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If = $$ a = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}\in \mathbb{R}^2\;\;\;\;T\left(\begin{bmatrix}x_1\\x_2\end{bmatrix}\right) = \begin{bmatrix} 1&2\\3&6\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$$ Identify all vectors in $\mathbb{R}^2$ that are taken by $T$ to the same vector $T(a)$.

Here's the full question linked here: screenshot of full problem

I figured out part A of the problem and found a basis for both $\ker(T)$ and $\operatorname{Im}(T)$.

$$ \ker(T) = \operatorname{span} \left\langle\begin{bmatrix} 2 \\ 1 \end{bmatrix}\right\rangle $$and$$ $$

$$ \operatorname{im}(T) = \operatorname{span}\left\langle \begin{bmatrix} 1 \\ 3 \end{bmatrix}\right\rangle $$

Both being $1$-dimensional.

I need some help with part B of the question.

Thanks.

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2 Answers 2

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Hint:

If $v \in $Ker$(T)$ then

$T(a+v)=T(a)+0=T(a)$

and, if $T(x)=T(a)$ then

$T(x)-T(a)=T(x-a)=0$

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$T(a + k\begin{bmatrix}-2\\1\end{bmatrix}) =T(a)+kT(\begin{bmatrix}-2\\1\end{bmatrix}) = T(a)$

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  • $\begingroup$ is my solution for ker(T) wrong? I did RREF on the matrix and solutions i got were {x1 = 2t, x2 =t) perhaps, it should be x1 = -2t? please correct me if i'm wrong. thank u.. $\endgroup$
    – A-Choo
    Mar 15, 2018 at 22:35
  • $\begingroup$ $\begin{bmatrix} 1&2\\3&6\end{bmatrix}\begin{bmatrix} 2\\1\end{bmatrix} = \begin{bmatrix}4\\12\end{bmatrix}$ $\endgroup$
    – Doug M
    Mar 15, 2018 at 22:37
  • $\begingroup$ thank you! [-2 1] would result in [0 0] which is what we want. Gotcha, thanks for ur help $\endgroup$
    – A-Choo
    Mar 15, 2018 at 22:48

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