Finding $T:V\to V$ By The Span Of $\ker(T)$ And $\operatorname{Im}(T)$

Question

Let $V=\mathbb{R}^3$ and let there be two subspaces of $V:$

$U=\operatorname{Span}\{\begin{pmatrix} 1 \\ 1\\ 1\ \end{pmatrix}\}$ and $W=\operatorname{Span}\{\begin{pmatrix} 0 \\ 1\\ 1\ \end{pmatrix}\,\begin{pmatrix} 1 \\ 0\\ 2\ \end{pmatrix}\}$

Find $T=\mathbb{R}^3\to \mathbb{R}^3$ such that $U=\ker(T)$ and $\operatorname{Im}(T)=W$

The vectors of $U+V$ is a basis of $\mathbb{R}^3$ so it is sufficient to show the linear transformation on the basis vectors

So $U=\ker(T)$ means that $\begin{pmatrix} a&b&c \\ d&e&f\\ g&h&k \end{pmatrix}\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}\iff \begin{cases} a=-b-c \\ d=-e-f\\ g=-h-k \end{cases}$

Now I have to find terms on $\begin{pmatrix} a&b&c \\ d&e&f\\ g&h&k \end{pmatrix}$ by the image of $T$ but how can I do it?

Answer 1

Now choose $b=0$, $e=1$, $h=1$, $c=1$, $f=0$, and $k=2$. Then your matrix becomes$$M_T=\begin{pmatrix}-1&0&1\\-1&1&0\\-3&1&2\end{pmatrix}$$and it is clear that $$M_T.\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\1\end{pmatrix}\text{ and that }M_T.\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}1\\0\\2\end{pmatrix}.$$

Answer 2

I would advise you to start exactly from the other direction. So since $W = span{(v_1, v_2)}$, calling your vectors $v_1$ and $v_2$, you directly can conclude that a possible way for $T$ to look like is the following: $$T = (v_1,v_2,a\cdot v_1 + b\cdot v_2)$$ with $a,b \in \mathbb{R}$. From the property with the kernel you can then conclude that $a = b = -1$.

Answer 3

Very simple and straightforward: the columns of the matrix of your linear map span its image.

Just stack the basis of $W$ on the first 2 columns and then have the matrix's third column be any linear combination of the first 2.