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So I've been given two diagonal matrices with non matching eigenvectors, A:$$ \begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \\ \end{matrix} $$ and B: $$ \begin{matrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 0 \\ \end{matrix} $$

Eigenvectors for A: (0, 0, 1), (-1, 0, 1), (0,1,0) Eigenvectors for B: (-1,1,0), (0,0,1), (-1,-1,1) AB=BA (I have proven this), but how do I go about making a joint basis for the two matrices so that I can have a matrix which diagonalizes both A and B?

Edit 1: So, I've figured out two of the three vectors which would make up the matrix I am looking for (that would simultaneously diagonalize A and B). (0,0,1) and (-1,1,1) both belong to the vector space spanned by A and B. Now I just need the third and I should be good. I'd be grateful for any input

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Note that the eigenspace for $A$ corresponding to $\lambda=1$ is the two-dimensional $\{s(0,0,1)+t(-1,0,1)\colon s,t\in\Bbb R\}$; in particular, there are other eigenvectors besides the ones you listed. Similarly for $B$ and $\lambda=0$: the eigenspace is $\{s(-1,1,0)+t(0,0,1)\colon s,t\in\Bbb R\}$. So you have more possibilities for the eigenvectors from which to form the joint basis.

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  • $\begingroup$ So @Greg Martin I've found two of the vectors (as listed in my edit). Would (-1,1,0) also be a valid vector lying in the joint vector space? Adding (-1,0,1) and (0,1,0) of A and subtracting (0,0,1) of A gives me (-1,1,0) I believe, although I am unsure if that is valid since that would not be a linearly independent vector to add... $\endgroup$ Commented Apr 24, 2019 at 2:23

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