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Does there exist a linear transformation $T : R^3 → R^3$ such that $ T \left(\begin{bmatrix} 6 \\ -3 \\ 3 \end{bmatrix}\right) $ = $ \begin{bmatrix} 1 \\2 \\ 3 \end{bmatrix} $ and $ T \left(\begin{bmatrix} 4 \\ -2 \\ 2 \end{bmatrix}\right) $ = $ \begin{bmatrix} 2 \\4 \\ 6 \end{bmatrix} $ How do I go about solving this problem? Looking for some help with my midterm review

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Observe that

$$\begin{pmatrix}4\\-2\\2\end{pmatrix}=\frac23\begin{pmatrix}6\\-3\\3\end{pmatrix}$$ , so it must be

$$T\begin{pmatrix}4\\-2\\2\end{pmatrix}=\frac23\;T\begin{pmatrix}6\\-3\\3\end{pmatrix}$$

because of linearity. Is this true?

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  • $\begingroup$ @geohelp Good. Then there cannot exist such... linear...transformation. $\endgroup$ – DonAntonio Oct 29 '16 at 21:57
  • $\begingroup$ @geohelp Precisely. And if asked, you must argue that a linear transformation fulfills, for any vector $\;v\;$ and any scalar $\;k\;$ , that $\;T(kv)=kTv\;$ , and since in the present case this isn't so the transformation cannot be linear. $\endgroup$ – DonAntonio Oct 29 '16 at 22:02
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Notice that the input vectors are multiples of each other and that the output vectors are multiples of each other. Since those multiples are not the same, such a transform doesn't exist.

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  • $\begingroup$ is that all I would need to say? is there a mathematical representation of what you just explained? $\endgroup$ – bjp409 Oct 29 '16 at 21:55
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If $T$ exists, we will have

$T(6,-3,3)=T(3(2,-1,1))=3T(2,-1,1)=(1,2,3)$

thus $T(2,-1,1)=\frac{1}{3}(1,2,3)$

and

$T(4,-2,2)=T(2(2,-1,1))=2T(2,-1,1)=(2,4,6)$

which gives

$T(2,-1,1)=(1,2,3)\neq \frac{1}{3}(1,2,3)$

so, such a linear application doesn't exist.

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Hint Both input vectors are multiples of $\mathbf{u}=\begin{bmatrix}2\\-1\\1\end{bmatrix}$. Then $T(\lambda\mathbf{u})=\lambda(T(\mathbf{u}))$.

Added explanation:

You are given that $T(3\mathbf{u})=\begin{bmatrix}1\\2\\3\end{bmatrix}$, this means $$T(3\mathbf{u})=3T(\mathbf{u})=\begin{bmatrix}1\\2\\3\end{bmatrix}.$$ But when you use the other condition, you get $$T(2\mathbf{u})=2T(\mathbf{u})=\begin{bmatrix}2\\4\\6\end{bmatrix}.$$

So you are getting two different values for $T(\mathbf{u})$. Hence such a T cannot exist.

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  • $\begingroup$ okay so where do I go from there? $\endgroup$ – bjp409 Oct 29 '16 at 21:56

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