# Invertibility of a linear transformation without knowing its matrix

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

Define $T:\mathbb{V} \rightarrow \mathbb{V}$ by $$T(\overrightarrow v)=\overrightarrow v + Proj_{W}\overrightarrow v$$

Show that $T$ is invertible.

My approach is to show that $T$ is injective and since $T$ goes from $\mathbb{V}$ to $\mathbb{V}$, that would imply that it is bjective and therefore invertible.

I let an arbitrary vector $\overrightarrow v \in KerT$.

$T(\overrightarrow v)=\overrightarrow 0$

$\Rightarrow$ $\overrightarrow v=-Proj_{W}\overrightarrow v$

This would mean that $\overrightarrow v$ is a linear combination of vectors which pertain to a basis of $\mathbb{W}$. $\therefore Proj_{W}\overrightarrow v=\overrightarrow v$

$\therefore \overrightarrow v=-\overrightarrow v \Rightarrow \overrightarrow v=\overrightarrow 0$

Therefore $T$ is bijective and invertible.

Is my approach correct?

• Careful: $\;\Bbb W\cap\Bbb V=\Bbb W\;$ . This doesn't say much. Commented May 6, 2016 at 23:09
• Thank you, I fixed that. I meant to say a linear combination of vectors of only the subspace W. Commented May 6, 2016 at 23:12

Your work is correct (though only in finitely many dimensions, as is the case). It would be less awkward to say that $\overrightarrow v=-Proj_{W}\overrightarrow v$ implies that $\overrightarrow v \in \mathbb W$ than to say that it is "a linear combination of vectors which pertain to a basis of $\mathbb W$", but either is correct.

One may also note that a more direct proof would be to note that the inverse is $$T^{-1}(\overrightarrow v)=\overrightarrow v - \frac{1}2Proj_W\overrightarrow v$$ which can be shown to be the inverse directly by composing it on either side of $T$. This works in any number of dimensions, which is a plus.

• Thank you for your answer. So $T^{-1}(\overrightarrow v + Proj_{W}\overrightarrow v)=I$? Commented May 6, 2016 at 23:24
• Well, $T^{-1}(\overrightarrow v + Proj_W \overrightarrow v) = \overrightarrow v$. Otherwise stated, $T^{-1}T = I$. (In infinite dimensions, one also needs to check that $TT^{-1}=I$, which is also true) Commented May 7, 2016 at 0:03

Looks good to me. You showed that is injective (which is enough for invertibility in the range of $T$). You didn't show though that is surjective; just because $T$ takes value in $\mathbb V$ does not mean it will take every value of $\mathbb V$

• Thanks for your answer. I'm not sure but if $dimV$ (dimension of domain of the linear transformation) is equal to $dimW$ (dimension of codomain of linear transformation) then $T$ is injective if and only if $T$ is surjective if and only if $T$ is bijective. Am I wrong? Commented May 6, 2016 at 23:27
• @Sadem You seem a little confused. First of all, $\mathbb W$ is not the range of the linear transformation; it's easy to see that $\mathbb W$ is comprised in the range of $T$, but you have to show it. Having said that, if $\dim \mathbb W = \dim \mathbb V$ then $\mathbb W = \mathbb V$ and $T(\vec v ) = 2\vec v$ which is clearly invertible, but useless (Also, this case has been eliminated because in the question we are assuming $\Bbb W \subset \Bbb V$.) You have to show that given any $\vec w \in \Bbb V$, you can find $\vec v \in \Bbb V$ such that $T(\vec v) = \vec w$
– Ant
Commented May 7, 2016 at 6:49