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Can someone please state the most known conditions for a Matrix $A$$\in \mathcal{M_n}(\mathbb{R})$ to be diagonalizable?

Without proof, just by shedding light on them please. In the means of being symmetric, or concerning eigenvalues, and so...

sara
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  • ${\bf R}^n$ having a basis consisting of eigenvectors of $A$. – Gerry Myerson Mar 21 '21 at 10:30
  • Equivalently one can define a matrix to be diagonalizable if there is an incredible matrix $S$ such that $S^{-1}AS$ is a diagonal matrix. – Lukas Mar 21 '21 at 11:22
  • @Lukas, I think you mean invertible, not incredible. – Gerry Myerson Mar 21 '21 at 11:58
  • All the answers state equivalent conditions. The most commonly mentioned sufficient condition for a matrix is usually the matrix being symmetric. (A related discussion can be found here: https://math.stackexchange.com/questions/255622/symmetric-matrix-is-always-diagonalizable?rq=1 ) – J. Becker Mar 21 '21 at 13:34
  • @Gerry Myerson, o wow - autocorrect did a great job there. You are right, I meant to write invertible. – Lukas Mar 21 '21 at 16:58

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If there are $n$ linearly independent eigenvectors, $$V1, V2, ..., Vn$$ then the matrix $P$ whose columns are these eigenvectors will satisfy $$ AP =PD $$ where D is the diagonal matrix of eigenvalues. We may solve for A or for D depending the application on hand. $$P^{-1}AP=D $$ or $$ A=PDP^{-1} $$

Mohammad Riazi-Kermani
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