Prove that $\operatorname{Trace}(A^2) \le 0$

Question

Let $A \in M_n(\mathbb{R})$ is a antisymmetric matrix such as $A^T=-A$. Prove that $\operatorname{Trace}(A^2) \le 0 $

I see that, for some matrix such as, their terms in diagonal are negative ?

I think this is wrong... Because if $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ (listed according to their algebraic multiplicities) then $\operatorname{Tr}(A^2)=\sum_{k=1}^n \lambda_k^2 \geq 0$. See http://en.wikipedia.org/wiki/Trace_(linear_algebra)#Eigenvalue_relationships — Surb, Jan 07 '15 at 08:11

score 8 · Answer 1 · answered Jan 07 '15 at 08:17

There are no doubt more sophisticated approaches, but every diagonal element of $A^2$ is obtained as $R_i\cdot C_i$ where $R_i$ and $C_i$ are row and column $i$ of $A$. Since $A$ is anti-symmetric one has $R_i=-C_i^T$, so $R_i\cdot C_i\leq0$ for each$~i$. With all diagonal elements of $A^2$ nonpositive, its trace certainly is so as well.

score 6 · Answer 2 · answered Jan 07 '15 at 08:28

6

Multiply both sides of your equality by $A$, you get

$$AA^T=-A^2$$

but then $AA^T$ is positive definite, so $-AA^T=A^2$ is negative semi-definite, hence its trace is negative.

answered Jan 07 '15 at 08:28

Adam Hughes

36,777

score 3 · Answer 3 · answered Jan 07 '15 at 08:43

Another approach:

Let $e_i, \; 1\leq i \leq n$ be the standard basis of $\mathbb{R}^n$ (or any orthonormal basis), then

$$\text{tr}(A^2) = \sum_{i=1}^n \langle A^2e_i, e_i \rangle = \sum_{i=1}^n \langle Ae_i, A^Te_i \rangle =-\sum_{i=1}^n \|Ae_i \|^2 \leq 0$$

score 2 · Answer 4 · answered Jan 07 '15 at 08:13

2

Without revealing too much: rewrite $A^2$ using the equality that's been given. Then take a closer look at the diagonal elements, now that you have written the matrix square in a new way.

answered Jan 07 '15 at 08:13

anon

151,657

Ofir Schnabel · Accepted Answer · 2015-01-07T09:01:59.713

2

Let $\lambda$ be an eigenvalue of $A$. Then using inner product we get $$\langle Av, Av\rangle=\langle \lambda v, \lambda v\rangle=\bar{\lambda} \lambda.$$ On the other hand $$\langle Av, Av\rangle =\bar{\lambda}\langle Av, v\rangle= \bar{\lambda}\langle v, A^t v\rangle=\bar{\lambda}\langle v, -A v\rangle= -\bar{\lambda} \bar{\lambda}\langle v, v\rangle.$$ Therefore $$\bar{\lambda} \lambda=-\bar{\lambda}\bar{\lambda}.$$ Now either $\lambda=0$ or by dividing by $\bar{\lambda}$ we get $\lambda=-\bar{\lambda}$. Either way we get that all eigenvalues are with real part zero and the result follows.

edited Jan 07 '15 at 09:01

answered Jan 07 '15 at 08:15

Ofir Schnabel

3,891

Can you tell me why $A^t\lambda v=\lambda^2v$, thank you. – Road Human Jan 07 '15 at 08:23
@RoadHuman an eigenvalue of $A$ is also an eigenvalue of $A^t$. For the proof see here – Bman72 Jan 07 '15 at 08:28
yes, whacka :D and if $\lambda^2=-\lambda^2$ then $\lambda=0$ i.e "both are zero" but not " eigenvalues are complex with real part zero and the result follows. " – Road Human Jan 07 '15 at 08:35
if $\lambda.\overline{\lambda}=0$ then $\lambda=0$ – Road Human Jan 07 '15 at 08:55
i think, if $\lambda_j=-i$ then we can't show that $\sum \lambda_i^2 \le 0$ – Road Human Jan 07 '15 at 08:58
some corrections were made, sorry for not being enough careful – Ofir Schnabel Jan 07 '15 at 09:04
@Road If you mean all eigenvalues are $-i$, well, that's not possible. Ofir's argument is now correct, so we can clean up the comment section. – anon Jan 07 '15 at 09:07
some are $-i$ not all. yeah. – Road Human Jan 07 '15 at 10:02

score 2 · Answer 6 · answered Jan 07 '15 at 08:54

2

Start with the Frobenius norm of A:

$$ \left\lVert A \right\rVert_F^2= \mathrm{Tr}(A^TA)=\mathrm{Tr}(-AA)=\mathrm{Tr}(-A^2)=-\mathrm{Tr}(A^2)\geq 0 \Leftrightarrow \mathrm{Tr}(A^2) \leq 0 $$

answered Jan 07 '15 at 08:54

eager2learn

2,799

Prove that $\operatorname{Trace}(A^2) \le 0$

6 Answers6

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