The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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

1 Comment »

  1. [...] 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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