Show that a linear transformation is invertible.

Question

Let $V$ be a finite-dimensional inner product space, and let $W \subset V$ be a subspace.

Define $T : V \to V$ by $T(v) = v + \operatorname{Proj}_W (v)$.

Show that $T$ is invertible.

Help, please?

Answer 1

As noted by @Abishanka Saha, we can find the inverse $T^{-1}$ explicitly.

Let $z \in V$. We have to find $v \in V$ such that $z = T(v) = v + \operatorname{Proj}_W v$.

We have:

$$\underbrace{\operatorname{Proj}_W(z)}_{\in W} + \underbrace{\operatorname{Proj}_{W^\perp } (z)}_{\in W^\perp} = z = v + \operatorname{Proj}_W (v) = 2\operatorname{Proj}_W(v) + \operatorname{Proj}_{W^\perp} (v) = \underbrace{\operatorname{Proj}_W(2v)}_{\in W} + \underbrace{\operatorname{Proj}_{W^\perp}(v)}_{\in W^\perp}$$

Since $W \oplus W^\perp = V$, the decomposition of $z$ is unique, so we can equate the summands:

$$\operatorname{Proj}_W(z) = \operatorname{Proj}_{W}(2v) \implies \operatorname{Proj}_W(v) = \frac12 \operatorname{Proj}_W(z)$$

$$\operatorname{Proj}_{W^\perp}(z) = \operatorname{Proj}_{W^\perp}(v) $$

So we conclude $v = \operatorname{Proj}_W(v) + \operatorname{Proj}_{W^\perp}(v) = \frac12 \operatorname{Proj}_{W}(z) + \operatorname{Proj}_{W^\perp}(z)$.

Hence, we conclude that the inverse is given by $T^{-1}(z) = \frac12 \operatorname{Proj}_{W}(z) + \operatorname{Proj}_{W^\perp}(z)$ for $z \in V$.

Answer 2

To show that a map is invertible, you have a few options:

1. Find an inverse (this might be hard here).

2. In the finite dimensional case, show that the map is surjective (also might be hard here, but easier than finding an inverse map).

3. In the finite dimensional case, show that the map is injective (this should be the right approach here).

One way to show that a map is injective (one-to-one) is to show that the kernel is trivial. In other words, the only vector so that $T(v)=0$ is the zero vector.

Assume that $T(v)=0$, then $v+\operatorname{Proj}_Wv=0$. Therefore, $-\operatorname{Proj}_Wv=v$, so $v$ is in $W$ and $\operatorname{Proj}_W(v)=v$. Therefore, $T(v)=0$ means $2v=0$, so $v=0$.

This proves the kernel is $0$, so $T$ is injective, so (since $V$ is finite dimensional), $T$ is an isomorphism.

Answer 3

$$T^{-1}(v)=\frac12\operatorname{Proj}_W(v) +\operatorname{Proj}_{W^{\top}}(v)$$