# Can performing row (column) operations to a set of dependent eigenvectors produce a basis of vectors that are again eigenvectors?

I am having some trouble finding a basis of eigenvectors that diagonalizes two matrices simultaneously.

I have found two bases of eigenvectors for two 3x3 matrices.

I can't seem to find an algorithm or any discussions on MSE on how to do this.

So, my idea is to arrange all 6 vectors into rows, place them in a matrix, and row-reduce (which is then equivalent to performing column operations on the 6 column vectors) until I find 3 linearly independent rows. These 3, new, row vectors form a new basis for $R^3$, but does not quite diagonalize my matrices anymore. So, they aren't eigenvectors anymore -- at least not all of them are.

But I feel this guess at an algorithm almost works -- using this new basis, my matrices A and B were almost diagonal. Off by one non-zero entry away from the main diagonal.

Do you know of an algorithm to find the explicit basis for simultaneously diagonalizability of matrices? I have only seen proofs on MSE of the existence of such a basis, which I already understand and have verified. But I want to compute this basis.

Thanks,

EDIT: the two matrices commute.

• are your matrices symmetric (or Hermitian)? – etothepitimesi Aug 10 '15 at 23:44
• hi @el.Salvador, no they are neither symmetric nor Hermitian ... – User001 Aug 10 '15 at 23:45

Consider for example

$$A = \pmatrix{4 & 6 & 0\cr -1 & -1 & 0\cr 3 & 6 & 1\cr}, B = \pmatrix{8 & 6 & -2\cr -2 & 1 & 1\cr 4 & 6 & 2\cr}$$

which do commute, and are diagonalizable $A$ has eigenvalues $1$ (with multiplicity $2$) and $2$, and $B$ has eigenvalues $3$ and $4$ (with multiplicity $2$).

Eigenvectors for the simple eigenvalues should be in the basis: say $$u = \pmatrix{3 \cr -1\cr 3\cr}, \ v = \pmatrix{1 \cr 0\cr 2\cr}$$ (for eigenvalue $2$ of $A$ and $3$ of $B$ respectively). $u$ is also an eigenvector of $B$ for eigenvalue $4$, and $v$ is also an eigenvector of $A$ for eigenvalue $1$.

The third member of the basis, say $w$, must be another eigenvector of $A$ for eigenvalue $1$ and of $B$ for eigenvalue $4$. Therefore it satisfies $(A - I) w = 0$ and $(B - 4I) w = 0$. Putting $A - I$ on top of $B-4I$ and row-reducing the resulting $6 \times 3$ matrix, we find a solution: $$w = \pmatrix{2\cr -1\cr 1\cr}$$

• Hi Professor Israel, thanks so much for this example. I have found the basis that puts both of my matrices in diagonal form. Can I ask you a couple of follow-up questions? Why does this method work? In your example, the vector u may not be an eigenvector of matrix B, and similarly the vector v may not be an eigenvector of matrix A. – User001 Aug 11 '15 at 2:03
• So, it doesn't seem like we are building a basis of common eigenvectors, except for the last eigenvector that we explicitly compute with row operations. Also, finding the last eigenvector was a bit tricky: the null space is of dimension = 2, as we know, but if I select an eigenvector that is in the span of only one of the eigenvectors, then this method will not work. – User001 Aug 11 '15 at 2:03
• In other words, writing out the third vector as a linear combination of the two eigenvectors that span the null space, the coefficients of the linear combination are not allowed to be zero, which seems like a peculiar restriction. What are your thoughts? Thanks @RobertIsrael. – User001 Aug 11 '15 at 2:03
• Oh, perhaps this restriction provides the guarantee that we have a basis for $R^3$ of eigenvectors of matrices A and B. If one of the coefficients were zero, then the their vector can be viewed as living in just a one-dimensional eigenspace, and we will not have collected enough eigenvectors from either matrix to form a basis. – User001 Aug 11 '15 at 2:12
• Yes, the restriction needs to be considered, since the algebraic and geometric multiplicities of the eigenvalues need to be the same, in order to achieve diagonalizability of the matrices. Thanks again, Professor Israel - have a great night :-) – User001 Aug 11 '15 at 3:45