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Let $T\colon M_{23} \to M_{33}$ be the linear transformation defined by $T(A)=\begin{pmatrix} 2 & -1 \\ 1 & 2 \\ 3&1 \end{pmatrix}\,A$, for $A\in M_{23}$. Find a basis for the kernel and range of $T$.

I don't know how to exactly approach this question. All I know is that the kernel of $T$ would be the nullspace. I row reduced $\begin{pmatrix} 2 & -1 \\ 1 & 2 \\ 3&1 \end{pmatrix}$ and got $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0&0 \end{pmatrix}$ but I don't know what to do from here.

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    @Gerry, yes but it doesn't matter: by checking the rank and nullity of the matrix definining the map is possible to deduce stuff...2012-06-27

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Let me give you a hint. If you write down a $2 \times 3$ matrix $A$ and compute $T(A)$ you can find what the range looks like and you can also compute the kernel from this. More explicitly, for an arbitrary $2 \times 3$ matrix

$ A = \begin{bmatrix} a & c & e\\ b & d & f\\ \end{bmatrix} $

you will get

\begin{align*} T(A) = \begin{bmatrix} 2 &-1\\ 1 & 2\\ 3 & 1 \end{bmatrix} \begin{bmatrix} a & c & e\\ b & d & f\\ \end{bmatrix} = \begin{bmatrix} 2a - b & 2c - d & 2e - f\\ a + 2b & c + 2d & e + 2f\\ 3a + b & 3c + d & 3e + f\\ \end{bmatrix} \end{align*}

Now, this tells you how a matrix in the range of $T$ looks like.

Can you find a basis for the range using this?

Also, for such a matrix $A$ to be in the kernel, you must have $T(A) = (0)_{3 \times 3}$. Then you can find conditions for $a, b, c, d, e, f$ so that you get the zero matrix and with that information you can find what a matrix in the kernel looks like and in that way find a basis for the kernel.

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Note that for $A$ to be in the kernel, each column of $A$ must be orthogonal to each row of that $3\times2$ matrix. What can you find that's orthogonal to both $(2,-1)$ and $(1,2)$?