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Studying for my Algebra exam, and this question popped out with no solution in a previous exam:

Given a matrix $A$ such that $A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix},\ A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix}$.

(I) Is $0$ an eigenvalue of the matrix?

(II) Find a matrix like that, where the sum of its' eigenvalues is $0$.

So I (think) solved (I) but have no clue for (II). Here's my solution for (I):

$A \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + A \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} = A \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ -4 \\ 6 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \\ -6 \end{pmatrix} = 0$, and then, the vector $v = \begin{pmatrix} 2 \\ -1 \\0 \end{pmatrix}$ supplies that $Av = 0v = 0$ meaning that $0$ is an eigenvalue of $A$ with an eigenvector $v$.

Community
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TheNotMe
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2 Answers2

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The answer for (I) looks good.

Good for (II) might be to note that $$ A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}2\\4\\-6\end{bmatrix} $$ and $$ A\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}1\\2\\-3\end{bmatrix} $$ which gives the first two columns of $A$: $$ A=\begin{bmatrix}1&2&x\\2&4&y\\-3&-6&z\end{bmatrix} $$ The trace is the sum of the eigenvalues. Thus, if the sum of the eigenvalues is $0$, we need $z=-5$. $x$ and $y$ are arbitrary.

robjohn
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Hint: If a matrix $A$ has column vectors $a_1,..,a_n$, then $$A\cdot e_i=a_i$$ where $e_1=\pmatrix{1\\0\\0\\ \vdots}$, $e_2=\pmatrix{0\\1\\0\\ \vdots}$, $e_3=\pmatrix{0\\0\\1\\ \vdots}$...

Berci
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