## 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 (of finite dimension ) to itself form an algebra over . 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

form a linear subspace of . 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 and . To make these upper-triangular, we’ll require that and for . What we need to check is that the matrix entry of the product for . But this matrix entry is a sum of terms as runs from to , and each term is a product of one matrix entry from each matrix. The first matrix entry can only be nonzero if , while the second can only be nonzero if . Thus their product can only be nonzero if . And this means that all the nonzero entries of the product are on or above the diagonal.

[…] 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 […]

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