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I'm working in my homework of linear algebra and there is this problem I'm strugling with

Let $V$ be a finite-dimensional vector space with an inner product, and let $T$ be a linear operator on $V$. I'm trying to prove the following result:

If $W$ is a subspace of $V$, then $W$ is $T$-invariant if and only if $W^\perp$ is $T^*$-invariant.

I know that $W$ is $T$-invariant means $T(w) \in W$ for all $w \in W$; also that $W^\perp$ is $T^*$-invariant means $T^*(v) \in W^\perp$ for all $v \in W^\perp$; and $T^*$ is the adjoint of $T$, satisfying $\langle T(x), y \rangle = \langle x, T^*(y) \rangle$ for all $x, y \in V$

For the forward direction ($\Rightarrow$), I tried assuming $W$ is $T$-invariant and taking $v \in W^\perp$. I need to show that $T^*(v) \in W^\perp$, which means I need to show $\langle w, T^*(v) \rangle = 0$ for all $w \in W$.

Using the adjoint property, I can write: $$\langle w, T^*(v) \rangle = \langle T(w), v \rangle$$

Since $W$ is $T$-invariant, $T(w) \in W$, and since $v \in W^\perp$, this should give me $\langle T(w), v \rangle = 0$.

However, I'm having trouble with the reverse direction ($\Leftarrow$) and would appreciate guidance on how to structure that proof properly.

How can I do the reverse direction ($\Leftarrow$)?

Any hints or suggestions or solutions would be greatly appreciated!

Dean Miller
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Jisbon
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3 Answers3

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In addition to the property $\left(T^*\right)^* = T$ mentioned in the previous answer. There's another property that may be helpful for your proof:

Let $V$ be an inner product space and $W$ be a finite-dimensional subspace of $V$. Then $W = \left(W^\perp\right)^\perp$.

andywho
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  • On 2nd thought, $W = {(W^{\bot})}^{\bot}$ and the definition of adjoint operator are all we need. Using ${(T^{})}^{} = T$ is overkill. – Duong Ngo Oct 15 '25 at 08:02
  • Yeah, I think you are right. Even though we want to use this property, it should also be used in the forward direction instead of the backward one. (However, it becomes trivial after we think of the uniqueness of one's adjoint.) – andywho Oct 15 '25 at 08:13
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Hint: Prove that ${(T^{*})}^{*} = T$ then the backward direction follows from the forward direction.

Duong Ngo
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For finite dimensional vector spaces, it holds that $(W^{\perp})^{\perp}=W$, so take $v\in W$, if you want to show $T(v)\in W$ then this is equivalent to show that $\forall w\in W^\perp,\ T(v)\cdot w=0.$

So take $w\in W^\perp$, then $T(v)\cdot w=v\cdot T^*(w)=0,$ this is because $T^*$ is $W^\perp$-invariant so $T^*(w)\in W^\perp$.

Gabriele
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