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Let $V$ be finite dimensional complex vector space, and $T:V\to V$ be a linear operator. Assume that there exists $0\neq v \in V$ such that $T^2v= \lambda v$ where $\lambda \geq 0$ and $T^2 = T \circ T.$

Question: Can we say that $Tv= \sqrt{\lambda}v$? If so, how I should justify?

Learner1
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  • Let $T \neq 0$ such that $T^2 = 0$ and a vector $v \neq 0$ with $Tv \neq 0$. –  Sep 06 '17 at 09:03

4 Answers4

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As we have $T^2\nu = \lambda \nu$ and $\nu\neq0$ we can say that $\lambda$ is an eigenvalue of $T^2$. Hence,$det(T^2-\lambda I) = 0$. Therefore, we have:

$$det(T^2-\lambda I) = det((T-\sqrt{\lambda}I)(T+\sqrt{\lambda}I)) = 0$$

As referenced here we can say:

$$det(T-\sqrt{\lambda}I)det(T+\sqrt{\lambda}I) = 0$$ So, $det(T-\sqrt{\lambda}I) = 0$ or $det(T+\sqrt{\lambda}I) = 0$. It means one of the $\sqrt{\lambda}$ and $-\sqrt{\lambda}$ is at least eigen value of $T$.

Anyhow, you can't say necessarily $\sqrt{\lambda}$ is eigen value of $T$. Also, If we know all eigen values of $T$ are positive and real, we can say the statement is true.

As mentioned by Jyrki, You should be aware that this analysis about $\lambda$ not about $\nu$ as eigen vector. So, if your question is also on $\nu$ you can't find anything in this analysis.

OmG
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  • thanks. What happens if we assume that $T$ is positive operator? – Learner1 Sep 06 '17 at 09:11
  • @Learner1 Also, If we know all eigen values of $T$ are positive and real, we can say the statement is true. – OmG Sep 06 '17 at 09:12
  • Thanks. Can you tell me how should I justify this? (I have assume that $\lambda \geq 0$) – Learner1 Sep 06 '17 at 09:14
  • You do get that one of $\pm\sqrt{\lambda}$ is an eigenvalue. But there's no need for $v$ to be an eigenvector. See Fred's post (+1) for an example. There the catch is that $v$ is a linear combination of eigenvectors belonging to $\pm\sqrt\lambda$. If $\lambda=0$ then there are more complicated possibilities. – Jyrki Lahtonen Sep 06 '17 at 09:35
  • @Learner1 It is so straightforward. As I said one of $\sqrt{\lambda}$ and $-\sqrt{\lambda}$ is eigen value of $T$, at least. So, If eigen values of $T$ are positive and real, $\sqrt{\lambda}$ is eigen value of $T$. – OmG Sep 06 '17 at 09:57
  • @OMG: Thanks. Can we argue without using determinant? (The determinant is not yet introduced in the classroom). Thanks... – Learner1 Sep 06 '17 at 10:08
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No. Let $T=diag(1,-1) \in \mathbb C^{2 \times 2}$. Then $T^2=I_2$. For

$v=(1,1)^t$ we have $T^2v=1*v$, but $Tv \ne \sqrt{1}v$.

Fred
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  • Thanks. If I assume $V$ is an inner product space, and $T$ is positive operator (that is $T=T^*$ , and $<Tv, v> \geq 0$ ), then can we expect? – Learner1 Sep 06 '17 at 09:06
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The most general (counter-)example: When $\dim V \geq 2$ take any two independent vectors $v$ and $w$ and define $T$ so that $Tv = w$ and $Tw=\lambda v$. On the complement (if any) of ${\rm Span} \{ v,w\}$ define $T$ as you like. You then have a counter-example.

To avoid the occurrence of the counter-example you should distinguish between zero and non-zero values of $\lambda$:

When $\lambda>0$ the linear-combinations $(\sqrt{\lambda}\; v)\pm w$ will have eigenvalues $\pm \sqrt{\lambda}$. Thus if you know that all eigenvalues have e.g. positive real part this example is excluded and the conclusion holds.

When $\lambda=0$ the example corresponds to a nil-potent part of the matrix. If you know that the matrix is diagonalizable then this example is excluded and the conclusion holds.

If e.g. you assume $T$ positive semi-definite both of the above examples are excluded and the conclusion holds.

H. H. Rugh
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  • @HHR: Thanks. How should I argue that if $T$ is positive, then the statement is true (without using determinant). (Using determinant, the answer is given by OMG ) – Learner1 Sep 06 '17 at 10:36
  • This fact has been used in the book " Linear Algebra Done Right" by sheldon Axler. 3rd Edition. Page No. 238. Proof of 7.52 (The author introduce determinant in the last chapter!) – Learner1 Sep 06 '17 at 10:39
  • @Learner1 It depends upon what you know about positive matrices? Do you know that such a matrix is diagonalizable with an orthonormal basis? If so it is easy (and I can add the few missing elements), but if not, it gets more messy. – H. H. Rugh Sep 06 '17 at 11:28
  • @HHRugh: Thanks. I know that positive matrices are diagonalizable with an orthonormal basis (by Spectral Theorem(ST), as positive operator is self adjoint and so ST is applicable). Please can you tell me how this fact is useful here? – Learner1 Sep 06 '17 at 11:41
  • Every eigenvalue of a positive operator is real and non-negative (which excludes then an eigenvalue of the form $-\sqrt{\lambda}<0$, the first case) and being diagonalizable implies e.g. $\ker T = \ker T^2$ (the second case). Is this sufficient for you? – H. H. Rugh Sep 06 '17 at 11:46
  • @HHRugh: Now it is clear to me that $Ker T = Ker T^2.$ But form this I am unable to see $Tv= \sqrt{\lambda} v$ if $T^2v= \lambda v$. Please can say bit more on this? Thanks – Learner1 Sep 06 '17 at 12:25
  • This only concerns the case $\lambda=0$. But then $T^2v=0$ implies $Tv=0$ by what we said about the kernels. In the case of a symmetric matrix (not even positive) here is a direct argument: when $T^2v=0$ then $0=(v,T^2 v) = (Tv,Tv) = |Tv|^2$ which then implies $Tv=0$ – H. H. Rugh Sep 06 '17 at 12:30
  • @HHRugh: Thanks. But then what about $\lambda >0$? How this case will follows? – Learner1 Sep 06 '17 at 12:35
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    For $\lambda>0$ the vector $u=Tv - \sqrt{\lambda}v$ verifies: $Tu = -\sqrt{\lambda} u$ which is not possible for a positive operator unless $u=0$. But this implies precisely that $Tv=\sqrt{\lambda} v$ as wanted. – H. H. Rugh Sep 06 '17 at 12:47
  • @HHR: Thanks a lot: – Learner1 Sep 06 '17 at 13:13
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No, you can't: the simplest example is a one-dimensional space $V$ and the map $T\colon V\to V$ defined by $T(v)=-v$; of course this extends to any dimension.

On the other hand, if $T$ is a positive definite hermitian operator, then the statement is true, because of the spectral theorem: write $$ T=\sum_{k=1}^{m}\lambda_k T_k $$ where $T_k$ are idempotent hermitian and $T_h\circ T_k=0$ for $h\ne k$. The scalars $\lambda_k$ are the eigenvalues of $T$, which are positive.

Then $$ T^2=\sum_{k=1}^m\lambda_k^2 T_k $$ and so $\lambda=\lambda_k^2$ for some $k$, that is, $\lambda_k=\sqrt{\lambda}$. Without loss of generality, we can assume $k=1$. Then $v=T_1(v)$ and $T_k(v)=0$ for $k>1$. Thus $$ T(v)=\lambda_1T_1(v)=\lambda_1v=\sqrt{\lambda}v $$

egreg
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