A Diagonalizable Matrix which is Not Diagonalized by a Real Nonsingular Matrix

Diagonalization Problems and Solutions in Linear Algebra

Problem 584

Prove that the matrix
\[A=\begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}\] is diagonalizable.
Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.
That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix.

 
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Proof.

We first find the eigenvalues of $A$ by computing its characteristic polynomial $p(t)$.
We have
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
-t & 1\\
-1& -t
\end{vmatrix}=t^2+1.
\end{align*}
Solving $p(t)=t^2+1=0$, we obtain two distinct eigenvalues $\pm i$ of $A$.
Hence the matrix $A$ is diagonalizable.


To prove the second statement, assume, on the contrary, that $A$ is diagonalizable by a real nonsingular matrix $S$.
Then we have
\[S^{-1}AS=\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}\] by diagonalization.
As the matrices $A, S$ are real, the left-hand side is a real matrix.
Taking the complex conjugate of both sides, we obtain
\[\begin{bmatrix}
-i & 0\\
0& i
\end{bmatrix}=\overline{\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}}=\overline{S^{-1}AS}=S^{-1}AS=\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}.\] This equality is clearly impossible.
Hence the matrix $A$ cannot be diagonalized by a real nonsingular matrix.


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