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Let \(A = PD{P^{ - {\bf{1}}}}\), where P is orthogonal and D is diagonal, and let \(\lambda \) be an eigenvalue of A of multiplicity k. Then \(\lambda \) appears k times on the diagonal of D.Explain why the dimension of the eigenspace for \(\lambda \) is k.

Short Answer

Expert verified

The dimension of eigenspace is k.

Step by step solution

01

Check, if p linearly independent eigenvectors

The orthogonal matrix P, which diagonalizes matrix A is formed by linearly independent eigenvectors.

For all eigenvalues of A, there is an eigenvector.

So, for p distinct eigenvalues of A, there are p linearly independent eigenvectors

02

Find the dimension of the eigenspace of \(\lambda \)

If there are n linearly independent vectors for matrix P, then \(n - p\) the eigenvector of Pcorresponds to the remaining eigenvalues.

The remaining eigenvalues are equal to \(\lambda \) and have multiplicity k.

So, for \({\lambda _k}\), there are k linearly independent vectors.

Thus, the dimension of eigenspace is k.

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