# Thread: Field, Subspace, Basis, Dim

1. ## Field, Subspace, Basis, Dim

Let $\mathbb{F}$ be a field and let n be a fixed positive integer. Define $U_n = \lbrace A \in M_n ( \mathbb{F}) | A_{ij} = 0 \ \forall \ i > j \rbrace$.

(1) Prove that $U_n$ is a subspace of $M_n ( \mathbb{F})$
(2) Give a basis for $U_n$
(3) What is dim $(U_n)$
(4)Let ${V_n}^{S-S}$ denote the subspace off all skew-symmetric matrices (satisfying $A^T = -A$). Show that $M_n ( \mathbb{F}) = {V_n}^{S-S} \oplus U_n$ if and only if $\mathbb{F}$ is not characteristic 2.

2. Originally Posted by Jimmy_W
Let $\mathbb{F}$ be a field and let n be a fixed positive integer. Define $U_n = \lbrace A \in M_n ( \mathbb{F}) | A_{ij} = 0 \ \forall \ i > j \rbrace$.

(1) Prove that $U_n$ is a subspace of $M_n ( \mathbb{F})$
show that for any two upper triangualr matrices $A,B$ and scalar $c,$ the matrix $cA+B$ is again upper triangular. this shouldn't take more than 2 minutes!

(2) Give a basis for $U_n$
let $\{e_{ij}: \ 1 \leq i,j \leq n \}$ be the standard basis for $M_n(\mathbb{F}).$ then $\{e_{ij}: \ 1 \leq i \leq j \leq n \}$ is a basis for $U_n.$

(3) What is dim $(U_n)$
by (2) it's clear that $\dim U_n= \frac{n(n+1)}{2}.$

(4)Let ${V_n}^{S-S}$ denote the subspace off all skew-symmetric matrices (satisfying $A^T = -A$). Show that $M_n ( \mathbb{F}) = {V_n}^{S-S} \oplus U_n$ if and only if $\mathbb{F}$ is not characteristic 2.
if $\text{char}(\mathbb{F}) = 2,$ then every diagonal matrix would be skew symmetric and upper triangular. thus $V_n \cap U_n \neq (0)$ and hence the sum wouldn't even be direct! however, since in this case

$\dim V_n = \dim U_n=\frac{n(n+1)}{2}$ and $\dim (V_n \cap U_n) = n,$ we'll have $\dim (V_n + U_n)=n(n+1)-n=n^2=\dim M_n(\mathbb{F}),$ and thus $V_n + U_n = M_n(\mathbb{F}).$ (note that the sum is not direct here!)

if $\text{char}(\mathbb{F}) \neq 2,$ then the diagonal of every element of $V_n$ would be all 0. thus in this case $V_n \cap U_n = (0).$ so the sum is direct. we also have $\dim V_n=\frac{n(n+1)}{2} - n = \frac{n(n-1)}{2}.$ therefore

$\dim (V_n \oplus U_n)=\frac{n(n-1)}{2} + \frac{n(n+1)}{2}=n^2=\dim M_n(\mathbb{F}),$ and so $V_n \oplus U_n = M_n(\mathbb{F}).$