Eigenvalues of transpose linear transformation

[Mr Davis 97]

Homework Statement

If $A$ is an $n \times n$ matrix, show that the eigenvalues of $T(A) = A^{t}$ are $\lambda = \pm 1$

Homework Equations

The Attempt at a Solution

First I assume that a matrix $M$ is an eigenvector of $T$. So $T(M) = \lambda M$ for some $\lambda \in \mathbb{R}$. This means that $M^t = \lambda M$. Then $\det (M^t) = \det (M)$. So $\det(M) = \lambda^n \det (M)$. But I can't seem to cancel the determinants since we don't know if $M$ is invertible or not. This is where I get stuck.

 

[fresh_42]

The statement is plain wrong: choose $A=0$ or any diagonal matrix $diag(\lambda_1,\ldots ,\lambda_n)$ with $\lambda_i$ of your choice. And these are only the simple counterexamples. Could it be, that $A\cdot A^t=1$ is the given condition?

 

[Mr Davis 97]

The statement is plain wrong: choose $A=0$ or any diagonal matrix $diag(\lambda_1,\ldots ,\lambda_n)$ with $\lambda_i$ of your choice. And these are only the simple counterexamples. Could it be, that $A\cdot A^t=1$ is the given condition?

But if $M^t = \lambda M$, and if $\lambda = \pm 1$, wouldn't the symmetric and anti-symmetric matrices work?

 

[fresh_42]

But if $M^t = \lambda M$, and if $\lambda = \pm 1$, wouldn't the symmetric and anti-symmetric matrices work?

What does this mean for a vector? A column vector is a multiple of a row vector? This would only mean $n=1$ and $\lambda (M-1)=0$, i.e. $\lambda = 0$ or $M=1$. Since the claim is obviously wrong, why bother an attempt to prove it? It cannot be proven. Perhaps I didn't understand your $T$. To me it looked like $T : A \longmapsto A^t$.

 

[Mr Davis 97]

I'll write out the problems exactly as it is written: Let $T$ be a linear operator on $M_{n \times n} (\mathbb{R})$ defined by $T(A) = A^t$. Show that $\pm 1$ are the only eigenvalues of 1.

The problem is from a standard textbook (Friedberg), so I don't think there's anything wrong with the problem

 

[fresh_42]

O.k., I thought it is about the eigenvalues of $A$. My fault, sorry. Then you might look at $T^2$ for the solution.

 

[Mr Davis 97]

O.k., I thought it is about the eigenvalues of $A$. My fault, sorry. Then you might look at $T^2$ for the solution.

That ends up working well. Why doesn't the determinant approach work well? I want to know so that next time I come across a similar problem I won't spend time going own that path

 

[fresh_42]

Why doesn't the determinant approach work well?

I think I know it now. It's the same thing as my error arose from. $M$ is a vector, and its determinant isn't important. We should only look at the determinant of $T$, which is clearly a regular mapping with $\det(T)^2 = 1$. The equation you found isn't wrong, but as you stated correctly, we don't know anything about $M$, only that $M \neq 0$ which isn't enough, if we pass to the determinant of $M$. I mean this reduces all $n^2$ information about $M$ to a single number, and more, it distracts from the properties of $T$.