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Remember that we've already proven the following, for any real symmetric $n\times n$ matrix $M$:
(i) Each eigenvalue of $M$ is real.
(ii) Each eigenvector can be chosen to be real.
(iii) Eigenvectors with different eigenvalues are orthogonal.
(b) Let $A$ be a real antisymmetric $n\times n$ matrix. Prove that each eigenvalue of $A^{2}$ is real and is less than or equal to zero.

The following continues my other question:

If $-\lambda^{2}$ and $-\mu^{2}$ are distinct, non-zero eigenvalues of $A^{2}$, show that there exist orthonormal vectors $u,\ u',\ w,\ w'$ with (I). $ Au= \lambda u'$, (II). $Aw= \mu w'$, (III). $Au'=-\lambda u$, (IV). $Aw'=-\mu w$.

Solution: Let $u, w$ be any unit eigenvectors such that $A^2 u = -\lambda^2 u$ and $A^2 w = -\mu^2 w$ $\qquad (*)$.

$1.$ What legitimises $u, w$ to be unit eigenvectors? Because any eigenvector can be normalised?

Then define $u' = \dfrac{1}{\lambda}Au$ and $w' = \dfrac{1}{\mu}Aw$.

$2.$ How would you determine to define $u', w'$ as such? This feels too clever and guileful.

Then u' and w' are also unit vectors, because $|u'|^2 = ... = |u'|$, and similarly, $|w'|^2 = ... = |w'|$. I skip the algebra.

$3.$ How would you determine to prove that $u', w'$ are unit vectors?

$4.$ How would you determine to compute $|u'|^2$, in order to prove that $u', w'$ are unit vectors? I would've calculated $|u'|$ instead? This also feels clever and guileful.

By inspection, we see that hypotheses (I)-(IV) as required by the question are satisfied. Due to $(*)$ as well, u' and w' are also eigenvectors for $A^2$. So from (iii), each of $\{u, u' \} \perp $ each of $\{w, w' \} $. We are left with proving $u \perp u'$ and $w \perp w'$.

$5.$ I see what (iii) says, but I still don't perceive why "each of $\{u, u' \} \perp $ each of $\{w, w' \} $" $?

So we want to prove $<u, u'> =0$. $<u, u'> = u^T u' = \lambda^{-1} \; \color{orangered}{ u^T Au } $. But because $\color{orangered}{ u^T Au } $ is a scalar, $\color{orangered}{ u^T Au } = ( \color{orangered}{ u^T Au } )^T = ... = -u^T Au \implies \color{orangered}{ u^T Au } = 0$.

$6.$ How would you determine/divine/previse the trick here, of transposing a scalar, to prove $<u, u'> = 0$?

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$(1)$ Yes any eigen vector can be normalised, i.e. if $v$ is an eigen vector, let $v'=\frac{v}{\|v\|}$

$(2)$ Take the two formulas you are given: $(I). Au=λu', (II). Aw=μw'$

$(I). Au=λu'\Rightarrow u'=\frac{1}{\lambda}Au$, and

$(II). Aw=μw'\Rightarrow w'=\frac{1}{\mu}Aw$.

So this seems to be the sensible inference (though I don't know if I would have though of it).

$(3)$ $|u'|^2=\langle u',u'\rangle=\langle \frac{1}{\lambda}Au,\frac{1}{\lambda}Au\rangle=\frac{1}{\lambda^2}(Au)^TAu=\frac{1}{\lambda^2}u^TA^TAu=-\frac{1}{\lambda^2}u^TA^2u=\frac{1}{\lambda^2}\lambda^2u^Tu=\langle u,u\rangle=|u|^2=1$

Thus $|u'|^2=1\Rightarrow |u|=1$ (the similar computation applies for $w'$)

$(4)$ Calculating $|u'|^2$ first is typically easier, since you do not have to deal with square roots in your computation, and then in the end $|u'|^2=1\iff|u|=1$.

$(5)$ All the vectors $u,u',w,w'$ are orthogonal, since:

$u\perp w,u\perp w',u'\perp w,u'\perp w'$, these all follow since they eigenvectors corresponding to distinct eigenvalues, and thus they must be orthogonal.

The other orthogonalities are coverd in $(6)$ below:

$(6)$ I would devise this because for any anti symmetric matrix, i.e. $A^T=-A$, we have $\langle Au,u\rangle=0$, this is because $\langle Au,u\rangle=u^TAu=(u^TAu)^T=u^TA^Tu=-u^TAu$.

So all in all we have $u^TAu=-u^TAu\iff u^TAu=0$

Whenever a useful property like that is at your disposal, you should try to use it.

So we have $\langle u',u\rangle=\langle \frac{1}{\lambda}Au,u\rangle=\frac{1}{\lambda}\langle Au,u\rangle=0$.

Please say if anything is unclear.

edit for part $(6)$ computation:

Okay so $\langle Au,u\rangle$ is a scalar, and for a scalar $a^T=a$ (since it is really a one dimensional matrix.)

So $\langle Au,u\rangle=u^TAu$, we know that if we transpose this we get the same number, so we have:

$(u^TAu)^T=(Au)^T(u^T)^T$ (this is due to the rule, for any two vectors $a,b$, $(ab)^T=b^Ta^T$.)

So $(u^TAu)^T=(u)^T(u^TA)^T=u^TA^Tu=-u^TAu$ ($A$ is antisymmetric, so $A^T=-A$)

$\Rightarrow u^TAu=(u^TAu)^T=-u^TAu\Rightarrow u^TAu=-u^TAu\Rightarrow u^TAu=0$

edit (transpose of a scalar)

The reason I thought to use the scalar transpose trick, is because for any proof/question it is generally a good idea to use all the information you are given, so I know the fact that $A^T=-A$ was going to be a part of it somewhere, and an easy way to show that $\langle Au,u\rangle=0$ for an antisymmetric matrix, is to use the scalar transpose trick.

Okay so I have proven that $A^T=-A\Rightarrow \langle Au,u\rangle =0$, but a crucial step in this is the scalar transpose.

So basically "$A^T=-A$ + scalar transpose trick" $\Rightarrow \langle Au,u\rangle =0$

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  • $\begingroup$ Thanks. I added an A; I think it was missing? For (3), thanks for the computations; I also wanted to know how you'd realise or determine that $u',w'$ are unit vectors, before calculating $|u'|^2$? For (6), I still don't perceive the trick of transposing a scalar. I think that's what you did in the equation that I edited? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 13:24
  • $\begingroup$ @LePressentiment, I have put in an edit, that I believe addresses this, and with regards to calculating $|u'|^2$, would calculate $|u'|^2$ before $|u'|$ as it is general a simpler computation :) $\endgroup$ – Ellya May 28 '14 at 13:36
  • $\begingroup$ @LePressentiment I would* $\endgroup$ – Ellya May 28 '14 at 13:47
  • $\begingroup$ @LePressentiment, see my edit :) $\endgroup$ – Ellya May 28 '14 at 17:37
  • $\begingroup$ Thanks again. I apologise for the many questions, but what's the intuition behind $A^T = -A$ and $<Au, u>$? I see that you proved $A^T = -A \implies <Au, u> = 0 $, so I'm trying to see why the scalar transpose trick is natural. $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 17:44
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1) Yes, eigenvectors are better thought of as directions that are invariant under the effect of the linear map. As directions, they do not have unique magnitudes.

2) I think this is poorly phrased but valid. I think part of the issue you're having is that, by calling these vectors $u'$ and so on, it seems like it must already be proven that $u'$ is part of the mutually orthonormal set the problem asks for. This is not yet proven, but the aim of the problem is to prove that property.

It would be quite valid merely to define a vector $a = A(u)/\lambda$ and so on and prove the relevant properties later. I think this is a clearer progression of logic.

3) The problem asks for an orthonormal set. Setting out to prove that $a$ is unit seems like a natural thing to do.

4) You would try to find $|a|^2$ because it follows more easily from the idea of using $A^2$. Observe:

$$|a|^2 = \frac{1}{\lambda^2} A(u) \cdot A(u) = \frac{1}{\lambda^2} A^TA(u) \cdot u = -\frac{1}{\lambda^2} A^2(u) \cdot u$$

But you know that $A^2(u) = -\lambda^2 u$, and you get that $|a|^2 = 1$, which implies that $|a| = 1$.

5) Remember that $A^2$ is symmetric. In particular, for any $b, c$:

$$A^2(b) \cdot c = A(b) \cdot A^T(c) = (A^T)^2(c) \cdot b = (-1)(-1) A^2(c) \cdot b = A^2(c) \cdot b$$

That proves symmetry. Since $u, w$ are eigenvectors of $A^2$, they must be orthogonal by (iii). $u', w'$ are also eigenvectors of $A^2$, so they must be mutually orthogonal.

6) The trick here is merely a way to prove that $A(b) \cdot b =0$ for any $b$. You might suspect it's the case by nature of the operator being antisymmetric.

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  • $\begingroup$ Thanks. For (4), I don't follow. $<Au, Au> = (Au)^TAu = u^TA^T \quad Au $? This isn't yourd 3rd equation though? $\endgroup$ – Greek - Area 51 Proposal May 28 '14 at 13:25
  • $\begingroup$ You're right; fixed. $\endgroup$ – Muphrid May 28 '14 at 14:12

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