# eigenvectors of an eigenvector matrix

Suppose I have a square matrix $$A\in \mathbb{R}^{d\times d}$$, with eigenvectors $$v_1,v_2,\ldots,v_n$$. Suppose I construct a new matrix $$V = [v_1\ v_2\ \cdots\ v_n]$$. Can anything be said about the eigenvalues or eigenvectors of this new matrix $$V$$. Do, $$A$$ and $$V$$ have same eigenvalues?

PS: I've not assumed A to be symmetric and $$d$$ can be greater than $$n$$. But if it helps derive something, please feel free to assume so or any other assumptions required.

PPS: I know that if $$V$$ is full rank, it diagonalizes $$A$$. I'm just wondering if there is any other relation.

Edit 1: Adding some special cases, that can be considered

1. $$A$$ is real-symmetric, so that $$v_1, v_2, \ldots$$ are orthogonal and $$d=n$$.
2. $$A$$ is normal (above holds).
• If $A$ is normal (for example symmetric or skew-symmetric) the matrix $V$ of eigenvectors will be orthogonal. This gives eigenvalues on the unit circle and again an orthogonal set of eigenvectors. Commented Jul 13, 2020 at 13:16
• Can you elaborate on why $V$ will have orthogonal eigenvectors again? Commented Jul 13, 2020 at 13:33
• If $A$ is symmetric, it is normal. All normal matrices are orthonormally diagonizable, meaning $V$ is a unitary matrix (orthogonal in the case of symmetric $A$). Unitary matrices are normal themselves, meaning they have an orthonormal basis of eigenvectors. Commented Jul 13, 2020 at 13:46
• This is a super interesting idea. One could repeat the process to get the eigenvectors of the eigenvectors of the eigenvectors, and so on. Then these "trajectories" (sequence of matrices starting from a given matrix) could be done for all invertible matrices, generating a partition of the set of invertible matrices into equivalence classes. Does this induce some algebraic structure? Does anyone know if this has been studied? Commented Jul 15, 2020 at 9:08

There is no reason for there to be any relation between $$A$$ and $$V$$ whatsoever. Of course $$V$$ is invertible since you are supposing the eigenvectors to form a basis (at least that is what I suppose, although the question does not actually say it), but apart from that it could be any matrix. And you can get loads of matrices satisfying the requirement that a given basis becomes one of eigenvectors: you can freely specify an eigenvalue independently for each of the vectors. This shows that the eigenvalues of $$A$$ cannot be determined even if $$V$$ is entirely known. Conversely, fixing $$A$$ you have quite some freedom for $$V$$, namely at least to independently scale each column, which completely messes up any eigenvalues that $$V$$ might have. If $$A$$ is a multiple of $$I$$, you can even choose $$V$$ to be any invertible matrix you want.
Not necessarily $$A$$ and $$V$$ will have same eigenvalues.
[ If $$A$$ is diagonalisable with an eigenvalue $$0$$ then one can construct $$V$$, invertible. Or, If $$A$$ is not diagonalisable with full rank then one can make $$V$$ with $$0$$ as an eigenvalue]