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$$T: \mathbb{R^2} \longrightarrow \mathbb{R^2}$$ $$Ker(T) = \mathbb{R}(1, 2)$$ $$T(2, 3) = (5, -2)$$

The matrix for the linear function is (this is the solution): $$\begin{vmatrix} 10 \quad -5 \\ -4 \quad 2 \end{vmatrix}$$ But, knowing that a linear function is uniquely determined by the vector on its basis (of $\mathbb{R^2}$ in this case), and that $\begin{vmatrix} 1 \\ 2 \end{vmatrix}$ and $\begin{vmatrix} 2 \\ 3 \end{vmatrix}$ are vectors of a $\mathbb{R^2}$ basis, I can only get $\begin{vmatrix}0 \quad 5 \\ 0 \quad -2 \end{vmatrix}$ as matrix of the $T$ linear function.

Can you help me, please?

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    I don't understand: **what's** the question here? Clearly the matrix, or the operator, have rank one here, and since $\;(1,2)\;$ is a basis for its kernel, then both matrices shown represent $\;T\;$ wrt different basis of $\;\Bbb R^2\;$ ...so?? – 2017-01-18
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    The question is: how can I get that solution? The matrix I got and the correct one have same rank, but can I simply change my matrix to the correct one? Or is there something else I missed? – 2017-01-18
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    Matrix of the operator $T$ in the solution is in respect to standard basis of $\mathbb R^2$ $$(1,0),(0,1)$$ not $(1,2), (2,3)$. – 2017-01-18
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    But I don't know how T is made: I only know that $T(2, 3) = (5, -2)$ and $T(t, 2t) = (0, 0)$, and I thought I could use a non-standard basis of $\mathbb{R^2}$ – 2017-01-18
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    You _can_ use another basis, but then you'll get a different matrix (as the matrix of a linear transformation depends on the basis chosen). Perhaps you are _supposed to_ use the standard basis, i.e. the question might (implicitly?) be that you need to find the matrix w.r.t. the standard basis. – 2017-01-18
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    So I should change my matrix to that of the standard basis using a "change of basis" matrix? – 2017-01-18
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    That's one option (where you can still use the matrix you set up so far); the other one would be to try and set it up in the standard basis immediately. – 2017-01-18

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Denote $e_1=(1,0)$ and $e_2=(0,1)$. So $(1,2)=e_1+2e_2$, $(2,3)=2e_1+3e_2$ and $(5,-2)=5e_1-2e_2$. We have

$$\begin{align} T(1,2)=&T(e_1)+2T(e_2)=(0,0)\\ T(2,3)=&2T(e_1)+3T(e_2)=(5,-2) \end{align}$$

Substituting $T(e_1)=-2T(e_2)$ in the second equation one gets

$$-T(e_2)=(5,-2)$$

And so $T(e_1)=(10,-4)$ and the matrix we're looking for is

$$\begin{bmatrix} 10 & -5\\-4 & 2\end{bmatrix}$$