Hey I have problems solving this exercise. Can someone help me?
I have the matrix $A=\begin{pmatrix} 3 & 4 &9 &-6 \\ 1&2 &-1 &2 \\ 2& 3& 4& -2 \end{pmatrix}$ and $F: \mathbb{R}^4 \rightarrow \mathbb{R}^3$, $x \rightarrow Ax$
I have to find a basis $W=(w_1,w_2,w_3w_4)$ of $\mathbb{R}^4$ and a basis $B=(b_1,b_2,b_3)$ of $\mathbb{R}^3$ so that the change-of-basis matrix is $M_B^W(F)=\begin{pmatrix} 1 & 0 &0 &0 \\ 0&1 &0 &0 \\ 0& 0& 0& 0 \end{pmatrix}$.
Now the problem is that I really don't know how to proceed.
My idea was to find a linear combination of
$1b_1+0b_2+0b_3=b_1=f(w_1) \\0b_1+1b_2+0b_3=b_2=f(w_2) \\ 0b_1+0b_2+0b_3=0=f(w_3) \\ 0b_1+0b_2+0b_3=0=f(w_4)$
So for $w_3,w_4$ we can take two vectors of $\ker(A)=(\left(\begin{matrix} -11 \\ 6 \\ 1 \\ 0 \end{matrix}\right),\left(\begin{matrix} 10 \\ -6 \\ 0 \\ 1 \end{matrix}\right))$
But now I really don't know how to proceed. Can someone help me?
Is there maybe an easy way to solve this problem?
As a first step, you can determine $\ker F$, which is$$\left\{(x,y,z,t)\in\Bbb R^4\,\middle|\,x=10t-11z\wedge y=6z-6t\right\}.$$Take two linearly independent vectors from $\ker F$, such as $w_3=(10,-6,1,0)$ and $w_4=(-11,6,0,1)$. Now let $w_1$ and $w_2$ be two linearly independent vectors such that no linear combination of them belongs to $\ker F$, such as $w_1=(1,0,0,0)$ and $w_2=(0,1,0,0)$. Finally, take $b_1=F(w_1)$, $b_2=F(w_2)$ and let $b_3$ be any vector form $\Bbb R^3$ such that the set $\{b_1,b_2,b_3\}$ is linearly independent.