Find a linear transformation that has specific two subspaces as kernel and image Let $V$ be a finite dimensional vector space. Let $U,W$ be subspaces of $V$ such that $\dim U + \dim W = \dim V$. Show that there exists a linear transformation $T:V\rightarrow V$ whose $\ker T=U$ and $Im \  T=W$.
It doesn't suppose that $U\cap W= \{0\}$ to make things easier.
Has someone any idea?
 A: Let $m = \dim U$ and $n = \dim W$. Let $U$ has a basis $\{\mathbf u_1, \mathbf u_2, \ldots , \mathbf u_m\}$ and $W$ has a basis $\{\mathbf w_1, \mathbf w_2,\ldots, \mathbf w_n\}$. Let $U^\perp$, the orthogonal complement of $U$ in $V$, has a basis $\{\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_n\}$. $\dim U^\perp = n$.
Let $A$ be the $(m+n)\times(m+n)$ matrix
$$A = \begin{pmatrix}\mathbf u_1&\mathbf u_2&\ldots& \mathbf u_m&\mathbf a_1&\mathbf a_2&\cdots&\mathbf a_n\end{pmatrix}$$
$A$ is invertible because its columns are linearly independent. Then construct $T$ as 
$$\begin{align*}
T &= \begin{pmatrix}0_{(m+n)\times m}& \mathbf w_1&\mathbf w_2&\cdots&\mathbf w_n\end{pmatrix} A^{-1}\\
 &= \begin{pmatrix}0_{(m+n)\times m}& \mathbf w_1&\mathbf w_2&\cdots&\mathbf w_n\end{pmatrix} \begin{pmatrix}\mathbf u_1&\mathbf u_2&\ldots& \mathbf u_m&\mathbf a_1&\mathbf a_2&\cdots&\mathbf a_n\end{pmatrix}^{-1}
\end{align*}$$
For the proof below, $\mathbf e_i$'s are the standard basis vectors in $V$.

Consider the kernel of $T$. For $i = 1,2,\ldots,m$,
$$\begin{align*}
T\mathbf u_i
&= \begin{pmatrix}0_{(m+n)\times m}& \mathbf w_1&\mathbf w_2&\cdots&\mathbf w_n\end{pmatrix} A^{-1} \mathbf u_i\\
&= \begin{pmatrix}0_{(m+n)\times m}& \mathbf w_1&\mathbf w_2&\cdots&\mathbf w_n\end{pmatrix} A^{-1} A \mathbf e_i\\
&= \begin{pmatrix}0_{(m+n)\times m}& \mathbf w_1&\mathbf w_2&\cdots&\mathbf w_n\end{pmatrix} \mathbf e_i\\
&= \begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}
\end{align*}$$

Consider the image of $T$. For $j = 1,2,\ldots,n$, there exists a vector $\mathbf v$ such that $T\mathbf v = \mathbf w_j$:
$$\begin{align*}
\mathbf w_j
&= \begin{pmatrix}0_{(m+n)\times m}& \mathbf w_1&\mathbf w_2&\cdots&\mathbf w_n\end{pmatrix} \mathbf e_{m+j}\\
&=\begin{pmatrix}0_{(m+n)\times m}& \mathbf w_1&\mathbf w_2&\cdots&\mathbf w_n\end{pmatrix} A^{-1} A \mathbf e_{m+j}\\
&= T A\mathbf e_{m+j}\\
&= T\mathbf a_j
\end{align*}$$

For general $\mathbf v = \sum_{i=1}^m c_i\mathbf u_i + \sum_{j=1}^n d_j\mathbf a_j$,
$$\begin{align*}
T\mathbf v
&= T\left(\sum_{i=1}^m c_i\mathbf u_i + \sum_{j=1}^n d_j\mathbf a_j\right)\\
&= \sum_{i=1}^m c_iT\mathbf u_i + \sum_{j=1}^n d_jT\mathbf a_j\\
&= 0 + \sum_{j=1}^n d_j\mathbf w_j\\
\end{align*}$$
Therefore:


*

*For vectors $\mathbf v \notin U$, $d_j$'s are not all zero, and so $T\mathbf v \ne 0$;

*For any vector $\mathbf v$, the result $T\mathbf v \in W$, and any vector $\mathbf w\in W$ can be mapped to.

