Is it possible to diagonalize a matrix without eigenvalues, or in other words, does the diagonal matrix of a diagonalization always have the eigenvalues of a matrix?

for example: if a 3x3 matrix has eigenvalues 1,2,3 can you diagonalize that matrix with a diagonal matrix that does not have the elements 1,2,3

my suspicion is the answer is no, but cant confirm my answer yet

  • $\begingroup$ I want to say it is not possible, $A$ diagonalizable implies just by definition that you have an invertible matrix $B$ $\ni$ $B^{-1}AB=I$ where the diagonal entries of $I$ are eigenvalues of the eigenvectors of $A$ which are column vectors of $B$ . I believe this is a definition so can't see another way. $\endgroup$ – Hossien Sahebjame Mar 28 '20 at 3:40
  • $\begingroup$ @HossienSahebjame no thats not the precise definition. the part where you say the diagonal entries and the columns of B consist of the eigenvectors of B are not included in the definition. that was the reason why i asked. the definition is: A is diagonalizable if there exists an invertible matrix P and a diagonal matrix D such that P^(-1)AP=D. it doesnt say anything about eigenvalues and eigenvectors, so my train of thought was, can you also diagonalize A without eigenvalues and eigenvectors, but i think the answer is no $\endgroup$ – user753933 Mar 28 '20 at 4:01
  • $\begingroup$ The characteristic polynomial of a matrix is invariant under base change. So it is the same for the initial matrix, and the diagonal matrix. In your example, the original matrix has the characteristic polynomial $(x-1)(x-2)(x-3)$, but after that it is $\prod(x-\text{"some other value"})$... (The "other values" are the diagonal entries of the "other diagonalization", well some of them may be among $1,2,3$, but we assume that at least one is really different...) $\endgroup$ – dan_fulea Mar 28 '20 at 4:20
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    $\begingroup$ $A = BDB^{-1}$ implies $AB = BD$. If the columns of $B$ are $b_1,\ldots,b_n$ and the diagonal elements of $D$ are $d_1,\ldots,d_n$, you can check that $AB = BD$ is equivalent to $Ab_i = d_i b_i$ for $i=1,\ldots,n$. As $B$ is invertible, the $b_i$ are all nonzero, so this means precisely that the $d_i$ are eigenvalues and $b_i$ are corresponding eigenvectors. $\endgroup$ – Bungo Mar 28 '20 at 4:47

If you can diagonalize a matrix, the diagonal matrix always has the eigenvalues on the diagonal. If your matrix is $A$ and you can find an invertible $B$ so that $B^{-1}AB=D$ with $D$ diagonal the eigenvectors of $D$ are the basis vectors with each one corresponding to the matching diagonal entry. If the upper left element of $D$ is $D_{11}$ we have $De_1=D_{11}e_1$ so $D_{11}$ is an eigenvalue. Then we have $ABe_1=BDe_1=D_{11}Be_1$ so $Be_1$ is an eigenvector of $A$ corresponding to eigenvalue $D_{11}$.


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