Answer to Question #150979 in Linear Algebra for Sourav Mondal

Question

Let V be the vector space of all n x n
matrices over R. What is the dimension of
V over R ? Further, let
Wn={Anxn belongs to V I Anxn is upper triangular}.
Check whether or not W is a subspace of V.

Answer 1

Let $V$ be the vector space of all $n \times n$ matrices over $\mathbb R$.  For $i,j\in\{1,2,...,n\}$ denote by $E_{ij}$ the matrix with a unigue 1 at the intersection of the row $i$ and the column $j$, and with the rest elements equal to 0: $e_{ij}=1$ and $e_{st}=0$ for all $s\ne i$ and $t\ne j$. Then $(E_{ij}\ |\ i,j\in\{1,..., n\})$ is a basis of the vector space $V$. Indeed, if $\sum\limits_{i=1}^n\sum\limits_{j=1}^na_{ij}E_{ij}=O$ where the matrix $O$ contains of only zeros, then $a_{ij}=0$ for all $i,j\in\{1,2,...,n\}$. Consequently, the matrices $E_{ij}$ are linear independent. Also each matrix $M$ with entries $m_{ij}$ can be represented as $M=\sum\limits_{i=1}^n\sum\limits_{j=1}^nm_{ij}E_{ij}$, and therefore, $(E_{ij}\ |\ i,j\in\{1,..., n\})$ indeed a basis of $V$. It follows that $\dim(V)=n^2.$

Further, let $W_n=\{A_{n\times n} \in V \ |\ A_{n\times n}\text{ is upper triangular}\}.$ Let  ${\displaystyle A_{n\times n}, B_{n\times n}}$ be elements of $W_n$  and ${\displaystyle \alpha ,\beta }$  be elements of $\mathbb R$. Since the elements under the main diagonal in matrices  ${\displaystyle A_{n\times n}, B_{n\times n}}$ are zeros, it follows that the elements under the main diagonal in the matrix ${\displaystyle\alpha A_{n\times n}+\beta B_{n\times n}}$ is zeros as well. Therefore,  ${\displaystyle\alpha A_{n\times n}+\beta B_{n\times n}}$  is in $W_n$, and we conclude that $W_n$ is a subspace of $V$.