Show that 1 and -1 are the only eigenvalues of this linear transformation

Define $T: M_{n\times n}\to M_{n\times n}$ by $T(A):= A^t$. Note that $T$ is a linear transformation.

Show that $1$ and $-1$ are the only eigenvalues of $T$.

Let $\lambda$ denote an eigenvalue of $T$. Then $T(A) = \lambda A = A^t$. We know that if $A$ is a symmetric matrix, then $A = A^t$ and if $A$ is a skew symmetric matrix then $-A = A^t$, ie. the only eigenvalues of $T$ are $1$ and $-1$ for the cases where $A$ is symmetric and skew symmetric.

Do I need to show these are the only possible times that $\lambda A$ can equal $A^t$? If so, how can I show that there are no other cases?

• Raising a matrix to a power is not a linear transformation. Does $(A+B)^t = A^t + B^t$? Does $(cA)^t = c(A^t)$ ? – DanielV May 25 '15 at 2:37
• @DanielV It means transpose. – Ian May 25 '15 at 2:38
• Hint: In general if $L : V\to V$ is linear and $L^2 = id$, what is the possible choice of eigenvalues? – user99914 May 25 '15 at 2:39
• @DanielV, no mention of powers in this problem, only transpose and eigenvalues. – Chad May 25 '15 at 2:40
• @John, hey, this is a serious question (it may seem sarcastic, but why do people always edit questions? It doesn't really change anything as to how it is viewed, so why the need? (Sorry, I'm new to the site and still trying to understand it. – Chad May 25 '15 at 2:42

if $\lambda$ is an eigen value and the corresponding eigenvector is $A \neq 0,$ then $$A^\top = \lambda A$$ taking transpose gives you $$A = \lambda A^\top=\lambda^2 A$$ which show you $$\lambda^2 = 1\implies \lambda = \pm 1.$$

• Genius! Thanks! – Chad May 25 '15 at 2:43
• This answer is really useful... – Tani May 25 '15 at 5:19

Note that $T^2$ is an identity transformation, so the minimal polynomial of $T$ is $\lambda^2-1.$

More generally, given a linear transformation $L$ such that there is a least positive integer $k$ for which $L^k$ is an identity transformation, we will always have that the minimal polynomial of $L$ divides $\lambda^k-1.$ (We may, in fact, have equality, but I'm not certain of that.)

• how can I conclude that because $T^2$ is the id transformation, the characteristic equation is $\lambda^2 = 1$? – Chad May 25 '15 at 2:54
• Because the identity transformation has $1$ as its only eigenvalue, and for any eigenvalue $\lambda$ of $T,$ we have that $\lambda^2$ is an eigenvalue of $T^2:=T\circ T.$ – Cameron Buie May 25 '15 at 2:59
• $T$ is a linear operator on an $n^2$-dimensional space: its characteristic polynomial has degree $n^2$; in fact, it is $(x - 1)^{n(n+1)/2} (x + 1)^{n(n-1)/2}$. Your approach is fine, of course; you just have the wrong description. – Hurkyl May 25 '15 at 11:36
• @Hurkyl: Good grief.... Thank you for bringing that to my attention. – Cameron Buie May 25 '15 at 14:43
• @Chad: Apologies for the confusion. I was misusing my terminology rather badly. – Cameron Buie May 25 '15 at 14:59