Let \(x,y \in \mathbb{C}\) and consider the matrix
\[M = \left[\begin{array}{ccc} 1 & 0 & x \\ 0 & 1 & 0 \\ y & 0 & 1 \end{array}\right]\]
Show that \([0, 1, 0]^t\) is an eigenvector of \(M\).
Compute the rank of \(M\) as a function of \(x\) and \(y\).
Find all values of \(x\) and \(y\) for which \(M\) is diagonalizable.
Solution
Let \(e_2=(0,1,0)^t\). Then \[ Me_2=e_2, \] so \(e_2\) is an eigenvector with eigenvalue \(1\).
The subspace \(\langle e_1,e_3\rangle\) is \(M\)-invariant, and in the ordered basis \((e_1,e_3)\) the restriction is \[ B=\begin{pmatrix}1&x\\ y&1\end{pmatrix}. \] Thus \(M\sim [1]\oplus B\).
1. Rank.Since\[ \det B=1-xy, \]we have\[ \rank M= \begin{cases} 3,&xy\ne1,\\ 2,&xy=1. \end{cases} \]Indeed, when \(xy=1\), the nonzero matrix \(B\) has determinant \(0\) and hence rank \(1\), while the \(e_2\) summand contributes one more dimension.
2. Diagonalizability.The characteristic polynomial of \(B\) is\[ (t-1)^2-xy. \]If \(xy\ne0\), its two roots \(1\pm\sqrt{xy}\) are distinct, so \(B\) and hence \(M\) are diagonalizable.
If \(xy=0\), then \[ B=I+N,\qquad N=\begin{pmatrix}0&x\\y&0\end{pmatrix},\qquad N^2=0. \] If \(x=y=0\), then \(N=0\) and \(M=I_3\), so \(M\) is diagonalizable. If exactly one of \(x,y\) is nonzero, then \(N\ne0\) is nilpotent, so \(B\) has a nontrivial Jordan block for the sole eigenvalue \(1\) and is not diagonalizable.
Therefore \[ \boxed{M\text{ is diagonalizable }\iff xy\ne0\text{ or }(x,y)=(0,0).} \]