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Let $\textsf{V}=\left\{ \begin{pmatrix} x_1\\x_2\\x_3\\x_4 \end{pmatrix} \in \mathbb{R}^4 :\, x_1=x_3+x_4 \textrm{ and } x_2=x_3-x_4 \right\}$. Find a basis for $\textsf V$ and for $\textsf{V}^\perp$.

My attempt : A basis for $\textsf V$ will be $(1, 0,-1 ,-1)$ and $(0,1,-1,1)$. And we know that $\dim (\textsf{V}^{\perp}) = \dim (\mathbb{R}^4) - \dim (\textsf V)= 4-2 =2$.

But I don't know how we can find the basis of $\textsf{V}^{\perp}$.

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    $\begingroup$ Have you seen the Gram-Schmidt process? $\endgroup$ – Luiz Cordeiro Sep 23 '19 at 16:20
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    $\begingroup$ Why don't you apply Gram-Schmidt? You have seen it for a reason, you know... $\endgroup$ – Luiz Cordeiro Sep 23 '19 at 16:22
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    $\begingroup$ How is $(1,0,-1,-1)$ in $V$? $\endgroup$ – Pink Panther Sep 23 '19 at 16:23
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    $\begingroup$ As @PinkPanther's mentioned, first fix your basis for $V$ (you want $x_1=-2$ in the first vector). Apply Gram-Schmidt to find an orthogonal basis of $V$. Now take a vector $e_1$ of the canonical basis of $\mathbb{R}^4$. Use Gram-Schmidt again to find the projection $p_V(e)$ of $e$ onto $V$. The vector $f_1=e_1-p_V(e_1)$ will be in the orthogonal $V^\perp$, and with some luck it will be nonzero. (cont.) $\endgroup$ – Luiz Cordeiro Sep 23 '19 at 16:29
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    $\begingroup$ Repeat this procedure with another vector $e_2$ of the canonical basis. With some luck, $f_2=e_2-p_V(e_2)$ will be LI with $e_1$. If not, try another vector of the canonical basis. As soon as you find two LI ones in $V^\perp$, you're done. $\endgroup$ – Luiz Cordeiro Sep 23 '19 at 16:29
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I'll do a similar one for you: Take the space $W=\left\{(x_1,x_2,x_3,x_4):x_1=3x_3, x_2=0, x_1=3x_4\right\}$.

Let us find a basis of $W^\perp$.

First, find a basis for $W$: The sole vector $w=(3,0,1,1)$ works. Note that $\langle w,w\rangle=3^2+0^2+1^2+1^2=11$

Now consider the first vector $e_1=(1,0,0,0)$ of the canonical basis of $\mathbb{R}^4$. The projection onto $W$ is $$p_W(e_1)=\frac{\langle e_1,w\rangle}{\langle w,w\rangle}w=\frac{3}{11}w$$ so the following vector is in $W^\perp$: \begin{align*} f_1&=e_1-p_W(e_1)\\ &=(1,0,0,0)-\frac{3}{11}\left(3,0,1,1\right)\\ &=(1,0,0,0)-\left(\frac{9}{11},0,\frac{3}{11},\frac{3}{11}\right)\\ &=\left(\frac{2}{11},0,-\frac{3}{11},-\frac{3}{11}\right) \end{align*}

Let's do the same for $e_2=(0,1,0,0)$: $$p_W(e_2)=\frac{0}{11}w=0$$ so I was lucky: $f_2=e_2-p_W(e_2)=e_2$ is in $W^\perp$.

Again, with $e_3=(0,0,1,0)$: $p_W(e_3)=\frac{1}{11}w$, so

$$f_3=e_3-\frac{1}{11}w=\left(-\frac{3}{11},0,\frac{10}{11},-\frac{1}{11},\right)$$

I'll leave it to you to verify that $\left\{f_1,f_2,f_3\right\}$ is LI, so it is a basis of $W^\perp$ (which has dimension $4-\dim W=4-1=3$).

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Note that $V= \{ (x_3+x_4, x_3-x_4,x_3,x_4)^T \}$ and since $(x_3+x_4, x_3-x_4,x_3,x_4)^T = x_3(1,1,1,0)^T + x_4(1,-1,0,1)^T $, we see that $v_1=(1,1,1,0)^T, v_2=(1,-1,0,1)^T$ spans $V$ and it is easy to check that they are linearly independent hence $V$ is a basis for $V$.

To compute a basis for $V^\bot$, we can start with $v_1,v_2,e_1,e_2,e_3,e_4$ and apply Gram Schmidt (ignoring the zero vectors that will result). This will produce an orthonormal basis for $\mathbb{R}^4$, and the first two will span $V$ and so the last two (non zero) will span $V^\bot$. (Actually, since $v_1 \bot v_2$, the process will just normalise $v_1,v_2$).

Alternatively, we can 'eyeball' $v_1,v_2$ and try to find two linearly independent vectors that are orthogonal to $v_1,v_2$. One pair is $(1,0,-1,-1)^T, (0,1,-1,1)^T$.

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  • $\begingroup$ Or, just read a basis for $V^\perp$ from the defining equations of $V$. $\endgroup$ – amd Sep 23 '19 at 18:23
  • $\begingroup$ I guess that would be 'eyeballing'? $\endgroup$ – copper.hat Sep 23 '19 at 18:25
  • $\begingroup$ Aye, but eyeballing the equations rather than the basis of $V$. $\endgroup$ – amd Sep 23 '19 at 18:26
  • $\begingroup$ I guess we are dualing now... $\endgroup$ – copper.hat Sep 23 '19 at 18:27
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Note that $x_1 = x_3 + x_4$ if and only if $x_1 - x_3 - x_4 = 0$, which is equivalent to $\langle x, (1, 0, -1, -1) \rangle = 0$. Similarly, $x_2 = x_3 - x_4$ if and only if $\langle x, (0, 1, -1, 1) \rangle = 0$. Therefore, $V$ is the set of vectors which are orthogonal to both $(1, 0, -1, -1)$ and $(0, 1, -1, 1)$.

Thus, if we set $W := \operatorname{span} \{ (1, 0, -1, -1), (0, 1, -1, 1) \}$, this implies that $V = W^\perp$, and therefore $V^\perp = (W^\perp)^\perp = W$. From here, it should be easy to find a basis for $W$.

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