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Let $V\subset \mathbb{R}^4$ be the subspace spanned by $e_1+e_2+e_3+e_4$ and $e_1+2e_2+3e_3+4e_4$. Find a basis of the subspace $V^{\circ}$ in the dual space $(\mathbb{R})^*$.

My attempt: Let $a = e_1+e_2+e_3+e_4$ and $b = e_1+2e_2+3e_3+4e_4$, then $a$ and $b$ are lin. independent by looking at the row reducing matrix $\bigl( \begin{smallmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 3 & 4 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 1 & 1 & 1 & 1\\ 0 & 1 & 2 & 3 \end{smallmatrix} \bigr)$. Thus we can extend the set $\{a,b\}$ to form the basis of $\mathbb{R}^4$. Example: $\{a,b,e_3,e_4\}$ forms a basis for $\mathbb{R}^4$ as $e_3,e_4$ are lin. independent of $a,b$. Let $\{f_1,f_2,f_3,f_4\}$ be the dual basis and we can say that $\{f_3,f_4\}$ forms the basis of $V^{\circ}$ because $f_3(a) = f_3(b) = f_4(a) = f_4(b) = 0$? Appreciate a hint.

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    $\begingroup$ Expand $f$ using the dual to the standard basis and use linearity to get a system of linear equations in the coordinates of $f$. $\endgroup$ – amd Mar 18 at 23:39
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Use the dual basis or (equivalently) the dot product to identify $(\Bbb{R}^4)^*$ with $\Bbb{R}^4$. Then the annihilator of a subspace is its usual orthogonal complement. Thus we can find a basis for the orthogonal complement by Gram-Schmidt. Anyway, that gives us an algorithm, but it's a bit tedious, so let's take a different route.

If you row reduce further you get $$ \newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}} \bmat 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ \emat $$ This means that to get an orthogonal vector, we can choose the last two coordinates freely and pick the first two such that we get something orthogonal. I.e., if our new vector is $\bmat a & b & c & d\emat$, then we need $a = c+2d$ and $b=-2c -3d$. Similarly, since the fourth vector doesn't need to be orthogonal to the third, merely linearly independent from it and orthogonal to the last two, we can also choose the last two coordinates freely (as long as they're linearly independent from the last two coordinates of the third).

What this boils down to is that we can choose the last two of the third vector to be $(1,0)$ and the last two of the fourth to be $(0,1)$, giving the final matrix $$ \bmat 1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 3 \\ 1 & -2 & 1 & 0 \\ 2 & -3 & 0 & 1 \\ \emat $$

Thus a basis for the annihilator of the subspace is $\{f_1-2f_2+f_3,2f_1-3f_2+f_4\}$ given in terms of the dual basis.

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  • $\begingroup$ Quick question. I get that the matrix is a basis of $\mathbb{R}^4$ but how do you conclude that the basis of $V^{\circ}$ is $\{f_1-2f_2+f_3, 2f_1-3f_2+f_4\}$? i.e. how can we get it in terms of the dual basis. The answer I got below is in terms of the coordinates of a point in $V$. Thanks. $\endgroup$ – manifolded Mar 19 at 4:31
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    $\begingroup$ @manifolded Under the standard identification of dual space with the original space via the dot product, the annihilator is identified with the orthogonal complement, and the dual basis is identified with the standard basis. Thus finding a basis for the orthogonal complement in terms of the standard basis and then changing the $e_i$s to $f_i$s gives the basis for the annihilator in terms of the dual basis. $\endgroup$ – jgon Mar 19 at 4:39
  • $\begingroup$ @anonymous_downvoter I don't care about the downvote itself, but I am interested if you think I've made an error. $\endgroup$ – jgon Mar 23 at 23:41
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Did it the long way here. To find the basis of $V^{\circ}$, we want linear functionals $f\in V^{*}$ s.t. $f(v)=0$, where $v = a_1v_1+a_2v_2$ ; $a_1,a_2\in \mathbb{R}$.

$$v_1 = e_1+e_2+e_3+e_4 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{pmatrix} \& \text{ }v_2 = e_1+2e_2+3e_3+4e_4 = \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \\ \end{pmatrix} $$

$$ f(v) = 0 \iff f(a_1v_1+a_2v_2)= a_1f(v_1)+a_2f(v_2) = 0 \iff f(v_1)=f(v_2)=0$$

For any $f\in V^{*}, \exists w\in R^{4}$, s.t. $f(v)= w.v$

$$f(v_1) = w.v_1 = 0\iff w_1+w_2+w_3+w_4 = 0$$ Similarly,

$$f(v_2) = w.v_2 = 0\iff w_1+2w_2+3w_3+4w_4 = 0 $$

$$ w = \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \\ \end{pmatrix} = \begin{pmatrix} t+2p \\ -2t-3p \\ t \\ p \\ \end{pmatrix} = t\begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \\ \end{pmatrix} + p\begin{pmatrix} 2 \\ -3 \\ 0 \\ 1 \\ \end{pmatrix}$$

So $V^{\circ} = \{tf_1+pf_2; t,p\in \mathbb{R}\}$, where $f_1(x,y,z,w) = x-2y+z$ and $f_2(x,y,z,w) = 2x-3y+w$. Thus the basis of $V^{\circ}$ is $\{f_1,f_2\}$

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