In general for all real (and complex ) 2 by 2 matrices, is it true that if there is a repeated eigenvalue (so all eigenvalues are the same), then we conclude that that matrice is non diagonalisable?

In general, for a n by n matrix, if ALL eigenvalues are repeated, then does this mean that matrix is non diagonalisable? Does the proof have to do with that if there was a diagonal matrix, then itd be a multiple of the n by n identity matrix, so there is no invertible matrix that satisfies the definition of similarity (diagonalisability)?

  • 2
    $\begingroup$ You might want to look up "algebraic multiplicity" and "geometric multiplicity". $\endgroup$
    – Batman
    May 15, 2017 at 19:32
  • $\begingroup$ This earlier question is closely related and got a number of answers. I suggest taking a look. $\endgroup$ May 16, 2017 at 4:50

3 Answers 3


No, there are plenty of matrices with repeated eigenvalues which are diagonalizable. The easiest example is $$ A=\begin{bmatrix}1&0\\0&1\end{bmatrix}. $$ The identity matrix has $1$ as a double eigenvalue and is (already) diagonal. If you want to write this in diagonalized form, you can write $$ A=I^{-1}AI=I^{-1}II $$ since $A$ is a diagonal matrix.

In general, $2\times 2$ matrices with repeated eigenvalue $\lambda$ are diagonalizable if and only if the eigenspace corresponding to $\lambda$ is two dimensional. In other words, if $$ A-\lambda I=\begin{bmatrix}a-\lambda&b\\c&d-\lambda\end{bmatrix} $$ has a two-dimensional null space. Using the rank-nullity theorem, we get that this happens exactly when the matrix has $0$ pivots. If $A-\lambda I$ has any nonzero entries, then it will have a pivot. Therefore, a $2\times 2$ matrix with repeated eigenvalues is diagonalizable if and only if it is $\lambda I$.

If $B$ is an $n\times n$ matrix, all of whose eigenvalues are $\lambda$, a similar result holds. A sneakier way to prove this is that if $B$ is diagonalizable, then $$ B=P^{-1}(\lambda I)P=\lambda P^{-1}IP=\lambda I, $$ where P is an invertible (basis changing) matrix.

Therefore, the only $n\times n$ matrices with all eigenvalues the same and are diagonalizable are multiples of the identity.

If only some of $B$'s eigenvalues have multiplicity, then the situation becomes more complicated and you really need to compute the dimensions of all the eigenspaces.

As the other posters comment, there are diagonal matrices which are not multiples of the identity, for example $$ \begin{bmatrix}1&0\\0&2\end{bmatrix} $$ and if all the eigenvalues of a matrix are distinct, then the matrix is automatically diagonalizable, but there are plenty of cases where a matrix is diagonalizable, but has repeated eigenvalues.


To be more precise than the previous answers: Repeated eigenvalues are necessary but not sufficient for a matrix to not be diagonalizable.

For example you can never have a $2\times 2$ matrix with two different eigenvalues which is not diagonalizable.

In other words: as soon as all eigenvalues are distinct then we can be sure to be able to diagonalize it.


A matrix with repeated eigenvalues can be diagonalized. Just think of the identity matrix. All of its eigenvalues are equal to one, yet there exists a basis (any basis) in which it is expressed as a diagonal matrix.

Furthermore, not all diagonal matrices are multiples of the identity. Example:

$$ \begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} $$

  • $\begingroup$ Not all matrices with repeated eigenvalues can be diagonalized, for example the matrix $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ can not be diagonalized, but has two eigenvalues equal to 1. $\endgroup$ Jun 13, 2023 at 13:10

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