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Show that if $A$ has $n$ linearly independent eigenvectors, then so does $A^{T} .[\text { Hint: Use the Diagonalization Theorem. }]$

Thus, $A^{T}$ is diagonalizable.

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 3

Diagonalization

Vectors

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Okay, So if the 80 very transpose that semen he DP interests transposed and then that's it equal to p inverse transposed the transformers and be transposed, that's able to pee transposed in verse, transpose and transpose. But this is a sequel to Q B G and cue in verse, where Q is equal to P. T and GT is a diagonal rejects. So based on this, we know that a T is diagonal Izabal.

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