(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that if an nxn matrix A has n linearly independent eigenvectors, then so does A^T

3. The attempt at a solution

Well, I understand the following:

(1) A is diagonalizable.

(2) A = PDP^-1, where P has columns of the independent eigenvectors

(3) A is invertible, meaning it has linearly independent columns, and rows that are not scalar multiples of each other.

(4) A^T has the same eigenvalues of A.

So, based mainly on (4), I can prove the "easy" case, in which there are n distinct eigenvalues.

IE: If A has n distinct eigenvalues, then A^T has those same distinct eigenvalues. Thus, If lambda_1 through lambda_n are distinct, then they each correspond to distinct eigenvectors v_1 through v_n for A and v_1T through v_nT for A^T.

In this case, the eigenvectors could be the same (in the case that A=A^T), but don't have to be.

My problem!

What if the eigenvalues are not distinct, IE there is some lambda_i with multiplicity =k.

I understand that in the case of A, the (A-lambda_i*I_n) matrix must have k free variables in order to give the (stated) n linearly independent eigenvectors.

So, how can I prove that the repeated eigenvalues in the A^T case ALSO correspond to (A^T-lambda_i*I_n) having k free variables and thus n linearly independent eigenvectors?

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# Linear Algebra: Diagonalization, Transpose, and Disctinct Eigenvectors.

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