# Does common eigenvectors between two matrices A,B implies some property for the vectors?

If there are two matrixes that they have common eigenvectors for some eigenvalues that implies that those two matrixes are identical? What can we say for those two matrices?

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Consider $\bigl(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\bigr)$ and $\bigl(\begin{smallmatrix} 1 & 2 \\ 0 & 1 \end{smallmatrix}\bigr)$, all their eigenspaces are identical, but the matrices aren't identical. – Najib Idrissi Sep 25 '12 at 9:36
What can we say about the geometry of those matrices if we represent them in the euclidean space? – curious Sep 25 '12 at 9:46
You can't "represent" a matrix in a vector space. The fact that it is Euclidian or not does not matter... "Euclidian" is a topological property more used in metric spaces where you need to calculate a distance for example. In the case of two matrices that share the same set of eigenvectors you can think of this as the matrices "deforming" the vector space in the same way. You can see it as a combination of simultaneous dilatations in each direction defined by the eigenvectors. – mak Nov 18 '13 at 2:31

If the two matrices have the same eigenvalues with the same multiplicity they have the same characteristic polynomial. If the multiplicity of these eigenvalues equal the dimension of the eigenspaces (vector sub-spaces) then the two matrices are similar to a diagonal matrix up to a change of base. This similarity is a transitive property and then you can say that the two matrices you have represent the same endomorphism but in two different bases of the same vector space. The underlying endomorphism is the same but the matrices are definitely NOT identical. Remember that a matrix is a way to represent an endomorphism after one has chosen a vector base.

Edit:

Similar matrices $A$ and $B$ are such that there exists an invertible matrix $P$ such that: $A = PBP^{-1}$

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What do you mean by similar matrices? – curious Nov 4 '13 at 10:20
Edited the answer ;) – mak Nov 18 '13 at 2:25

If the two matrices are diagonalizable, then they must be equal.

Explanation:

Let $P$ be the matrix whose columns are a basis of eigenvectors of $A,B$. Then $P^{-1}AP=D_1,P^{-1}BP=D_2$ where $D_1,D_2$ are daigonal matrices with the (mutual) eigenvalues on the diagonal. Now we just need to make sure that the orders in which they are placed on the diagonal are the same. But these orders are in correspondence to the order of the columns of $P$ so we finished.

In particular this is true for any two symmetric positive-definite (s.p.d) matrices. This uniqueness result is used to prove the uniqueness of the positive square root of a s.p.d matrix.

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If two matrices have the same set of eigenvectors but different eigenvalues, then they can be simultaneously diagonalized, which means that the two matrices commute which each other, that is if the two matrices are A and B, AB = BA.

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can you elaborate more on "simultaneously diagonalized"? – curious Nov 4 '13 at 10:20
It means that they are similar to diagonal matrices written over the basis formed by the eigenvectors. – mak Nov 18 '13 at 2:27