# Does “identical eigenvalues” mean that there is no diagonal matrix?

If I have a 2 x 2 matrix and eigenvalues for that matrix are 3,3, does this mean that there exits no diagonal matrix?

If I have 2 distinct eigenvalues, then eigenvectors corresponding to two distinct

eigenvalues are independent and form a basis. Then I can find out a diagonal matrix having

eigenvalues as entries. But, I am not sure whether having identical eigenvalues means that

there is no diagonal matrix.

• Boring counterexample, but consider the case that the matrix $A$ is already diagonal. If it is not already diagonal, then there does not exists a diagonal matrix. – Nigel Overmars Oct 31 '14 at 22:25
• No necessarily; the eigenvalues of $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right]$ are $1$ and $1$, but it is most certainly a diagonal matrix. – Hayden Oct 31 '14 at 22:25
• If original matrix is not diagonal form and eigenvalues are the same, then there is no diagonal matrix? – dkim Oct 31 '14 at 22:30
• @dkim Yes, that is true. – Nigel Overmars Oct 31 '14 at 22:31

The eigenvalues for the diagonal matrix $\left(\begin{smallmatrix} x & 0 \\ 0 & x\end{smallmatrix}\right)$ are $x$ and $x$... so having repeated eigenvalues certainly doesn't mean non-diagonalizability.

What are the eigenvalues of the identity matrix?

When you diagonalize the rest (the distinct eigenvalues), the subspace with the degenerate eigenvalue can behave in two ways:

*) Diagonalizable. Geometric multiplicity (the number of eigenvectors) is the same as the multiplicity of the eigenvalues. In this case, when you diagonalize the rest, the remaining block is already diagonal.

*) Not enough eigenvectors: you can put the degenerate block into a Jordan normal form, but nondiagonal terms will be nonzero.

For some matrices, you can be sure it's diagonalizable (for instance, symmetric matrices always are).

Well, you can consider the identity matrix, $I$, which has unique eigenvalue $1$ - and it is already in diagonal form.

What we want here is that the $2\times2$ matrix is diagonalisable iff:

either there are two eigenvalues with corresponding eigenbases of dimension $1$ or there is a unique eigenvalue with an eigenbasis of dimension $2$.

More generally, a $n\times n$ matrix is diagonalisable if the dimension of the space, which is the direct sum of all eigenspaces, is $n$.

Now, the eigenbasis corresponding to eigenvalue $1$ for $I$ is of dimension $2$. However, consider $M=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$. It only has one eigenvalue, namely $\lambda = 1$, but its eigenbasis for $M$, namely the span of $e_{1}$, is of dimension $1$. And so it is not diagonalisable.

I mention for posterity's sake that if the eigenvalues are the same, then the matrix is diagonalisable iff it is already in diagonal form, since $P^{-1}DP = P^{-1}(nI)P = nP^{-1}P = nI$.