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I want to diagonalize the Matrix $A$, so that $A = V*\Lambda*V^{-1}$. Where $V$ is the Matrix of Eigenvectors of $A$ and $\Lambda$ is the diagonal Matrix of Eigenvalues of $A$. However inverting $V$ can be a pain, so I thought it would be easier to use the Gram-Schmidt process to make my eigenvectors orthogonal, so that I can just transpose $V$ to invert it, but that gives me the wrong solution. Why? What would be an easier way to diagonalize $A$ without having to invert Matrices manually?

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  • $\begingroup$ if A is symmetric there exists an orthogonal basis (and you can use Graham -Schmidt) But if A is not symmetric, there is no guarantee of an orthogonal basis. $\endgroup$
    – Doug M
    Commented Aug 29, 2017 at 20:20

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Not all diagonalizable matrices can be put in diagonal form in a orthonormal basis, however, this can be achieved for symmetric ones.

If $\{e_1,\ldots,e_n\}$ is a basis of eigenvectors of $A$, then if $\{v_1,\ldots,v_n\}$ is the basis obtained by Gram-Schmidt orthonormalization, then $v_i$ needs not to be an eigenvector of $A$, all one has is the following:

  • For all $1\leqslant k\leqslant n$, $\textrm{Span}(e_1,\ldots,e_k)=\textrm{Span}(v_1,\ldots,v_k)$.

  • For all $1\leqslant k\leqslant n$, $\langle e_k,v_k\rangle>0$.

In all generality, a linear combination of eigenvectors is not an eigenvector.

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  • $\begingroup$ Thanks! Why the emphasis on orthonormal and not just orthogonal? $\endgroup$
    – macco
    Commented Aug 29, 2017 at 20:23
  • $\begingroup$ One can always choose unitary eigenvectors, so the emphasis is really on orthogonal as you notice. $\endgroup$
    – C. Falcon
    Commented Aug 29, 2017 at 20:25
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Not all matrices are orthogonally diagonalizable. If you just throw all of the eigenvectors into the Gram-Schmidt machine, you will end up with outputs that aren’t eigenvectors. The best you can do in general is to find an orthogonal basis for each eigenspace, but unless the eigenspaces are mutually orthogonal, there’s no orthonormal eigenbasis for the entire space.

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