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I am trying to prove that a matrix A is diagonaizable. The matrix is 4x4, with a diagonal of a and the rest b, where b is not zero. I have found the determinant, and am going to find the eigenvectors, however, is it true that if $A=A^T$ then A is diagonalizable? If so, how can I prove this?

Thanks!

Liz Muir
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    It is indeed true. Check this post Why symmetric matrices are diagonalizable?. – yakobyd Jun 28 '18 at 19:26
  • It is based on property of normal matrices. You just need to study Schur's tridiagonalization theorem, which is very basic theorem , to know that matrices with property $A^HA= AA^H$ are diagonalizable. – Shak Jun 28 '18 at 19:58
  • All eigenvalues of $A$ are real. And you can show that the kernel of $A-\lambda I$ is the same as the kernel of $(A-\lambda I)^2$ for real $\lambda$ because $\langle (A-\lambda I)^2x,x\rangle = \langle (A-\lambda I)x,(A-\lambda I)x\rangle$ for real $\lambda$ and symmetric $A$. So the minimal polynomial of such $A$ has no repeated factors, which is equivalent to knowing that $A$ is diagonalizable. – Disintegrating By Parts Jun 28 '18 at 21:48

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