Find all complex triples $(x, y, z) $such that the following matrix is diagonalizable
$A = \begin{bmatrix}1&x&y \\ 0&2 & z \\0&0&1\end{bmatrix}$
my attempts : matrix A is an upper triangular matrix. so the the eigenvalues of A are diagonal entries $1,2,1$
This implies that $A$ is diagonalizable.
Case 1: if $x=y= z= 0$
case 2 : if $x\neq y \neq z \neq 0$
Now Im confused that How can i find all complex triples $(x, y, z) $such that the following matrix is diagonalizable
On
Guide:
The eigenvalues are $1$ and $2$.
We want the algebraic multiplicity to be equal to the geometric multiplicity for each eigenvalue. Hence for eigenvalue $1$, we have to make sure the geometric multiplicity is $2$.
Make sure $A-1\cdot I=A-I$ has nullity $2$.
On
As you say, the eigenvalues are 1, 2, and 1. For $A$ to be diagonalizable, it must be the case that the $\lambda = 2$ eigenspace has dimension 1 and the $\lambda = 1$ eigenspace has dimension 2. Let’s find the eigenvectors with eigenvalue 1. $Av =\lambda v$ becomes $$ \begin{bmatrix} v_1 + v_2 x + v_3 y \\ 2 v_2 + v_3 z \\ v_3 \end{bmatrix} = \lambda \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}. $$ If $\lambda = 1$, this implies no restriction on $v_1$ (accounting for one dimension of the $\lambda= 1$ eigenspace) and the relation $$ \begin{bmatrix} x & y \\ 1 & z \end{bmatrix} \begin{bmatrix} v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ among $v_2, v_3$. The latter matrix must have a one-dimensional kernel to account for the other dimension of the $\lambda = 1$ eigenspace. By the rank-nullity theorem, this implies a one-dimensional image (or column space). That linear dependence among the columns implies $x = a y$ and $1 = az$ for some $a$. Solving, we get $a = 1/z$, hence $x = y / z$, or $xz = y$.
On
This is equivalent to the fact that $$ A-I_3 = \begin{bmatrix} 0 &x&y \\ 0&1 & z \\ 0&0&0 \end{bmatrix} $$ is such that $dim(ker(A-I_3))=2$ which is, in turn, equivalent to $$ dim(Im(A-I_3))=1 $$ and the fact that the two columns $$ \begin{bmatrix} x&y \\ 1 & z \\ 0&0 \end{bmatrix} $$ are proportional. You finally find $$ xz=y $$
Diagonalizable means the minimal polynomial is squarefree. The full characteristic polynomial is $(\lambda - 1)^2 (\lambda - 2).$
Diagonalizable if and only if $$ (A - I)(A - 2I) = 0. $$ This gives a restriction on the triple $x,y,z.$ And, when we do have $y = xz,$ we have $$ \left( \begin{array}{ccc} 1&-x&-xz \\ 0&1&z \\ 0&0&1 \\ \end{array} \right) \left( \begin{array}{ccc} 1&x&xz \\ 0&2&z \\ 0&0&1 \\ \end{array} \right) \left( \begin{array}{ccc} 1&x&0 \\ 0&1&-z \\ 0&0&1 \\ \end{array} \right) = \left( \begin{array}{ccc} 1&0&0 \\ 0&2&0 \\ 0&0&1 \\ \end{array} \right) $$
On the other hand, if $y = xz + t$ with $t \neq 0$ we have $$ \frac{1}{t} \left( \begin{array}{ccc} 0&t&zt \\ 1&-x&-xz \\ 0&0&t \\ \end{array} \right) \left( \begin{array}{ccc} 1&x&xz +t \\ 0&2&z \\ 0&0&1 \\ \end{array} \right) \left( \begin{array}{ccc} x&t&0 \\ 1&0&-z \\ 0&0&1 \\ \end{array} \right) = \left( \begin{array}{ccc} 2&0&0 \\ 0&1&1 \\ 0&0&1 \\ \end{array} \right) $$