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
${\begin{array}{rcl}\displaystyle {T{\bigg (}{\begin{bmatrix}1\\0\end{bmatrix}}+{\begin{bmatrix}0\\1\end{bmatrix}}{\bigg )}}&=&\displaystyle {T{\bigg (}{\begin{bmatrix}1\\1\end{bmatrix}}{\bigg )}}\\&&\\&=&\displaystyle {{\begin{bmatrix}0\\2\end{bmatrix}}.}\end{array}}$

Step 2:
On the other hand, notice
${\begin{array}{rcl}\displaystyle {T{\bigg (}{\begin{bmatrix}1\\0\end{bmatrix}}{\bigg )}+T{\bigg (}{\begin{bmatrix}0\\1\end{bmatrix}}{\bigg )}}&=&\displaystyle {{\begin{bmatrix}1\\1\end{bmatrix}}+{\begin{bmatrix}1\\1\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}2\\2\end{bmatrix}}.}\end{array}}$
So, now we know
$T{\bigg (}{\begin{bmatrix}1\\0\end{bmatrix}}+{\begin{bmatrix}0\\1\end{bmatrix}}{\bigg )}\neq T{\bigg (}{\begin{bmatrix}1\\0\end{bmatrix}}{\bigg )}+T{\bigg (}{\begin{bmatrix}0\\1\end{bmatrix}}{\bigg )}.$
Therefore,  $T$  is not a linear transformation.

(b)

Step 1:
Using the row-column rule for multiplication, we have
${\begin{array}{rcl}\displaystyle {AB}&=&\displaystyle {{\begin{bmatrix}1&-3&0\\-4&1&1\end{bmatrix}}{\begin{bmatrix}2&1\\1&0\\-1&1\end{bmatrix}}}\\&&\\&=&\displaystyle {\begin{bmatrix}1(2)+-3(1)+0(-1)&1(1)+-3(0)+0(1)\\-4(2)+1(1)+1(-1)&-4(1)+1(0)+1(1)\end{bmatrix}}\\&&\\&=&\displaystyle {{\begin{bmatrix}-1&1\\-8&-3\end{bmatrix}}.}\end{array}}$

Step 2:
Now,  $B$  and  $A^{T}$  are both  $3\times 2$  matrices.
Hence,  $BA^{T}$  is undefined.

Step 3:
For  $A-B^{T},$  we have
${\begin{array}{rcl}\displaystyle {A-B^{T}}&=&\displaystyle {{\begin{bmatrix}1&-3&0\\-4&1&1\end{bmatrix}}-{\begin{bmatrix}2&1&-1\\1&0&1\\\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}-1&-4&1\\-5&1&0\end{bmatrix}}.}\end{array}}$

Final Answer:
(a)     $T$  is not a linear transformation
(b)     $AB={\begin{bmatrix}-1&1\\-8&-3\end{bmatrix}},$  $BA^{T}$  is undefined and  $A-B^{T}={\begin{bmatrix}-1&-4&1\\-5&1&0\end{bmatrix}}$

Return to Review Problems