Confused by the point of this linear algebra question

Question

The question reads: Let $T : \mathbb{R^4} → \mathbb{R^3}$ be given by $T(v) = Av$, where A = $$ \begin{pmatrix} 1 & 1 & 1 & 2 \\ 1 & 1 & 1 & 3 \\ 0 & 0 & 0 & 4 \\ \end{pmatrix} $$ Find the matrix $B$ of $T$ with respect to the basis {$v_1, v_2, u_1, u_2$} of $\mathbb{R^4}$ and the basis {$w_1, w_2, z_1$} of $\mathbb{R^3}$. Verify that $B$ is in canonical form.

Now,

the vectors $u_1, u_2$ are the basis for $ker(T)$ which are $(-1, 1, 0, 0), (-1, 0, 1, 0)$.

The vectors $v_1, v_2$ are two vectors which extend the basis of $ker(T)$ to $\mathbb{R^4}$, which I chose to be $(0,0,0,1), (1,0,0,0)$.

The vectors $w_1 = T(v_1) = (2,3,4)$ and $w_2 = T(v_2) = (1,1,0)$, and finally the vector $z_1$ is the vector which extends the basis of {$w_1, w_2$} to $\mathbb{R^3}$, which I chose to be $z_1 = (0,0,1)$.

Can anyone help me put the pieces of the puzzle of this question together for me? I started the question by trying to form the matrix $B$ by applying the transformation to the vectors of the given basis for $\mathbb{R^4}$, but since two of those basis vectors are that of the kernel of T, they just give me back the zero vectors. This is why I'm confused and unsure of the point of the question. Am I missing something here? Help would be appreciated. Thank you.

Answer 1

You are on the right way. Indeed, the columns of $B$ will be $T(u_1),T(u_2),T(v_1),T(v_2)$, expanded in the basis $\{w_1,w_2,z_1\}$.

Since $T(u_1)=T(u_2)=0$, the first two columns of $B$ will be zero.

Next, $T(v_1)=w_1$, so the third column of $B$ will be $(1,0,0)^T$. Finally, $T(v_2)=w_1$, so the fourth column of $B$ is $(0,1,0)^T$. All in all, we get $$ B=\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix} $$ If you decide that $v_1,v_2$ should be the first two basis vectors of $\Bbb R^4$ instead, then the corresponding two columns of $B$ will move to the left side, giving you $$ \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\end{pmatrix} $$ It's up to you which one you prefer.

Answer 2

I am giving a general result which might help you.

Let $\beta = \{v_1, v_2, v_3, v_4 \}$ and $\gamma = \{u_1, u_2, u_3 \}$ be the basis for $\mathbb{R}^4$ and $\mathbb{R}^3$ respectively.

Then, define $[T]_\beta^\gamma = \begin{bmatrix} | & | & | & | \\ [T(v_1)]_\gamma & [T(v_2)]_\gamma & [T(v_3)]_\gamma & [T(v_4)]_\gamma\\ | & | & | & | \\ \end{bmatrix}$ as a $3 \times 4$ matrix.

This is the matrix which

1. takes a vector in $\mathbb{R}^4$ with coordinates with respect to the basis $\beta$.
2. applies $T$ to it
3. returns a vector in $\mathbb{R}^3$ with coordinates with respect to the basis $\gamma$.

$[T(v_i)]_\gamma$ is the image of $v_i$ under $T$ with coordinates with respect to the basis $\gamma$, where $v_i \in \{v_1, v_2, v_3, v_4 \} = \beta$.