# The Similarity of a Matrix with its Diagonal Matrix of Eigenvalues

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?

• You want to know a criterion of diagonalizability. Are you working in an algebraically closed field? Jun 3, 2022 at 0:19
• See what you can do with $$\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$$ Jun 3, 2022 at 0:28

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 $$\begin{bmatrix}0 \\ 1 \end{bmatrix}$$.