031 Review Part 2, Problem 7

(a) Let  $T:\mathbb {R} ^{2}\rightarrow \mathbb {R} ^{2}$  be a transformation given by

$T{\bigg (}{\begin{bmatrix}x\\y\end{bmatrix}}{\bigg )}={\begin{bmatrix}1-xy\\x+y\end{bmatrix}}.$

Determine whether  $T$  is a linear transformation. Explain.

(b) Let  $A={\begin{bmatrix}1&-3&0\\-4&1&1\end{bmatrix}}$  and  $B={\begin{bmatrix}2&1\\1&0\\-1&1\end{bmatrix}}.$  Find  $AB,~BA^{T}$  and  $A-B^{T}.$

Foundations:
A map  $T:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}$  is a linear transformation if
$T({\vec {x}}+{\vec {y}})=T({\vec {x}})+T({\vec {y}})$
and
$T(a{\vec {x}})=aT({\vec {x}})$
for all  ${\vec {x}},{\vec {y}}\in \mathbb {R} ^{n}$  and all  $a\in \mathbb {R} .$

Solution:

(a)

Step 1:
We claim that  $T$  is not a linear transformation.
Consider the vectors  ${\begin{bmatrix}1\\0\end{bmatrix}}$  and  ${\begin{bmatrix}0\\1\end{bmatrix}}.$
Then, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{T\bigg(\begin{bmatrix} 1\\ 0 \end{bmatrix}+\begin{bmatrix} 0\\ 1 \end{bmatrix}\bigg)} & = & \displaystyle{T\bigg(\begin{bmatrix} 1\\ 1 \end{bmatrix}\bigg)}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 0\\ 2 \end{bmatrix}.} \end{array}}

Step 2:
On the other hand, notice
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{T\bigg(\begin{bmatrix} 1\\ 0 \end{bmatrix}+\begin{bmatrix} 0\\ 1 \end{bmatrix}\bigg)} & = & \displaystyle{T\bigg(\begin{bmatrix} 1\\ 1 \end{bmatrix}\bigg)}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 0\\ 2 \end{bmatrix}.} \end{array}}

(b)

Step 1:

Step 2:

Final Answer:
(a)
(b)

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