I have the following problem consisting of 3 parts of which I'm not being able to figure out the last.

Notation: $T^*$ is the adjoint operator of $T$. $Im(T) = \{T(v) : v \in V\}$. $Ker(T) = \{v \in V : T(v) = \vec{0} \}$

Let $V$ be a finite vector space with inner product over $\mathbb{R}$, let $S$ be a subspace of $V$ and $T: V \to V$ a linear operator.

  1. Prove that $S$ is $T-invariant \iff S^{\bot}$ is $T^*-invariant$.

  2. Suppose $dim(S) = 1$. Prove that $S$ is $T-invariant \iff \exists \lambda_0$ eigenvalue of $T$ such that $S \subset Ker(T - \lambda_0 I)$.

  3. Now suppose that $dim(S) = 2$ and $dim(V) = 3$. Prove that $S$ is $T-invariant \iff \exists \lambda_0$ eigenvalue of T such that $Im(T - \lambda_0 I) \subset S$.

I thought about using the fact that $Im(T)\ \bot\ Ker(T^*)$ and part 2 but I haven't been able to make it work right.

  • $\begingroup$ I don't get why you call those corrections "LaTeX errors". In my course's notes it's all in the way I did it in the first place. $\endgroup$
    – Otomeram
    Aug 16, 2020 at 5:28
  • $\begingroup$ The so-called '$\text{LaTeX}$ errors ' are called errors(My apologies) cause I just fixed dim to \dim and Ker to \ker and Im to \text{im}. The cause was, these are the standard LaTeX codes for the symbols. Again, I'm sorry if it hurt you. -Regards, Ralph $\endgroup$ Aug 16, 2020 at 7:47
  • $\begingroup$ No hurt feelings. My question was basically if these were actually standards. Again, my textbook writes it exactly the way I did, so I naturally assumed this was a correct way of doing it. Furthermore, I'm very used to reading things like $T-invariant$ in cursive instead of plain text as to emphasize that it's a specific defined term. $\endgroup$
    – Otomeram
    Aug 16, 2020 at 7:50
  • 1
    $\begingroup$ the cursive can be done using \textit{} or \emph{} in any online LaTeX editor. Making them inside $ $ doesn't seem right and in MSE, we use MathJax. You can make the cursive using _ _ the bold one can be made using ** **. In any LaTeX editor, the bold can be done using \textbf{}. You can visit TeX SE for more info. I hope it'll help :) $\endgroup$ Aug 16, 2020 at 8:08

2 Answers 2


You already have a lot of the crucial ideas laid out! The relation between part 3 and part 2 is seems to be that you are exchanging everything with a dual-ish concept, so to speak: we're looking at images instead kernels, and rather than $S$ being one-dimensional, the orthogonal complement $S^\bot$ is one-dimensional. So, somehow once we know part 2, part 3 should not be too far away.

So, assume the conditions of part 3 and that $S$ is $T$-invariant. By part 1, this means $S^\bot$ is $T^*$-invariant. But since $\dim(S^\bot) = \dim V - \dim S = 1$, we can apply part 2 to $T^*$ and $S^\bot$, and we find that this is equivalent to there being an eigenvalue $\lambda$ of $T^*$ so that $S^\bot \subset \ker (T^* - \lambda I)$. But if you know the relation $$ W \subset U \implies U^\bot \subset W^\bot,$$ then we can deduce $\ker(T^* - \lambda I)^\bot \subset (S^\bot)^\bot = S.$ And you've already indicated that you know that $\text{im}(T - \lambda I) \bot \ker((T - \lambda I)^*)$. Hence $$ \text{im}(T - \lambda I) \subset \ker((T - \lambda I)^*) = \ker(T^* - \lambda I) \subset S.$$

This proves the one direction, and the other direction goes similarly by working backwards if you know that $\ker(T)$ and $\text{im}(T^*)$ are not only orthogonal, but that they are indeed even orthogonal complements of one another, i.e. applying $\bot$ to one gives you the other.

All of these relations I am using are fairly elementary properties of orthogonal complements in finite dimensions, so I'm kinda hoping you already know about them, but if that's not the case, it might be a good exercise to try and prove them yourself! Hope that helps, let me know if something is still unclear!

EDIT: I indeed left out an argument why, in the above, $\lambda$ is an eigenvalue of $T$, we only know it is an eigenvalue of $T^*$. This part can be tricky if you approach it from the wrong angle; an eigenvector of $T^*$ will in general not be an eigenvector of $T$, so there is no way to directly compare things. However, there is still hope for an elementary proof.

We know that $\text{im}(T - \lambda I) \subset S$. But that means $T - \lambda I$ can be understood as a map from the 3-dimensional space $V$ into the 2-dimensional space $S$. Hence, simply by the rank-nullity theorem, it must have a nontrivial kernel, so there is a $v \in V$ with $Tv - \lambda v = 0$. But that must be exactly your eigenvector!

  • $\begingroup$ I did get all of this on my own, and tried to put it in a comment but it was too big. The problem is that you only showed that there exists an eigenvalue of $T^*$ that satisfies that, and the problem asks for an eigenvalue of $T$. I haven't been able to prove that this is also an eigenvalue of $T$, and until you do that this solution is wrong. $\endgroup$
    – Otomeram
    Aug 16, 2020 at 5:10
  • $\begingroup$ Ah, it may have been worthwhile to mention that this was the part you're struggling with. Even if you can't put it in a comment, it might have been good to edit your original post. That way I wouldn't have spent so much time on proving something you already know :) But okay, I added an elementary proof for why $\lambda$ is an eigenvalue of $T$, and from the right angle, it's not all that bad. $\endgroup$ Aug 16, 2020 at 7:14
  • $\begingroup$ Definitely should've edited my original post. I didn't think of that, I'm still new here. Thanks! $\endgroup$
    – Otomeram
    Aug 16, 2020 at 7:20

Given that I had already done what Lukas wrote, but a little clearer in my opinion, I'm going to lay it out. Still, this is not a valid proof because it lacks one last step which I will explain further below.

First, let's find a more useful way of writing the right-hand side of the equivalence:

$$Im(T - \lambda_0 I) = Ker((T - \lambda_0 I)^*)^{\bot} = Ker(T^* - \lambda_0 I)^{\bot}$$


$$Ker(T^* - \lambda_0 I)^{\bot} \subset S \iff S^{\bot} \subset Ker(T^* - \lambda_0 I)$$

By part 1. we have that

$$S \text{ is } T-invariant \iff S^{\bot} \text{ is } T^* - invariant$$

$$\text{Since } dim(V) = 3 \text{ and } dim(S) = 2 \implies dim(S^{\bot}) = 1$$

so by part 2. we have

$$S^{\bot} \text{ is } T^* - invariant \iff \exists \lambda_0 \text{ eigenvalue of } T^* \text{ such that } S^{\bot} \subset Ker(T^* - \lambda_0 I)$$

The problem with this as I said in a comment above is that here $\lambda_0$ is a value of $T^*$, and I need to prove that there exists an eigenvalue of $T$ that satisfies the property.

I can't use that the eigenvalues of $T$ are the conjugates of the eigenvalues of $T^*$ without giving a proof of it because such theorem is not in the course's notes where this problem came from. I did see some proofs of it (e.g. Do T and T* have the same eigenvalues with the same algebraic multiplicity?) but both proofs use theorems that are not in the book either (i.e. Schur decomposition and $\overline{det(A)} = det(A^*)$) so I can't imagine that they would expect you to prove either of these for a problem that should come almost inmediately from the previous parts.

In conclusion, I'm looking for a different way to prove that this $\lambda_0$ that appears at the end of the proof is also an eigenvalue of $T$.


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