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On my exam today there's this question:

A is a real n by n matrix and it is its own inverse. Prove that A is diagonalizable.

It seems very few students solved it if any.

All I know is that it's eigenvalue has to be 1 or -1. But how do I know the dimension of the eigenspace is enough? I've searched through internet and the solutions I found is all about minimal polynomial which I haven't learnt. Is there any method using only properties of eigenvectors?

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  • $\begingroup$ What conditions do you know of for diagonalisability? For instance, do you know a matrix is diagonalisable if and only if $$\operatorname{ker}(A - \lambda I)^2 = \operatorname{ker}(A - \lambda I)$$ for each $\lambda$? $\endgroup$ – Theo Bendit Jun 15 '18 at 1:51
  • $\begingroup$ @Theo Bendit the method we use through this class is to find a basis consisting of eigenvectors. $\endgroup$ – Fluffy Skye Jun 15 '18 at 1:53
  • $\begingroup$ @Theo Bendit I actually don't know that. Is that easy to show? $\endgroup$ – Fluffy Skye Jun 15 '18 at 1:54
  • $\begingroup$ I admit, I don't really know a nice direct method for showing this. It's a result that falls out of of the Jordan Basis theory. There are some really excellent tools for describing diagonalisability, but a bit of work needs to be done previously. $\endgroup$ – Theo Bendit Jun 15 '18 at 1:57
  • $\begingroup$ @Theo Bendit Well, since this is on my linear algebra final exam. We should be able to solve it using knowledge we have. Especially all other problems on my exam are exceptionally easy compared to this. Maybe there's some smart argument? $\endgroup$ – Fluffy Skye Jun 15 '18 at 1:58
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If you haven't covered minimal polynomials and related topics this was a hard question. Here's a simple proof, using nothing but the definitions. (The proof looks like magic - I don't see how anyone would think of it if they hadn't learned about minimal polynomials etc.)

First:

If $x$ is any vector then $x+Ax$ and $x-Ax$ are eigenvectors of $A$.

Proof: Say $z=x+Ax$. Then $$Az = A(x+Ax)=Ax+A^2x=Ax+x=z.$$Similarly if $z=x-Ax$ then $Az=-z$..

Hence:

Any vector is a linear combination of eigenvectors of $A$.

Proof: $x=\frac12(x+Ax)+\frac12(x-Ax)$.

Now say $E$ is the set of eigenvectors of $A$. We've shown that $E$ spans $\Bbb R^n$. Since any spanning set contains a basis, $E$ contains a basis for $\Bbb R^n$. That is, there is a basis consisting of eigenvectors, so $A$ is diagonalizable.

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  • $\begingroup$ Yes! This is really skillful! In fact our score came out and the highest is full mark! I guess some people are just smart lol. $\endgroup$ – Fluffy Skye Jun 16 '18 at 22:35
  • $\begingroup$ @FluffySkye I can finally delete my incorrect answer. $\endgroup$ – Kenny Lau Jun 17 '18 at 0:43
  • $\begingroup$ @Kenny Lau Is it incorrect? The idea is the same though. $\endgroup$ – Fluffy Skye Jun 17 '18 at 4:57

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