# True or False. Every Diagonalizable Matrix is Invertible

## Problem 439

Is every diagonalizable matrix invertible?

Contents

## Solution.

The answer is No.

### 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:

### 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.

- Diagonalizable, but not invertible.

Example: \[\begin{bmatrix}

0 & 0\\

0& 0

\end{bmatrix}.\] - Invertible, but not diagonalizable.

Example: \[\begin{bmatrix}

1 & 1\\

0& 1

\end{bmatrix}\] - Not diagonalizable and Not invertible.

Example: \[\begin{bmatrix}

0 & 1\\

0& 0

\end{bmatrix}.\] - Diagonalizable and invertible

Example: \[\begin{bmatrix}

1 & 0\\

0& 1

\end{bmatrix}.\]

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