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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).
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    $\begingroup$ 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. $\endgroup$ – Laray Jul 13 at 13:16
  • $\begingroup$ Can you elaborate on why $V$ will have orthogonal eigenvectors again? $\endgroup$ – Nagabhushan S N Jul 13 at 13:33
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    $\begingroup$ 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. $\endgroup$ – Laray Jul 13 at 13:46
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    $\begingroup$ 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? $\endgroup$ – Nick Alger Jul 15 at 9:08
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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.

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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]

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