3
$\begingroup$

This is problem #11, Section 6.6 from Hoffman & Kunze, Linear Algebra (p. 213).

Let $V$ be a vector space, let $W_1, \ldots, W_k$ be subspaces of $V$, and let $$V_j = W_1 + \cdots + W_{j-1} + W_{j+1} + \cdots + W_k.$$ Suppose that $V = W_1 \oplus \cdots \oplus W_k$. Prove that the dual space $V^*$ has the direct-sum decomposition $V^* = V_1^0 \oplus \cdots \oplus V_k^0$.

Here, $V_i^0$ indicates the annihilator of the subspace $V_i$.

My initial efforts are to say, we know, $$ \dim V_j + \dim V_j^0 = \dim V$$ and from the definition of $V_j$ we have $$ \dim V_j = \dim V - \dim W_j$$ So, $$ \dim V_j^0 = \dim W_j \qquad j = 1, \ldots, k$$

Then, since $\dim V = \dim V^*$ and $\dim V = \dim W_1 + \cdots + \dim W_k$ we have $$ \dim V^* = \dim V_1^0 + \cdots \dim V_k^0$$.

Now, if we can show $V_1^0 + \cdots + V_k^0 = V^*$, then we can conclude that in fact $V^* = V_1^0 \oplus \cdots \oplus V_k^0$, as desired.

I'm not sure how to get this last piece, though it seems that it should be true.

$\endgroup$

2 Answers 2

1
$\begingroup$

For any linear functional $f$ in $V^*$, Let $g_i(\alpha)=f(E_i\alpha)$, where $range(E_i)=W_i$, and $E_i$ is projection for direct sum.

Each $g_i$ belongs to $V_i^0$. Any $v\in V_i$, $v=E_1x_1+..+E_{i-1}x_{i-1}+E_{i+1}x_{i+1}+..E_kx_k$,
$g_i(v)=f(E_iv)=0$ since $E_iE_j=0$ for all $i\neq j$.

Also, $f(\alpha)=f(E_1\alpha+...+E_k\alpha)=f(E_1\alpha)+..f(E_1\alpha)=g_1(\alpha)+...+g_k(\alpha)$ for all $\alpha \in V$.

Therefore, We get $g_i\in V_i^0$ such that $f=g_1+..+g_k$.

$\endgroup$
2
  • $\begingroup$ How does one verify that $g_i^2=g_i$ and $g_i g_j= 0$ when $i\neq j$? I tried showing $g_i^2=g_i$ but ended up with $f(E_i f(w_i))$ which is a functional evaluated on a scalar. Same happened when I tried to show $g_i g_j= 0$ when $i\neq j$. $\endgroup$
    – taue2pi
    Sep 14, 2017 at 0:02
  • $\begingroup$ @taue2pi The $g_i$ are not projections. You need to find another way to show that the $V_i^0$ are independent, a step missing (or assumed) in rudgn55's answer. $\endgroup$
    – user0
    Jul 18, 2021 at 17:35
1
$\begingroup$

Let $B_j=\{\beta_1^j,...\beta_{n_j}^j\}$ be a basis for $W_j$, then $$B=(B_1,...,B_k)=\{\beta_1^1,...\beta_{n_1}^1,\beta_1^2,...\beta_{n_2}^2,...,\beta_1^k,...\beta_{n_k}^k\}$$ is a basis for $V$. Let $B^*=\{f_1^1,...f_{n_1}^1,f_1^2,...f_{n_2}^2,...,f_1^k,...f_{n_k}^k\}$ be the basis for $V^*$ which is dual to $B$. Let us fix $j$ and show that $B_j^*=\{f_1^j,...f_{n_j}^j\}$ is a basis for $V_j^0$, this will give us the desired result.

For $i=1,...,n_j$, by the definition of the dual basis we have $f_i^j(\beta_l^r)=0$ whenever $r\neq j$. So if $\alpha \in V_j$, $\alpha$ is a linear combination of the $\beta_l^r$, $r\neq j$, and thus $f_i^j(\alpha)=0$. This shows that $f_i^j \in V_j^0$.

Now the $f_i^j$ are linearly independant, so we must only show that they span $V_j^0$. Let $f \in V_j^0$, then $$f=\sum_{r=1}^k \sum_{l=1}^{n_r}f(\beta_l^r)f_l^r = \sum_{l=1}^{n_j}f(\beta_l^j)f_l^j $$ and we can conclude.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .