Proof Hint: Show +-1 are Eigenvalues of T on nxn Matrices with Real Entries

[eckiller]

Hi everyone,

Let T be a linear operator on nxn Matrices with real entries defined by
T(A) = transpose(A).

Show that +-1 are the only eigenvalues of T.


Any tips on how to start this. I thought about writing the matrix representation relative to the standard basis, but it seemed really messy/tedious to write that out in general. Is there an easier way, or is that the only way to go?

 

[Don Aman]

use the fact that T^2 = id

 

[M98Ranger]

To start, we can write out the matrix representation of T with respect to the standard basis. This will give us a better understanding of the linear transformation and allow us to see how the eigenvalues behave. Let's consider a generic nxn matrix A = (a_ij) with real entries. Then, T(A) = (b_ij) where b_ij = a_ji (the transpose of A).

Now, let's consider the eigenvector equation T(A) = cA where c is an eigenvalue and A is an eigenvector. Substituting in our expression for T(A), we get (b_ij) = c(a_ij). This means that for each entry in the matrix, we have b_ij = ca_ij.

We can see that this is only possible if c = 1 or c = -1. In other words, the only possible eigenvalues for T are +1 and -1. To prove that these are indeed eigenvalues, we can choose specific matrices A and see what happens when we apply T to them.

For example, let's take A = (1 0; 0 1). Applying T, we get T(A) = (1 0; 0 1) = A. This means that A is an eigenvector with eigenvalue +1. Similarly, if we take A = (0 1; 1 0), we get T(A) = (0 1; 1 0) = -A. This shows that A is an eigenvector with eigenvalue -1.

Thus, we can conclude that the only possible eigenvalues for T are +1 and -1. This is because the linear transformation T(A) = transpose(A) essentially flips the matrix about its main diagonal, which corresponds to multiplying by +1 or -1. Therefore, +1 and -1 are the only eigenvalues of T on nxn matrices with real entries.