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Let $M$ be the vector space of $2\times 2$ matrices with entries in $\Bbb{R}$, and define a map \begin{align*} T: M&\to\Bbb{R}^2\\ A&\mapsto Aw, \end{align*} where $w=\begin{pmatrix}1\\3\end{pmatrix}$.

a)Is $T$ a linear transformation?

b)Find a basis for $\ker T$

I know how to do a, but for b, I have no clue how to do it. Please help.

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please check that the edit is what you intended –  Stahl Apr 21 '13 at 1:18

1 Answer 1

Hint: You can show that for every $A,B\in M_{2\times 2}(\mathbb R)$; $T(aA+B)=aT(A)+T(B)$ by taking two $2\times 2$ matrices $A,B$. Note that if $A\in ker(T)$, so $T(A)=0$ and then we have $$a+3b=0, c+3d=0$$ where $A=\begin{pmatrix} a & b \\ c&d \end{pmatrix}$. Solve the system above to find a proper vector.

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I am confuced, i try to do it, and i got the basis is (-3,1) –  Prince Zhao Apr 21 '13 at 2:20
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You want to find the kernel of the transformation, so you take an arbitrary matrix whose entries Babak denoted by $a,b,c$ and $d$ and find all such quantities that transform the vector $(1,3)^T$ to zero. Solving the associated linear system yields a two-parameter family of solutions, for example if you let $b=t$ and $d=s$ then your solution space is of the form $a=-3t,b=t,c=-3s,d=s$, meaning that plugging in any value for $s$ and $t$ will yield a solution of your system. –  dezign Apr 21 '13 at 2:25
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The "vectors" in your space are 2x2 matrices, so how could $(-3,1)$ be a basis? –  dezign Apr 21 '13 at 2:29
    
So the basis should be (-3,1 -3,1)? –  Prince Zhao Apr 21 '13 at 2:34
    
@PrinceZhao: As dezign noted you have a solution space generated as pointed above, so the matrix you asked is one of them. Think of another matrices which the linear combination of them could build the whole space. –  B. S. Apr 21 '13 at 7:58

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