# Thread: Field, Subspace, Basis, Dim

1. ## Field, Subspace, Basis, Dim

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

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

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

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

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

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

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

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

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

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