Let \[ A = \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix}. \] Is \(A\) similar over the field \(\RR\) to a diagonal matrix? Is \(A\) similar over the field \(C\) to a diagonal matrix?
Solution
For \[ A=\begin{bmatrix}6&-3&-2\\4&-1&-2\\10&-5&-3\end{bmatrix}, \] a determinant computation gives \[ \chi_A(x)=(x-2)(x^2+1). \] Over \(\mathbb R\), the only characteristic value is \(2\). A real \(3\times3\) matrix similar over \(\mathbb R\) to a diagonal matrix must have three real characteristic values counted with multiplicity, namely its diagonal entries. Since \(x^2+1\) has no real root, \(A\) is not similar over \(\mathbb R\) to a diagonal matrix.
Over \(\mathbb C\), the characteristic values are \(2,i,-i\), which are distinct. Corresponding characteristic vectors are, for example, \[ (1,0,2),\qquad (3+i,3+i,5),\qquad (3-i,3-i,5). \] Eigenvectors belonging to distinct characteristic values are linearly independent. Hence these three vectors form a basis of \(\mathbb C^3\), so \(A\) is similar over \(\mathbb C\) to \[ \operatorname{diag}(2,i,-i). \]