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Suppose that $A$ is a real symmetric $n\times n$ matrix whose eigenvalues are not all distinct. Show that

  1. If an eigenvalue $\lambda_1$ has multiplicity $2$, it has two orthonormal eigenvectors $b_1,\,b_2$.

  2. There cannot be more than two linearly independent eigenvectors with that eigenvalue $\lambda_1$. Construct an orthonormal basis of $\mathbb{R}^n$ using the eigenvectors.

(I have deleted parts (c) and (d) of my question.)

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    $\begingroup$ Too many questions at once. $\endgroup$ – Git Gud Sep 15 '13 at 13:50
  • $\begingroup$ (c) is obviously wrong: think about the identity matrix of size $N$. $\endgroup$ – Bazin Sep 15 '13 at 14:06
  • $\begingroup$ What do you mean when you write $Av_1v_2$? This makes no sense to me. But I think you mean some inner product, and so you're on the right track. $\endgroup$ – Ted Shifrin Sep 15 '13 at 14:43
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    $\begingroup$ (b) is nonsense, unless the eigenvalue in (b) is the one in (a), the one with multiplicity 2. Can you prove that an eigenvalue with multiplicity 2 has at most 2 linearly independent eigenvectors? The other hypotheses on $A$ are not needed for this. $\endgroup$ – Gerry Myerson Sep 16 '13 at 10:22
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Since $A$ is real symmetric, $A=Q^*DQ$ for some orthogonal $Q$. We have

$(A-\lambda I)v=0$ if and only if $Q^*(D-\lambda I)Qv=0$.

Since $Q$ is invertible, there is one to one correspondence between the eigenvectors of $A$ and $D$, corresponding to eigenvalue $\lambda$. In particular, the dimensions of the null spaces of $A-\lambda I$ and $D-\lambda I$ must then be the same. If the algebraic multiplicity of $\lambda$ is $2$, then $D-\lambda I$ has rank equal to $n-2$. So the dimension of $ker(D-\lambda I)=2$. This corresponds with the dimension of eigenspace of $\lambda$.

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