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