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Let $\ f:\mathbb R^3 \to \mathbb R^3$ be a linear transformation such that:

$\ker (f) = \operatorname{span} \{ (1,0,0) , (0,1,0) \}$ and $\operatorname{Im} (f) \subseteq \operatorname{span} \{ (0,1,0),(0,0,1) \}.$

Find all associated matrices of $f$ with respect to canonical basis.

Can anybody help me with this?

What I tried is basically useless, the only think I concluded that the matrix should have rank $1$ since dimension of kernel space is $2$, this is an old exam question with which I have a problem and have been breaking my head for already more than half an hour...

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    $\begingroup$ Please provide more details, for instance, what are your thoughts about the question, what you have tried, where you are stuck. $\endgroup$ Jun 26 '17 at 21:14
  • $\begingroup$ Can you provide a generic matrix with kernel as above? $\endgroup$
    – Doug M
    Jun 26 '17 at 21:15
  • $\begingroup$ Please edit that into your question. $\endgroup$ Jun 26 '17 at 21:20
  • $\begingroup$ The part with the image is also confusing me $\endgroup$ Jun 26 '17 at 21:20
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    $\begingroup$ Yes a generic matrix would be any such that first and second columns are zero, at least i think so, and yes there should be a span, i forgot to put the sign $\endgroup$ Jun 26 '17 at 21:24
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(I'm writing up the answer in the comments.)

Note the following fact about $3 \times 3$ matrices (which generalises to higher dimensions): \begin{align*} \left(\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)\left(\begin{array}{c}1 \\ 0 \\ 0\end{array}\right) &= \left(\begin{array}{c}a_{11} \\ a_{21} \\ a_{31}\end{array}\right) \\ \left(\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)\left(\begin{array}{c}0 \\ 1 \\ 0\end{array}\right) &= \left(\begin{array}{c}a_{12} \\ a_{22} \\ a_{32}\end{array}\right) \\ \left(\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)\left(\begin{array}{c}0 \\ 0 \\ 1\end{array}\right) &= \left(\begin{array}{c}a_{13} \\ a_{23} \\ a_{33}\end{array}\right). \end{align*} Put simply, multiplying a matrix by the $k$th (column) vector in the standard basis, returns the $k$th column in the matrix.

Fix some $f$ satisfying the conditions in the question, and let $M$ be its standard matrix. Since we know $(1, 0, 0), (0, 1, 0) \in \operatorname{ker} f$, we know that $$M\left(\begin{array}{c}1 \\ 0 \\ 0\end{array}\right) = M\left(\begin{array}{c}0 \\ 1 \\ 0\end{array}\right) = \left(\begin{array}{c}0 \\ 0 \\ 0\end{array}\right).$$ So, using the above fact, the first two columns of $M$ must contain only zeros. We similarly know the third column of $M$ must be non-zero, otherwise, $$(0, 0, 1) \in \operatorname{ker} f = \operatorname{span}\lbrace (1, 0, 0), (0, 1, 0), (0, 0, 1) \rbrace = \mathbb{R}^3.$$ We are further restricted in what the third column can contain by the restriction on the image of $f$. Since it must be contained in $\operatorname{span}\lbrace (0, 1, 0), (0, 0, 1) \rbrace$, we must have $$M\left(\begin{array}{c}0 \\ 0 \\ 1\end{array}\right) = \left(\begin{array}{c}0 \\ a \\ b\end{array}\right)$$ for some $a, b \in \mathbb{R}$, which forms the third column. Thus, $M$ must take the form, $$M = \left(\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & a \\ 0 & 0 & b\end{array}\right)$$ for some $a, b \in \mathbb{R}$, with $a \neq 0$ or $b \neq 0$. It is not difficult to verify that every such matrix satisfies the conditions, so we have our characterisation.

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