# A Diagonalizable Matrix which is Not Diagonalized by a Real Nonsingular Matrix ## 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. Add to solve later

Contents

## 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. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Linear Algebra ##### Diagonalize the Upper Triangular Matrix and Find the Power of the Matrix

Consider the $2\times 2$ complex matrix $A=\begin{bmatrix} a & b-a\\ 0& b \end{bmatrix}.$ (a) Find the eigenvalues of $A$. (b)...

Close