Why does the standard basis vector $e_2$ lie in the kernal of this matrix? Doesn't $e_1$ also lie in it too? $$\pmatrix{0&0&1\\0&0&0\\0&0&0}$$
Simple matrix problem basis vectors
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matrices
ordinary-differential-equations
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0Because when you multiply them you get the zero vector. – 2012-11-28
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0@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? – 2012-11-28
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1$e_3$ isn't in the kernel, so maybe that's what they wanted to get at. – 2012-11-28
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1Yes. $e_1$ is in the kernel, too. – 2012-11-28
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0Thanks guys. My book is terrible at explaining. – 2012-11-28
1 Answers
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(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
$\pmatrix{0&0&1\\0&0&0\\0&0&0} \pmatrix{0\\0\\1}=\pmatrix{1\\0\\0}\neq \pmatrix{0\\0\\0}$.