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Summary:: Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.
Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.
We know that if ##A## is diagonalizable then ##g(\lambda)=a(\lambda)##. So all we have to show is that ##A## is diagonalizable.
I found a proof by contradiction: Assuming ##A## is not diagonalizable we have
$$(A- \lambda_i I)^2 v=0, \ (A- \lambda_i I) v \neq 0$$
Where ##\lambda_i## is some repeated eigenvalue. Then
$$0=v^{\dagger}(A-\lambda_i I)^2v=v^{\dagger}(A-\lambda_i I)(A-\lambda_i I) \neq 0$$
Which is a contradiction (where ##\dagger## stands for conjugate transpose).
But I do not really understand it; what is meant by 'generalized eigenvalues of order ##2## or higher'?
I asked about this proof a while ago but the answer I got did not really convince me...
We could go over an alternative proof of course, the aim is understand how to show that real symmetric matrices are diagonalizable.
Thank you!
Let ##A \in \Bbb R^{n \times n}## be a symmetric matrix and let ##\lambda \in \Bbb R## be an eigenvalue of ##A##. Prove that the geometric multiplicity ##g(\lambda)## of ##A## equals its algebraic multiplicity ##a(\lambda)##.
We know that if ##A## is diagonalizable then ##g(\lambda)=a(\lambda)##. So all we have to show is that ##A## is diagonalizable.
I found a proof by contradiction: Assuming ##A## is not diagonalizable we have
$$(A- \lambda_i I)^2 v=0, \ (A- \lambda_i I) v \neq 0$$
Where ##\lambda_i## is some repeated eigenvalue. Then
$$0=v^{\dagger}(A-\lambda_i I)^2v=v^{\dagger}(A-\lambda_i I)(A-\lambda_i I) \neq 0$$
Which is a contradiction (where ##\dagger## stands for conjugate transpose).
But I do not really understand it; what is meant by 'generalized eigenvalues of order ##2## or higher'?
I asked about this proof a while ago but the answer I got did not really convince me...
We could go over an alternative proof of course, the aim is understand how to show that real symmetric matrices are diagonalizable.
Thank you!