Producing for a direct sum $V = U \oplus W$ an endomorphism $P$ of $V$ s.t. $Im(P) = U$ and $\ker(P) = W$

Question

I've been asked to prove this:

Let $V$ be a vector space over a field $F$ and suppose that $V = U \oplus W$, where $U$ and $W$ are subspaces of $V$.

Show that $Im(P)=U$ and $\ker(P)=W$ for some $P \in End(V )$.

Sorry if I come across as being a bit dense but doesn't the direct sum hold the property of being commutative?

How can we specifically say that $Im(P) = U$? Say we took the direct sum as $W \oplus U$. Then $Im(P) = W$?

Could someone explain to me where I'm going wrong and maybe point me in the right direction? Cheers!

And sorry again if the answer is very obvious.

Answer 1

The direct sum is commutative in the sense that there is a canonical isomorphism $U \oplus W \cong W \oplus U$. But that doesn't mean that a given endomorphism $P$ behaves w.r.t. the summands $U$ and $W$ in the same way.

As for the problem itself:

Hint We can decompose the matrix for $P$ (w.r.t. any adapted basis of $V$ adapted to the decomposition $V = U \oplus W$) into blocks, as $$P := \begin{pmatrix} A & B \\ C & D \end{pmatrix},$$ so that for any $(u, w) \in U \oplus W$ we have $$P\begin{pmatrix}u \\ w\end{pmatrix} =\begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix}u \\ w\end{pmatrix}.$$

Now, what restrictions do our conditions $\ker P = W$ and $\text{im }P = U$ impose on the matrix blocks $A, B, C, D$?

Since $\ker P = W$, for all $w \in W$ we must have $$0 = P\begin{pmatrix} 0 \\ w\end{pmatrix} = \begin{pmatrix}Bu \\ Dw\end{pmatrix},$$ and so $B = 0$, $D = 0$. On the other hand, since $\text{im }P = U$, for all $(u, w) \in U \oplus W$ we must have that $$P\begin{pmatrix}u \\ w\end{pmatrix} = \begin{pmatrix}Au + Bw \\ Cu + Dw\end{pmatrix}$$ has the form $\begin{pmatrix}\ast \\ 0\end{pmatrix}$, and so $C = 0$ and $D = 0$, and altogether we have $B = 0$, $C = 0$, $D = 0$. This last condition guarantees that $\text{im }P \subseteq U$, but we must ensure that equality holds, or equivalently, that $\text{rank }P = \dim U$. Now, $\text{rank }P = \text{rank }A$, so our condition is that $\text{rank }A = \dim U$; but $A$ is a square matrix of size $\dim U$, so $A$ must be invertible. Thus, the transformations with the given properties are exactly the ones of the form $$\begin{pmatrix}A & 0 \\ 0 & 0 \end{pmatrix},$$ where $A$ has full rank. There is a special choice here, namely where $A$ is the identity matrix (of size $\dim U$). In this case, $P$ is nothing more than the canonical projection of $U \oplus W$ onto $U$.

Answer 2

If $V = U \oplus W$, then not only can we find

• $P \in \operatorname{End}(V)$ such that $\operatorname{Im}(P) = U$ and $\ker P = W$, but it's also true, like you pointed out, that there's some

• $Q \in \operatorname{End}(V)$ such that $\operatorname{Im}(Q) = W$ and $\ker Q = U$.

It doesn't really matter which summand they wanted as the image and which as the kernel; just that there are certain maps that have certain properties, with respect to direct sums.

As for proving the theorem, I would look for a map of the form

$$Pv = \begin{cases}??, & v\in U \\ ??,& v \not\in U\end{cases}.$$

Then, what's the most natural place to send things in $U$?