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Why does the standard basis vector $e_2$ lie in the kernal of this matrix? Doesn't $e_1$ also lie in it too? $$\pmatrix{0&0&1\\0&0&0\\0&0&0}$$

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Because when you multiply them you get the zero vector. – ՃՃՃ Nov 28 '12 at 4:16
@use so that means $e_1$ is also in the kernal as well, right? Because the way my book discussed this problem made it seem as if $e_1$ wasn't in the kernal? – Q.matin Nov 28 '12 at 4:18
$e_3$ isn't in the kernel, so maybe that's what they wanted to get at. – Katie Dobbs Nov 28 '12 at 4:19
Yes. $e_1$ is in the kernel, too. – ՃՃՃ Nov 28 '12 at 4:19
Thanks guys. My book is terrible at explaining. – Q.matin Nov 28 '12 at 4:23
up vote 3 down vote accepted

(Just to avoid leaving an unanswered question.)

A vector $\vec{v}$ is in the kernel of a matrix $A$ if and only if $A\vec{v}=\vec{0}$. Since

$\pmatrix{0&0&1\\0&0&0\\0&0&0} \pmatrix{1\\0\\0}=\pmatrix{0\\0\\0}=\pmatrix{0&0&1\\0&0&0\\0&0&0} \pmatrix{0\\1\\0}$,

$e_1$ and $e_2$ belong to the kernel; while $e_3$ does not because

$\pmatrix{0&0&1\\0&0&0\\0&0&0} \pmatrix{0\\0\\1}=\pmatrix{1\\0\\0}\neq \pmatrix{0\\0\\0}$.

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