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Calculate the dimension of the image of $\tau: \mathbb{Q^{3,4}}\rightarrow\mathbb{Q^{3,3}}, X \mapsto AX$

$A := \begin{pmatrix} -1&3&2&1\\ 4&-12&-7&-5 \\ 3&-9&-4&-5 \end{pmatrix}$

My attempt:

1) Reduce A

$A' := \begin{pmatrix} -1&-3&0&-3\\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$

2) Dimension(A) = $n \cdot m = 3 \cdot 4 = 12$

3) Dimension(Kernel(A)) = $n - rank(A)$

$4 - rank(A) = 4 - 2 = 2$ /Edited

4) Dimensionformula: $Dim(A) - Dim(Kernel(\phi)) = Dim(Image(\phi))$

$12 - 2 = 10$

So the dimension of the image is supposed to be 11, which is weird to me, if you can point out missconceptions, I'd appreciate that very much.

1 Answers 1

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The kernel of $A$ is $2$-dimensional. (It should be $4-2=2$.) After fixing this, I think the rest of the work is ok.

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    First of all, thanks! I've got a followup question: Would the solutionspace be $\begin{pmatrix} 0\\0\\0\\0 \end{pmatrix}+ lin\begin{pmatrix} \begin{pmatrix} 3\\1\\0\\0 \end{pmatrix}, \begin{pmatrix} 3\\1\\0\\1 \end{pmatrix}\end{pmatrix} $? – 2017-01-28
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    @WhatAMesh To be clear, I mean that you need to change part (4) to $12-2 = 10$ as well. I also do not understand your followup question. – 2017-01-28
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    Oh yea, I forgot to do that. I mean the Kernel of A, so Ax = 0 – 2017-01-28