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Finding the basis for the kernel of:

\begin{pmatrix} a & b \\c & d\end{pmatrix}

$which$ $maps$ $to:$

\begin{pmatrix} a \\a\\3a + b \end{pmatrix}

It's all complex, but I'm not sure if that's relevant!

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I think it is about time you go to the FAQ section and read about how to use LaTeX to properly write mathematics in this site. –  DonAntonio Nov 28 '12 at 17:48
    
I can figure out how you (probably) intended the matrix to be formatted, but what do you mean by "C^3(a, a, 3a+d)"? As DonAntonio points out, we won't run into such troubles if you format the posts, yourself. –  Cameron Buie Nov 28 '12 at 17:55
    
Honest: try to get out of that hectic rythm for some minutes and learn how to type properly mathematics, so that people won't have to guess what you meant. The time of us all is important. –  DonAntonio Nov 28 '12 at 19:18
    
@DonAntonio I've given it a go for you ;) –  Becky Nov 28 '12 at 20:34
    
Many thanks, @Becky...but I really think you've given a go for yourself. :) –  DonAntonio Nov 28 '12 at 22:42

1 Answer 1

up vote 0 down vote accepted

So you have the map $F:\mathbb C^{2\times 2} \longrightarrow \mathbb C^3, \ \begin{pmatrix}a&b\\c&d \end{pmatrix} \mapsto\begin{pmatrix}a\\a\\3a+b \end{pmatrix}$ and you want to find $\ker F$.
By definition $$\ker F=\left\{ \begin{pmatrix}a&b\\c&d \end{pmatrix}\in\mathbb C^{2\times 2}:F\left(\begin{pmatrix}a&b\\c&d \end{pmatrix}\right)=\begin{pmatrix}0\\0\\0 \end{pmatrix}\right\}\\ =\left\{ \begin{pmatrix}a&b\\c&d \end{pmatrix}\in\mathbb C^{2\times 2}:\begin{pmatrix}a\\a\\3a+b \end{pmatrix}=\begin{pmatrix}0\\0\\0 \end{pmatrix}\right\}\\ =\left\{ \begin{pmatrix}a&b\\c&d \end{pmatrix}\in\mathbb C^{2\times 2}:a=0,3a+b=0\right\}=\cdots$$ After that find the basis.

An answer is

$\left\{\begin{pmatrix}0&0\\1&0 \end{pmatrix},\begin{pmatrix}0&0\\0&1 \end{pmatrix}\right\}$

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So is the basis \begin{pmatrix} 0 & 0 \\r & s\end{pmatrix} ? Where r and s are just constants which don't affect it? –  Becky Nov 28 '12 at 21:19
    
The basis must contain $2$ elements. The $\left\{\begin{pmatrix}0&0\\ r & s \end{pmatrix} \right\}$ is not a basis for any $r,s$. The $\left\{\begin{pmatrix}0&0\\ r&0\end{pmatrix}, \ \begin{pmatrix}0&0 \\ 0 & s \end{pmatrix}\right\}$ is a basis for every $r,s\neq0$. –  P.. Nov 28 '12 at 21:40
    
Oh sorry, I meant that's the kernel. –  Becky Nov 28 '12 at 21:58

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