# How to find the basis of $\textsf V$ and for $\textsf{V}^\perp$?

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 $$\textsf{V}^{\perp}$$.

• Have you seen the Gram-Schmidt process? – Luiz Cordeiro Sep 23 '19 at 16:20
• Why don't you apply Gram-Schmidt? You have seen it for a reason, you know... – Luiz Cordeiro Sep 23 '19 at 16:22
• How is $(1,0,-1,-1)$ in $V$? – Pink Panther Sep 23 '19 at 16:23
• 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.) – Luiz Cordeiro Sep 23 '19 at 16:29
• 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. – Luiz Cordeiro Sep 23 '19 at 16:29

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 $$4-\dim W=4-1=3$$).

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 $$(1,0,-1,-1)^T, (0,1,-1,1)^T$$.

• Or, just read a basis for $V^\perp$ from the defining equations of $V$. – amd Sep 23 '19 at 18:23
• I guess that would be 'eyeballing'? – copper.hat Sep 23 '19 at 18:25
• Aye, but eyeballing the equations rather than the basis of $V$. – amd Sep 23 '19 at 18:26
• I guess we are dualing now... – copper.hat Sep 23 '19 at 18:27

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 $$W$$.