How to find the basis of a transformation matrix?

Question

I need help with b. Lets call the column vectors of the transformation matrix $w_1, w_2, w_3$. I can already see that $w_3 = \begin{bmatrix} 1\\ 2\\ 2 \end{bmatrix}$ or simply the norm. But I am struggling with $w_1$ and $w_2$. I got a hint that they are supposed to be parallel but I just simply can't see it. I mean $n*v=0$ gives me a random vector $v$ that is parallel to the plane, for example $\begin{bmatrix} -2\\ 1\\ 0\end{bmatrix}$. But when projected it is the same and not equal to $\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$

Also in this new basis? Wouldn't the plane's equation change, and thus its normal? I mean, don't we have to first find the plane in the new basis to do anything?

Answer 1

Take any two linearly independent vectors $w_1$ and $w_2$ from $>H$, such as $w_1=(-2,1,0)$ and $w_2=(-2,0,1)$. Then, if $w_3=(1,2,2)$, $\{w_1,w_2,w_3\}$ will be a basis such that $T(w_1)=w_1$, $T(w_2)=w_2$ and $T(w_3)=0$. So, the matrix of $T$ with respect to this basis will be$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}.$$

Answer 2

Too long for a comment. I am writing for a little connection between part $(a)$ and part $(b)$ since they seem as if they are independent.

By using the projection of a vector onto a hyper plane formula, $T(v)=v-\frac{\large {n\cdot v}}{\Vert n\Vert^2}n$, where $n=(1,2,2)$, we find $T(e_1)=(\frac{8}{9},-\frac{2}{9},-\frac{2}{9})$, $T(e_2)=(-\frac{2}{9},\frac{5}{9},-\frac{4}{9})$, $T(e_2)=(-\frac{2}{9},-\frac{4}{9},\frac{5}{9})$.Hence, the standard matrix of $T$ is $[T]=\left[\begin{matrix} \frac{8}{9}&-\frac{2}{9}&-\frac{2}{9} \\ -\frac{2}{9}&\frac{5}{9} & -\frac{4}{9}\\ -\frac{2}{9}& -\frac{4}{9}&\frac{5}{9} \end{matrix}\right]$. This was part $(a)$.

The reduced row echelon form of $[T]$ is $[T]_{red}=\left[\begin{matrix} 1&0&-\frac{1}{2} \\ 0&1& -1\\ 0& 0&0 \end{matrix}\right].$ Now, let $w_1=(1,0,-\frac{1}{2})$ and $w_2=(0,1,-1).$ Notice that $T(w_1)=w_1$ and $T(w_2)=w_2$ as expected. Let $\mathcal B=\{w_1,w_2,n\}$. Then with respect to this basis $[T]_{\mathcal B}=\left[\begin{matrix} 1&0&0 \\ 0&1& 0\\ 0& 0&0 \end{matrix}\right]$ because $T(n)=(0,0,0)$. $n$ does not show up in the matrix $[T]_{red}$, it makes the basis of the null space.