031 Review Part 2, Problem 4

Suppose  $T$  is a linear transformation given by the formula

$T{\Bigg (}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\\end{bmatrix}}{\Bigg )}={\begin{bmatrix}5x_{1}-2.5x_{2}+10x_{3}\\-x_{1}+0.5x_{2}-2x_{3}\end{bmatrix}}$

(a) Find the standard matrix for  $T.$

(b) Let  ${\vec {u}}=7{\vec {e_{1}}}-4{\vec {e_{2}}}.$  Find  $T({\vec {u}}).$

(c) Is  ${\begin{bmatrix}-1\\3\end{bmatrix}}$  in the range of  $T?$  Explain.

Foundations:
1. The standard matrix of a linear transformation  $T:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}$  is given by
${\begin{bmatrix}T({\vec {e_{1}}})&T({\vec {e_{2}}})&\cdots &T({\vec {e_{n}}})\end{bmatrix}}$
where  $\{e_{1},e_{2},\ldots ,e_{n}\}$  is the standard basis of  $\mathbb {R} ^{n}.$
2. A vector  ${\vec {x}}$  is in the image of  $T$  if there exists  ${\vec {x}}$  such that
$T({\vec {x}})={\vec {v}}.$

Solution:

(a)

Step 1:
Notice, we have
$T({\vec {e_{1}}})={\begin{bmatrix}5\\-1\end{bmatrix}},T({\vec {e_{2}}})={\begin{bmatrix}-2.5\\0.5\end{bmatrix}},T({\vec {e_{3}}})={\begin{bmatrix}10\\-2\end{bmatrix}}.$

Step 2:
So, the standard matrix of  $T$  is
$[T]={\begin{bmatrix}5&-2.5&10\\-1&0.5&-2\end{bmatrix}}$

(b)

Step 1:
Since  $T$  is a linear transformation, we know
${\begin{array}{rcl}\displaystyle {T({\vec {u}})}&=&\displaystyle {T(7{\vec {e_{1}}}-4{\vec {e_{2}}})}\\&&\\&=&\displaystyle {T(7{\vec {e_{1}}})-T(4{\vec {e_{2}}})}\\&&\\&=&\displaystyle {7T({\vec {e_{1}}})-4T({\vec {e_{2}}}).}\end{array}}$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {T({\vec {u}})}&=&\displaystyle {7{\begin{bmatrix}5\\-1\end{bmatrix}}-4{\begin{bmatrix}-2.5\\0.5\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}35\\-7\end{bmatrix}}+{\begin{bmatrix}10\\-2\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}45\\-9\end{bmatrix}}.}\end{array}}$

(c)

Step 1:
To answer this question, we augment the standard matrix of  $T$  with this vector and row reduce this matrix.
So, we have the matrix
$\left[{\begin{array}{ccc|c}5&-2.5&10&-1\\-1&0.5&-2&3\end{array}}\right].$

Step 2:
Now, row reducing this matrix, we have
${\begin{array}{rcl}\displaystyle {\left[{\begin{array}{ccc|c}5&-2.5&10&-1\\-1&0.5&-2&3\end{array}}\right]}&\sim &\displaystyle {\left[{\begin{array}{ccc|c}5&-2.5&10&-1\\-5&2.5&-10&15\end{array}}\right]}\\&&\\&\sim &\displaystyle {\left[{\begin{array}{ccc|c}5&-2.5&10&-1\\0&0&0&14\end{array}}\right].}\end{array}}$
From here, we can tell that the corresponding system is inconsistent.
Hence, this vector is not in the range of  $T.$

Final Answer:
(a)     $[T]={\begin{bmatrix}5&-2.5&10\\-1&0.5&-2\end{bmatrix}}$
(b)     ${\begin{bmatrix}45\\-9\end{bmatrix}}$
(c)     See above

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