True or False. Every Diagonalizable Matrix is Invertible
Problem 439
Is every diagonalizable matrix invertible?
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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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