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Math Help - Field, Subspace, Basis, Dim

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Jimmy_W View Post
    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}).
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