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For an $a \in \mathbb{R}$ let $\phi_a: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear mapping such that $\phi_a(x) := \begin{pmatrix} 1 & 2 & 2 \\1 & 3 & 5 \\ 1 & -1 & a \\ \end{pmatrix}$.

If $a = -7$ $\phi$ is not bijective.

Now I am looking for a matrice $B$ such that a linear mapping $\psi : \mathbb{R}^3 \rightarrow \mathbb{R}^3$, $x \rightarrow B\cdot x$ guarantees that$Ker(\psi) = Im(\phi_a)$ where $a=-7$

I am looking for a general approach rather then a solution.

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  • $\begingroup$ Think about the properties of the kernel. What is required for it to be in 1-1 correspondence? How can we relate this to the dimension of the kernel? Of the image? $\endgroup$ – Dohleman Sep 20 '13 at 16:25
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For $a=7$ the last column is linearly dependent with the first two, so it does not affect the image of $\phi_7$. Now you are looking for a nonzero matrix $B$ satisfying $$ B\cdot \begin{pmatrix} 1 & 2 \\1 & 3 \\ 1 & -1\\ \end{pmatrix} =0. $$ That should not be hard, a $1\times 3$ matrix $B=(b_1~b_2~b_3)$ will do. The two entries of the $1\times 2$ product that should vanisg give you $2$ linear homogeneous equations in $3$ unknowns for which you want a non trivial solution; standard equation solving should easily provide such a solution. If you want a square matrix, just as null rows.

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  • $\begingroup$ One more question: Why can we input the image of $\phi$ as $x$ in the kernel $Bx=0$? This step is not yet clear to me. $\endgroup$ – Laura Sep 20 '13 at 16:44
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    $\begingroup$ You question asks to make $\ker(\psi)$ equal to $Im(\phi_7)$, so it seems natural to take the output of $\phi_7$ as input to $\psi$ (i.e., to $B$) and to ask that the result be zero. $\endgroup$ – Marc van Leeuwen Sep 20 '13 at 17:08

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