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Let $V$ be a vector space, let $W_1,...W_k$ be subspaces of V such that $V=W_1\oplus...\oplus W_k$. Let $$V_j = W_1 + \cdots + W_{j-1} + W_{j+1} + \cdots + W_k.$$

Prove the dual space $V^*$ has the direct sum decomposition $V^* = V_1^0 \oplus \cdots \oplus V_k^0$ where $V_j^0$ is the annihilator of $V_j$.

This question is already answered here. The operators given are $g_i(\alpha)=f(E_i\alpha)$ where $E_i\alpha=\alpha _i \in W_i$. It seems to work fine, each $g_i$ belongs to $V^0_i$ and $$f(\alpha)=f(E_1\alpha+...+E_k\alpha)=f(E_1\alpha)+..f(E_1\alpha)=g_1(\alpha)+...+g_k(\alpha)$$

but in order to conclude $V^* = V_1^0 \oplus \cdots \oplus V_k^0$ one needs to also verify $g_i$ are projections, that is

  • $g^2_i= g_i$
  • $g_i (g_j)=0$ for $i\neq j$
  • range of $g_i = W_i$

I'm stuck veryfing this. For the first requisite $g^2_i(\alpha)=g_i[g_i(\alpha)]=g_i[f(\alpha _i)]$ but this is a functional evaluated by a scalar. Same problem happens for the second. Any help with this or any suggestion of other projections is greatly appreciated.

taue2pi
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1 Answers1

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Here it does not make sense to talk about projections and orthogonality. There is no inner product given, the decomposition of $V$ into these $W_j$ is not orthogonal.

In order to show that the sum is direct, you have to show that $$ \sum_{i=1}^k g_i=0 $$ for $g_i\in V_i^o$, $i=1\dots k$, implies $g_i=0$ for all $i$. Or, in other words, the decomposition of $f$ into these $g_i$'s is unique.

Now take $g_i\in V_i^o$ such that $$ \sum_{i=1}^kg_i=0. $$ Let us take arbitrary $w_i\in W_i$, $i=1\dots k$, and set $v=\sum_{i=1}^k w_i$. Since $v-w_j \in V_j$ this implies $g_j(v-w_j)=0$ or $g_j(v)=g_j(w_j)$. This implies $$ 0 = \sum_{j=1}^kg_j(v) = \sum_{j=1}^k g_j(w_j). $$ This holds for all $w_j\in W_j$, $j=1\dots k$, hence $g_j=0$ for all $j$ and the sum $V^* = V_1^o + \dots + V_k^o$ is direct.

daw
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  • Thank you, much more simple than what I had in mind. My confusion comes from the chapter of the book this excercise is in. The main result of the chapter is: If $E_1,...,E_k$ are linear operators on $V$ such that $E_i$ is a projection, $E_iE_j=0$ if $i\neq j$, $I=E_1+...+E_k$ and $range(E_i)=W_i$ then $V=W_1\oplus ...\oplus W_k$. I thought $g_1,...,g_k$ were these operators but, as you said, didn't make sense because they weren't operators on $V^*$, just functionals. Thanks again. – taue2pi Sep 14 '17 at 08:25