- #1
JTemple
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Homework Statement
Show that if an nxn matrix A has n linearly independent eigenvectors, then so does A^T
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?