# True or False. Every Diagonalizable Matrix is Invertible ## Problem 439

Is every diagonalizable matrix invertible? Add to solve later

## Solution.

### Counterexample

We give a counterexample. Consider the $2\times 2$ zero matrix.
The zero matrix is a diagonal matrix, and thus it is diagonalizable.
However, the zero matrix is not invertible as its determinant is zero.

### More Theoretical Explanation

Let us give a more theoretical explanation.
If an $n\times n$ matrix $A$ is diagonalizable, then there exists an invertible matrix $P$ such that
$P^{-1}AP=\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix},$ where $\lambda_1, \dots, \lambda_n$ are eigenvalues of $A$.
Then we consider the determinants of the matrices of both sides.
The determinant of the left hand side is
\begin{align*}
\det(P^{-1}AP)=\det(P)^{-1}\det(A)\det(P)=\det(A).
\end{align*}
On the other hand, the determinant of the right hand side is the product
$\lambda_1\lambda_2\cdots \lambda_n$ since the right matrix is diagonal.
Hence we obtain
$\det(A)=\lambda_1\lambda_2\cdots \lambda_n.$ (Note that it is always true that the determinant of a matrix is the product of its eigenvalues regardless diagonalizability.
See the post “Determinant/trace and eigenvalues of a matrix“.)

Hence if one of the eigenvalues of $A$ is zero, then the determinant of $A$ is zero, and hence $A$ is not invertible.

The true statement is:

a diagonal matrix is invertible if and only if its eigenvalues are nonzero.

### Is Every Invertible Matrix Diagonalizable?

Note that it is not true that every invertible matrix is diagonalizable.

For example, consider the matrix
$A=\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix}.$ The determinant of $A$ is $1$, hence $A$ is invertible.
The characteristic polynomial of $A$ is
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
1-t & 1\\
0& 1-t
\end{vmatrix}=(1-t)^2.
\end{align*}
Thus, the eigenvalue of $A$ is $1$ with algebraic multiplicity $2$.
We have
$A-I=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ and thus eigenvectors corresponding to the eigenvalue $1$ are
$a\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ for any nonzero scalar $a$.
Thus, the geometric multiplicity of the eigenvalue $1$ is $1$.
Since the geometric multiplicity is strictly less than the algebraic multiplicity, the matrix $A$ is defective and not diagonalizable.

### Is There a Matrix that is Not Diagonalizable and Not Invertible?

Finally, note that there is a matrix which is not diagonalizable and not invertible.
For example, the matrix $\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ is such a matrix.

## Summary

There are all possibilities.

1. Diagonalizable, but not invertible.
Example: $\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}.$
2. Invertible, but not diagonalizable.
Example: $\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix}$
3. Not diagonalizable and Not invertible.
Example: $\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}.$
4. Diagonalizable and invertible
Example: $\begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}.$ Add to solve later

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