Return to Answer

Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that for generic $\lambda_i$ that element generates your algebra.

First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by $$ A:=\left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$ and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).

Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$. The first operation is to consider $N:=A^2-aA$, which is now nilpotent with non-zero terms on the subdiagonal. Then you take powers of $N$. Then you take powers of $N^*$, etc.

Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that for generic $\lambda_i$ that element generates your algebra.

First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by $$ A:=\left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$ and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).

Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$. The first operation is to consider $N:=A^2-aA$, which is nilpotent with non-zero terms on the subdiagonal. Then you take an appropriate linear combination of powers of $N$ to generate the element $$ \left(\begin{matrix} 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ \end{matrix}\,\right). $$ Then you take the transpose of that element, and you're essentially done.

Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that for generic $\lambda_i$ that element generates your algebra.

First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by $$ \left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$ and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).

Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$.

Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that for generic $\lambda_i$ that element generates your algebra.

First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by $$ A:=\left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$ and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).

Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$. The first operation is to consider $N:=A^2-aA$, which is now nilpotent with non-zero terms on the subdiagonal. Then you take powers of $N$. Then you take powers of $N^*$, etc.

Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that this element generates your algebra.

First of all, $(X-\lambda_i)^{n_i}$ has zero for its $i$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand. The remaining term is upper triangular with non-zero entries on the diagonal.

Then it's a matter of playing around to see that this generates everything.

Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that for generic $\lambda_i$ that element generates your algebra.

First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by $$ \left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$ and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).

Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$.