# Orthogonal complement and norm

Can someone point out what I am fundamentally doing wrong in this question?

Consider the vector space $$\mathbb{R}^3$$ with the standard inner product (dot product) and let $$H=span\left\{(2,1,0),(0,1,2)\right\}$$ be a subspace of $$\mathbb{R}^3$$. If $$(4,12,8)=u+v$$, with $$u$$ in H and $$v\epsilon$$H$$^\perp$$, then $$||v||$$ equals:

$$(A)$$ $$\sqrt6$$

$$(B)$$ $$1$$

$$(C)$$ $$3$$

$$(D)$$ $$\sqrt{24}$$

$$(E)$$ $$\sqrt3$$

So my reasoning was that I needed what $$||v||$$ was. $$v$$ is the orthogonal complement of $$H$$. Since the vectors in $$H$$ are written as rows, we can find the Null space of $$H$$ in order to determine the orthogonal complement and hence $$v$$.

Thus:

$$\left[ \begin{array}{ccc|c} 2 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ \end{array} \right]$$

$$...$$

$$\left[ \begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ \end{array} \right]$$

Therefore, $$x_1=t,x_2=-2t,x_3=t$$ and then:

$$v=span\left\{(1,-2,1)\right\}$$

Thus, $$||v||=\sqrt{1^2+(-2)^2+1^2}$$ $$||v||=\sqrt6$$

However, the answer says it should be $$(D)$$, $$\sqrt{24}$$. Why is that?

• Your $v=span\left\{(1,-2,1)\right\}$ is a subspace, not a vector! You are searching a vector in this subspace. – Emilio Novati Dec 21 '18 at 9:44

## 4 Answers

You don't get $$\|v\|$$ by just finding some vector orthogonal to the two given vectors. The formula for $$v$$ is $$v=\frac {|\langle (4,12,8), (1,-2,1) \rangle |} {\ {\|(1,-2,1\|}}$$ which works out to $$\sqrt {24}$$.

When you write the given vector $$(4,12,8)$$ in terms of the span of $$(2,1,0), (0,1,2)$$ and its orthogonal complement you will get $$(4,12,8)=a(2,1,0)+b(0,1,2)+c(1,-2,1)$$ and $$v =c((1,-2,1)$$. You are missing the coefficient $$c$$ in your calculation. To find the value of $$c$$ all you have to do is take the inner product of both sides with $$(1,2,1)$$. That gives you the formula for $$v$$.

• Where is this formula from? – Future Math person Dec 21 '18 at 9:44
• @FutureMathperson I have added some more steps to my answer. – Kavi Rama Murthy Dec 21 '18 at 9:49
• Can't I just solve the system of equations instead to find the value of $a,b,c$ simultaneously? I get the same answer doing it that way. – Future Math person Dec 21 '18 at 9:57

You found already a basis $$w = (1,-2,1)$$ for $$H^{\perp}$$.

Now,

• $$(4,12,8) = u + \underbrace{sw}_{=v} \Rightarrow (4,12,8)\cdot w \stackrel{u \perp w}{=} s||w||^2 \Rightarrow -12 = 6s \Rightarrow \boxed{s=-2}$$
• $$||v|| = ||sw|| = |s|\cdot||w|| = 2 \sqrt{6} = \boxed{\sqrt{24}}$$
• I like this too! I ended up doing a system of equations to solve for $s$ but this is a good way too. I think that's what the others were trying to tell me as well. – Future Math person Dec 21 '18 at 10:08

$$\bf v$$ is not the span of $$(1,-2,1); H^\perp$$ is the span of $$(1,-2,1).\bf v$$ is the component of $$(4,12,8)$$ in $$H^\perp$$ and $$\|\bf v\|$$ is the norm of $$\bf v$$, that is, the absolute value of the projection of $$(4,12,8)$$ along $$(1,-2,1)$$. You may recall this is given by$$\|\mathbf v\|=\begin{vmatrix}\displaystyle\frac{\langle(1,-2,1),(4,12,8)\rangle}{\|(1,-2,1)\|}\end{vmatrix}=\frac{12}{\sqrt6}=\sqrt{24}$$

Just to suggest another way.

We have the vectors $$\vec a=(2,1,0)$$ and $$\vec b=(0,1,2)$$, so the vector $$\vec c=\vec a \times \vec b=(2,-4,2)$$ spans the orthogonal complement of the subspace spanned by $$\vec a$$ and $$\vec b$$.

Normalizing this vector we have $$\hat c=\frac{\vec c}{|\vec c|}=\frac{1}{\sqrt{24}}(2,-4,2)$$

so the component of $$\vec d=(4,12,8)$$ in the subspace spanned by $$\hat c$$ is: $$u=(\vec d \cdot \hat c)=\frac{8-48+16}{\sqrt{24}}=\frac{-24}{\sqrt{24}}$$ and $$|u|=\sqrt{24}$$.