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I know that for a symmetric matrix, its eigenvector are orthogonal. But for a repeated eigenvalue for a symmetric matrix, why still its eigenvalue must be still orthogonal.

I read somewhere that symmetric matrix can not have defective eigenvalue, so for a repeated eigenvalue with m multiplicity, we will still have m orthogonal eigenvector for it.

Kindly please for your answer, can you show by taking a 3 by 3 matrix and show it.

Anshul Sharma
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1 Answers1

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If a symmetric matrix $A$ has an eigenvalue $\lambda$ with multiplicity $k$, then the dimension of $\ker(A-\lambda I)$ is equal to $k$. Therefore, you can find an orthogonal basis for $\ker(A-\lambda I)$ (because all subspaces have an orthogonal basis).

Now note that

  1. The dimension of the kernel is $k$, therefore the basis has $k$ elements.
  2. All elements of $\ker(A-\lambda I)$ are eigenvectors of $A$.

From the points above, you can conclude that the orthogonal basis of $\ker(A-\lambda I)$ is comprised of $k$ orthogonal eigenvectors of $A$.

5xum
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  • Say if eigenvalue are like -1,-1 and 8 for N=3 dimension. Here multiplicity is 2? So for k multiplicity i.e for -1 and -1 both will correspond to different eigenvector and for eigenvalue=8 will have another eigenvector. I am confused . Also i am not familiar with ker. is the ker used for only repeated eigenvalue. – Anshul Sharma Apr 19 '21 at 06:02
  • @AnshulSharma In your case, the space $\ker(A+I)$, i.e. the eigenspace for the eigenvalue $-1$, will have a dimension of $2$. You can then go and find an orthogonal basis of the space, and the orthogonal basis will be two orthogonal vectors. Both will be eigenvectors. $\ker$ is the kernel (also called the nullspace) of $A$. If you do not know about nullspaces, then you are not yet ready to learn about eigenvalues. – 5xum Apr 19 '21 at 06:07