I have been trying to prove the following result: If $A$ is real symmetric matrix with an eigenvalue $\lambda$ of multiplicity $m$ then $\lambda$ has $m$ linearly independent e.vectors.
Is there a simple proof of the following result?
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linear-algebra
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0Can you use the theorem that a real symmetric matrix can be diagonalized? That makes it easy but may be out of order. – 2012-12-21
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0Do you know that a symmetric matrix is diagonalizable? – 2012-12-21
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1@RossMillikan: you did beat me by 2 seconds ;-) – 2012-12-21
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0If $A$ has an eigenvector $v$, then $A$ maps $\text{sp} \{v \}$ into $\text{sp} \{v \}$, and $(\text{sp} \{v \})^\bot$ into $(\text{sp} \{v \})^\bot$. You can use this to show that $A$ must have a set of eigenvectors that spans the space in question. – 2012-12-21