Exercise · ZW-DIAGONALIZATION-010 · ★★★ Advanced
Is the \(90^\circ\) rotation, \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), diagonalizable over \(\mathbb{R}\)?
Choose one option.
Show answer
- B. No: \(\lambda^2+1=0\) has no real roots
Explanation
The characteristic polynomial is \(\lambda^2+1\), with no real roots, so there are no real eigenvectors. It is diagonalizable over \(\mathbb{C}\).
Review the concept: Diagonalization
A square matrix \(A\) is diagonalizable if it is similar to a diagonal matrix \(D\), that is, if it can be written as an invertible matrix \(P\) times \(D\) times the inverse of \(P\).
\[A=PDP^{-1}\]
More practice problems: diagonalization
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