Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
$e_3$ isn't in the kernel, so maybe that's what they wanted to get at. –  Katie Dobbs Nov 28 '12 at 4:19
1  
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

1 Answer 1

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.