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What is the meaning of the multiplication of matrix B(composed of eigenvectors) and the transpose of B (eigenvectors are of a matrix A)? So, B's column vectors are eigenvectors of A, and I want to know what is the meaning of B*transpose(B)?

why there is no one answering this question...

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Can you clarify your question a bit? I find it hard to follow. – Alex Becker Apr 10 '11 at 19:39
Hope I am clear enough now. – Rn2dy Apr 10 '11 at 19:46
it's the matrix of the inner product in the basis given by the eigenvectors of $A$ - is it enough? – user8268 Apr 10 '11 at 19:50
@baboon: Yes, it makes sense now. – Alex Becker Apr 10 '11 at 19:55
@user8268: it is definitely inner product, but I'd like to know the properties of this matrix, I guess it is the projection matrix , it is orthogonal, and if I want to project a new sample vector to this matrix, so to do dimensionality reduction, correct me. – Rn2dy Apr 10 '11 at 19:58

If $A$ has a complete set of eigenvectors, $B$, (i.e. the eigenvectors of $A$ form a basis), then $A$ is self-adjoint with respect to an appropriate inner product. I.e., $\langle Ax, y \rangle_M = \langle x,Ay \rangle_M$. This inner product is often convenient to use when working with $A$. It is defined by a matrix $M$, i.e., $\langle x, y \rangle_M = x^T M y$, and $M$ is called the mass matrix in finite element applications.

We have that $AB = B\Lambda$, where $\Lambda$ is the diagonal matrix of eigenvalues of $A$. You can easily verify that we can write $A = M^{-1}S$, where $M^{-1} = BB^T$ is positive definite (as will be its inverse, the mass matrix), and $S = (B^{-1})^T\Lambda B^{-1}$ is symmetric (I think it might be called the stiffness matrix in some contexts, but I'm not sure about that). You can check directly that this definition of the mass matrix does indeed yield an inner product for which $A$ is self-adjoint.

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