - $15xy^{2}$\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=15\cdot $4^{(2x)}$ln(3y) - $15xy^{2}$\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[sin(u)]=cos(u)du\&d[a^u$$]=a^uduln$$(a)\&d[ln(u)]=\frac{du}{u}$\&d[e^u$$]=e^udu$\&\nabla f=\begin{bmatrix}30\cdot $4^{(2x)}$ln(3y)ln(4) - $15y^{2}$\$\frac{(15\cdot $4^{(2x)}$$)}{y}- 30xy\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Determine the gradient, }$\nabla$\text{, of the function}$\&f(x,y,z)=12ln(3y)ln(2z)tan(2x)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y,z)=12ln(3y)ln(2z)tan(2x)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&d[ln(u)]=\frac{du}{u}$\&\nabla f=\begin{bmatrix}12ln(3y)ln(2z)\cdot $(2tan(2x)^{2}$ + 2)\$\frac{(12ln(2z)tan(2x))}{y}$\$\frac{(12ln(3y)tan(2x))}{z}$\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Find the vector }$\nabla f $\text{ for $}$f(x,y)=9x^{2}$tan(2y)\end{align}$

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Answer

$\begin{align}&$\text{In asking for }$\nabla$\text{, the question asks for the gradient.}$\&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y)=9x^{2}$tan(2y)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&\nabla $f=\begin{bmatrix}18xtan(2y)\9x^{2}$\cdot $(2tan(2y)^{2}$ + 2)\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Determine the gradient, }$\nabla$\text{, of the function}$\&f(x,y,z)=- $xcos(3z)e^{(5y)}$ - 20\cdot $2^{(3z)}$$y^{2}$ln(4x)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y,z)=- $xcos(3z)e^{(5y)}$ - 20\cdot $2^{(3z)}$$y^{2}$ln(4x)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[e^u$$]=e^udu$$\&d[cos(u)]=-sin(u)du\&d[ln(u)]=\frac{du}{u}$\&d[a^u$$]=a^uduln$(a)\&\nabla f=\begin{bmatrix}- $cos(3z)e^{(5y)}$ -$\frac{ (20\cdot $2^{(3z)}$$$y^{2}$)}{x}\- $5xcos(3z)e^{(5y)}$ - 40\cdot $2^{(3z)}$$yln(4x)\3xsin(3z)e^{(5y)}$ - 60\cdot $2^{(3z)}$$y^{2}$ln(4x)ln(2)\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Determine the gradient, }$\nabla$\text{, of the $function}$\&f(x,y)=-18e^{(5x)}$tan(y)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y)=-18e^{(5x)}$tan(y)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[e^u$$]=e^udu$$\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&\nabla $f=\begin{bmatrix}-90e^{(5x)}$$tan(y)\-18e^{(5x)}$\cdot $(tan(y)^{2}$ + 1)\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Calculate the gradient of the $function}$\&f(x,y,z)=14tan(5y)e^{(4z)}$tan(x) - 13\cdot $5^{(4x)}$ln(2z)tan(2y)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y,z)=14tan(5y)e^{(4z)}$tan(x) - 13\cdot $5^{(4x)}$ln(2z)tan(2y)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$$(u)+1)du\&d[e^u$$]=e^udu$$\&d[a^u$$]=a^uduln$(a)\&d[ln(u)]=\frac{du}{u}$\&\nabla $f=\begin{bmatrix}14tan(5y)e^{(4z)}$\cdot $(tan(x)^{2}$ + 1) - 52\cdot $5^{(4x)}$$ln(2z)tan(2y)ln(5)\14e^{(4z)}$tan(x)\cdot $(5tan(5y)^{2}$ + 5) - 13\cdot $5^{(4x)}$ln(2z)\cdot $(2tan(2y)^{2}$ + $2)\56tan(5y)e^{(4z)}$tan(x) -$\frac{ (13\cdot $5^{(4x)}$$tan(2y))}{z}\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Determine the gradient, }$\nabla$\text{, of the function}$\&f(x,y)=11ysin(5x) + $11x^{2}$tan(y)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=11ysin(5x) + $11x^{2}$tan(y)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&d[sin(u)]=cos(u)du\&\nabla f=\begin{bmatrix}55ycos(5x) + 22xtan(y)\11sin(5x) + $11x^{2}$\cdot $(tan(y)^{2}$ + 1)\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Calculate the gradient of the function}$\&f(x,y)=4ycos(5x) - 13ln(3y)sin(3x)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=4ycos(5x) - 13ln(3y)sin(3x)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[sin(u)]=cos(u)du\&d[ln(u)]=\frac{du}{u}$\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&d[cos(u)]=-sin(u)du\&\&\nabla f=\begin{bmatrix}- 20ysin(5x) - 39cos(3x)ln(3y)\4cos(5x) -$\frac{ (13sin(3x))}{y}$\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Determine the gradient, }$\nabla$\text{, of the function}$\&f(x,y)=13\cdot $2^{(6y)}$ln(3x)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y)=13\cdot $2^{(6y)}$ln(3x)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[ln(u)]=\frac{du}{u}$\&d[a^u$$]=a^uduln$(a)\&d[sin(u)]=cos(u)du\&\nabla f=\begin{bmatrix}$\frac{(13\cdot $2^{(6y)}$$)}{x}\78\cdot $2^{(6y)}$ln(3x)ln(2)\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Calculate the gradient of the $function}$\&f(x,y,z)=-9xsin(6y)e^{(3z)}$\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y,z)=-9xsin(6y)e^{(3z)}$\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[sin(u)]=cos(u)du\&d[e^u$$]=e^udu$\&\nabla $f=\begin{bmatrix}-9sin(6y)e^{(3z)}$$\-54xcos(6y)e^{(3z)}$$\-27xsin(6y)e^{(3z)}$\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Calculate the gradient of the $function}$\&f(x,y,z)=-18x^{2}$yln(2z)\end{align}$

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Answer

$\begin{align}&$\text{The gradient of a function is a vector of first derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$f$}{d_{x_$1}$$}$\frac{d^$f$}{d_{x_2}$$$}\...$\frac{df}{d_{x_n}$$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y,z)=-18x^{2}$yln(2z)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot g'\&d[ln(u)]=\frac{du}{u}$\&\nabla $f=\begin{bmatrix}-36xyln(2z)\-18x^{2}$$ln(2z)\-$\frac{(18x^{2}$$y)}{z}\end{bmatrix}\end{align}$

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Question

$f(x,y ) = (x+2y+3 $)^{4}$$

Evaluate the gradient vector of at .

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Answer

The gradient of is the vector of partial first derivatives

$\left ( $f_{x}$, $f_{y }$ \right )$ . Find these derivatives:

$= $\frac{\partial }{\partial x}$ (x+2y+ 3) \cdot4 \cdot (x+2y+3) ^{3}$

$=1 \cdot4 \cdot (x+2y+3) ^{3}$

$=4 (x+2y+3) ^{3}$

$= $\frac{\partial }{\partial y}$ (x+2y+ 3) \cdot 4 \cdot (x+2y+3) ^{3}$

$=2 \cdot 4 \cdot (x+2y+3) ^{3}$

$=8(x+2y+3) ^{3}$

Evaluate and :

$= 4 \cdot $3^{3 }$$

$= 4 \cdot 27$

$= 8 \cdot $3^{3 }$$

$= 8 \cdot 27$

The gradient vector is

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Question

Define $\textbf{f} : $\mathbb{R}^{2}$ \rightarrow $\mathbb{R}^{2}$$ as follows:

$\textbf{f} (x,y) = ( \cos (xy), \cos (x+y ))$

Evaluate the Jacobian matrix of $\textbf{f}$ at .

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Answer

The Jacobian matrix of a function $\textbf{f}(x,y) = $(f_{1}$(x,y), $f_{2}$(x,y))$ is the matrix of partial first derivatives

$\begin{bmatrix} $f_{1 x}$ & $f_{1 y}$ \ \ $f_{2 x}$ & $f_{2 y}$ \end{bmatrix}$

Find each partial first derivative, and evaluate the expression at .

The Jacobian matrix at is $\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$ .

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Question

$\begin{align}&$\text{Find the matrix }$Hf $\text{ for $}$f(x,y,z)=13y^{2}$tan(2x)tan(4z)\end{align}$

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Answer

$\begin{align}&$\text{In asking for H, the question asks for the Hessian.}$\&$\text{The Hessian of a function is a matrix of second derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$2f$}{d_{x_$1}$$^2$$}&...&$\frac{d^$2f$}{d_{x_1}$$$d_{x_$n}$}\...&...&...$\frac{d^$2f$}{d_{x_n}$$$d_{x_$1}$}&...&$\frac{d^$2f$}{d_{x_$n}$$^2$}\end{bmatrix}\&$\text{Considering our function: $}$f(x,y,z)=13y^{2}$tan(2x)tan(4z)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$$(u)+1)du\&Hf=\begin{bmatrix}52y^{2}$tan(2x)tan(4z)\cdot $(2tan(2x)^{2}$ + 2)&26ytan(4z)\cdot $(2tan(2x)^{2}$ + $2)&13y^{2}$\cdot $(2tan(2x)^{2}$ + 2)\cdot $(4tan(4z)^{2}$ + 4)\26ytan(4z)\cdot $(2tan(2x)^{2}$ + 2)&26tan(2x)tan(4z)&26ytan(2x)\cdot $(4tan(4z)^{2}$ + $4)\13y^{2}$\cdot $(2tan(2x)^{2}$ + 2)\cdot $(4tan(4z)^{2}$ + 4)&26ytan(2x)\cdot $(4tan(4z)^{2}$ + $4)&104y^{2}$tan(2x)tan(4z)\cdot $(4tan(4z)^{2}$ + 4)\end{bmatrix}\end{align}$

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Question

$\begin{align}&$\text{Determine the Hessian, }$H$\text{, of the function}$\&f(x,y,z)=-cos(6x)tan(5y)sin(5z)\end{align}$

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Answer

$\begin{align}&$\text{The Hessian of a function is a matrix of second derivatives taken with}$\&$\text{respect to its constituent variables. The form is as $follows:}$\&\begin{bmatrix}$\frac{d^$2f$}{d_{x_$1}$$^2$$}&...&$\frac{d^$2f$}{d_{x_1}$$$d_{x_$n}$}\...&...&...$\frac{d^$2f$}{d_{x_n}$$$d_{x_$1}$}&...&$\frac{d^$2f$}{d_{x_$n}$$^2$}\end{bmatrix}\&$\text{Considering our function: }$f(x,y,z)=-cos(6x)tan(5y)sin(5z)\&$\text{And utilizing derivative rules:}$\&(f\circ g )'=(f'\circ g )\cdot $g'\&d[cos(u)]=-sin(u)du\&d[tan(u)]=\frac{du}{cos^2$$(u)}$=(tan^2$(u)+1)du\&d[sin(u)]=cos(u)du\&Hf=\begin{bmatrix}36cos(6x)tan(5y)sin(5z)&6sin(6x)sin(5z)\cdot $(5tan(5y)^{2}$ + 5)&30cos(5z)sin(6x)tan(5y)\6sin(6x)sin(5z)\cdot $(5tan(5y)^{2}$ + 5)&-10cos(6x)tan(5y)sin(5z)\cdot $(5tan(5y)^{2}$ + 5)&-5cos(6x)cos(5z)\cdot $(5tan(5y)^{2}$ + 5)\30cos(5z)sin(6x)tan(5y)&-5cos(6x)cos(5z)\cdot $(5tan(5y)^{2}$ + 5)&25cos(6x)tan(5y)sin(5z)\end{bmatrix}\end{align}$

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