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Thread: Field, Subspace, Basis, Dim

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Jimmy_W View Post
    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 $\displaystyle V_n \oplus U_n = M_n(\mathbb{F}).$
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