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Let $V = \text{span}(\{\vec{v}_1,\vec{v}_2,\vec{v}_3\})$ be a $3$ dimensional subspace of $\mathbb{R}^4$. Prove that the orthogonal complement of $V$ has dimension $1$

My approach:

  • Set $A = \left[v_1|v_2|v_3\right]$, $\{v_1, v_2, v_3\}_{3\times1}$ column vectors.
  • $V^{\perp} = N_A$ where $N_A$ is the null space of $A$.

I am uncertain of where to go from here (or if this approach is appropriate to begin with).

MSL
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  • Do you know the rank-nullity theorem? – Cheerful Parsnip Nov 23 '15 at 01:02
  • When you construct the matrix you want $v_1, v_2, v_3$ to be the rows, not the columns of the matrix. –  Nov 23 '15 at 01:04
  • @GrumpyParsnip $\text{Rank}(A)+\text{Null}(A)=\text{Dim}(V)$ where $A = \left[\vec{v_1}|...|\vec{v_n}\right]$ and ${\vec{v_1},...,\vec{v_n}}_{n\times1} \in V$? Something along those lines? – MSL Nov 23 '15 at 01:06
  • @Bye_World Why's that? Apologies, my knowledge on these concepts are a tad shaky. – MSL Nov 23 '15 at 01:08
  • @MSL See if you can figure it out from the link. –  Nov 23 '15 at 01:09
  • @Bye_World I'll give it a look. Thanks for your assistance! – MSL Nov 23 '15 at 01:09
  • @Bye_World I've still not quite gotten it yet; unfortunately the material you provided didn't precipitate any enlightenment. I understand that the rank of a given matrix $A$ plus its nullity is equal to number of columns in the matrix which here represents dimension of $V$, but I'm not seeing how to tie it all together to get the result (also, a sidenote: we should totally get together and play a game of DND sometime). – MSL Nov 23 '15 at 02:25
  • @Bye_World It'd be a little easier with some numbers, but to prove it generally is a bit more challenging, because it's just given that $\vec{v}_1, \vec{v}_2, \vec{v}_3$. I know that the nullity of $A$ is equal to the number of free variables determined by the matrix $A$ and the rank of $A$ is the number of rows with leading zeroes when the matrix is reduced to row echelon form, but I don't know how to reduce $A$ and show this analytically, because any way I do it wouldn't work if it so happened that some component turned out to be zero. – MSL Nov 23 '15 at 02:54

3 Answers3

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Use the fact that for any $n\times m$ matrix $A$ $$\operatorname{row}(A) = \operatorname{null}(A)^\bot$$

Here's why:

Let $a_i$ be the $i$th row of $A$ and $v$ be an $m\times 1$ matrix. Then $$Av = \pmatrix{a_1 \\ a_2 \\ \vdots \\ a_n}v = \pmatrix{a_1 \cdot v \\ a_2 \cdot v \\ \vdots \\ a_n \cdot v}$$

Confirm this for yourself.

Now consider the case $Av=0$. From the above you can see that this implies that $a_i\cdot v=0$ for all $i$. Thus $a_i$ is orthogonal to $v$ for all $i$. This implies that the set of all $x$ such that $Ax=0$ is the orthogonal complement of the space spanned by the rows of $A$, i.e. $\operatorname{row}(A)$. But the set of all $x$ such that $Ax=0$ is exactly the definition of the $\operatorname{null}(A)$. Thus $\operatorname{row}(A) = \operatorname{null}(A)^\bot$.

Now let's use that. Construct a $3\times 4$ matrix from the vectors $v_1, v_2, v_3$: $\pmatrix{v_1 \\ v_2 \\ v_3}$. Then from the above the nullspace of this matrix is the orthgonal complement of $\operatorname{span}(v_1, v_2, v_3) = V$.

Then we use the rank nullity theorem which states that $$\dim(\operatorname{rank}(A)) + \dim(\operatorname{null}(A)) = m$$

Thus in this case $3+\dim(V^\bot)=4$. Thus $\dim(V^\bot)=1$.$\ \ \ \ \square$

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Let $w_1,\ldots,w_k$ be an orthonormal basis of the orthogonal complement of $V$. Then $$ v_1,v_2,v_2,w_1,\ldots,w_k $$ is an orthonormal basis of $\mathbb R^4$.$ $

Martin Argerami
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    @Bye_World - in what way does it fail to answer the question? Bases for $V$ and $V^\perp$ providing a basis for $\Bbb R^4$ means that $\dim V + \dim V^\perp = \dim \Bbb R^4$, So $3 + \dim V^\perp = 4$. – Paul Sinclair Nov 23 '15 at 03:50
  • @Bye_World: and why do you think it is "useful" to make the OP use the rank-nullity theorem when she/he doesn't understand dimension? – Martin Argerami Nov 23 '15 at 04:57
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Your notation is very confusing. Initially you say that $\def\v#1{{\bf #1}}\v v_1,\v v_2,\v v_3$ are vectors in $\Bbb R^4$: this means that they should have $4$ components. But further down you say that they are $3\times1$ vectors.

To understand this sort of problem you should remember that basically everything comes down to linear equations. When you understand this, you can leave out the equations and jump straight to matrices or other techniques, but I really suggest that for a start you always write out appropriate equations.

So, say $\v v_1=(a,b,c,d)$. To have $\v x=(x_1,x_2,x_3,x_4)\in V^\perp$ you need $\v v_1\cdot\v x=0$, that is, $$ax_1+bx_2+cx_3+dx_4=0\ .$$ Doing the same for $\v v_2$ and $\v v_3$ and putting the results in matrix form, $$\pmatrix{a&b&c&d\cr .&.&.&.\cr .&.&.&.\cr}\v x=\v 0\ .$$ That is, $V^\perp=\ker A$, where $A$ is the matrix with (NB) rows which are the given vectors. Now for $A$ we have $${\rm rank}(A)+{\rm nullity}(A)=4\ ,$$ and ${\rm rank}(A)=3$ (because the rows are independent, because they span $V$ which is given to be $3$-dimensional). So $$\dim(V^\perp)=\dim\ker(A)={\rm nullity}(A)=1\ .$$

David
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  • The quoted text is what I was given. I do not understand it either. I assume it being a "three dimensional subspace" implies that it contains vectors of the form $u=(a,b,c,0)$. – MSL Nov 23 '15 at 01:16
  • First: there is nothing wrong with the quoted text. The problem is with what you have written just below it concerning "$3\times1$ vectors". Second: vectors in $V$ need not be $(a,b,c,0)$, they could be any vectors at all in $\Bbb R^4$, provided that they are linear combinations of three independent vectors. If you are not clear what is meant by "independent", "basis", "dimension" etc, you need to put this problem aside, revise earlier concepts, and come back to this later. I'm sure you can also ask another question on MSE if you need help with the fundamentals. – David Nov 23 '15 at 01:20
  • Vectors in $S={\vec{v}_1,\vec{v}_2,\vec{v}_3}$ are linearly dependent $\iff$ for $c_1,c_2,c_3 \in \mathbb{R}, c_1\vec{v}_1+c_2\vec{v}_2+c_3\vec{v}_3=\vec{0} \implies c_1,c_2,c_3 = 0$ such that no vector $v_i$ in $S$ can be written as a linear combination of the vectors in $S\setminus{v_i} ,\ 1 \leq i \leq 3$. A basis held within a set $W$ for a vector space of dimension $V$ holds $n$ vectors such that $W \subset V$, $n = \text{dim}(V)$, $\text{span}(W)=V$, and $W$ is linearly independent. I know all three of those things. It is the problem itself to which I am confused. – MSL Nov 23 '15 at 01:33