When is a matrix $A$ similar to its diagonal matrix $D$ of eigenvalues? Evidently the two matrices have the same eigenvalues, but this by itself is not enough to conclude similarity. If the eigenvalues are distinct, the matrices are similar. If $A$ is diagonalizable, then we have such similarity by definition. Can we say anything better?
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$\begingroup$ You want to know a criterion of diagonalizability. Are you working in an algebraically closed field? $\endgroup$– Sassatelli GiulioJun 3, 2022 at 0:19
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1$\begingroup$ See what you can do with $$ \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) $$ $\endgroup$– Will JagyJun 3, 2022 at 0:28
2 Answers
You need a basis for the vector space consisting of eigenvectors of the matrix. Of course, if all eigenvalues are different then their corresponding eigenvectors are independent so form a basis. But two or more eigenvalues can be the same as long as their corresponding eivenvectors are independent.
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$\begingroup$ This is what I expected. This question is labeled as a challenge problem in Strang's text, so I thought I had missed something. $\endgroup$ Jun 3, 2022 at 1:56
For Will Jagy's example, $\begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ the eigenvalue equation is $\left|\begin{array}{cc} 1- \lambda & 0 \\ 0 & 1- \lambda \end{array}\right|= (1- \lambda)^2= 0$ which has the double root $\lambda= 1$. What are its eigenvectors? If $\begin{bmatrix}x \\ y\end{bmatrix}$ is an eigenvector corresponding to eigenvalue 1 then we must have $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix} x+ y \\ y\end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix}$. We must have x+ y= x and y= y. The first equation gives y= 0 while the second equation does not give a value for x. Any vector of the form $\begin{bmatrix} x \\ 0 \end{bmatrix}$ is an eigenvector. Clearly there is no basis for $R^2$ consisting of eigenvectors so this matrix cannot be diagonalized.
On the other hand, the matrix $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ already is* diagonal even though its only eigenvalue is 1. There are two independent eigenvectors coorespondng t that eigenvalue, $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix}0 \\ 1 \end{bmatrix}$.