The Algebra of Upper-Triangular Matrices

Here’s another little result that’s good over any field, algebraically-closed or not. We know that the linear maps from a vector space $V$ (of finite dimension $d$ ) to itself form an algebra over $\mathbb{F}$ . We can pick a basis and associate a matrix to each of these linear transformations. It turns out that the upper-triangular matrices form a subalgebra.

The easy part is to show that matrices of the form

$\displaystyle\begin{pmatrix}\lambda_1&&*\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}$

form a linear subspace of $\mathrm{End}(V)$ . Clearly if we add two of these matrices together, we still get zero everywhere below the diagonal, and the same goes for multiplying the matrix by a scalar.

The harder part is to show that the product of two such matrices is again upper-triangular. So let’s take two of them with entries $s_i^j$ and $t_i^j$ . To make these upper-triangular, we’ll require that $s_i^j=0$ and $t_i^j=0$ for $i>j$ . What we need to check is that the matrix entry of the product $s_i^jt_j^k=0$ for $i>k$ . But this matrix entry is a sum of terms as $j$ runs from ${1}$ to $d$ , and each term is a product of one matrix entry from each matrix. The first matrix entry can only be nonzero if $i\leq j$ , while the second can only be nonzero if $j\leq k$ . Thus their product can only be nonzero if $i\leq k$ . And this means that all the nonzero entries of the product are on or above the diagonal.

February 5, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

1 Comment »

[…] diagonal matrices are themselves diagonal. Thus, diagonal matrices form a further subalgebra inside the algebra of upper-triangular matrices. This algebra is just a direct sum of copies of , with multiplication […]

Pingback by Diagonal Matrices « The Unapologetic Mathematician | February 9, 2009 | Reply

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