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

Question

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).

Answer 1

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

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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 $\operatorname{row}(A) = \operatorname{null}(A)^\bot$.

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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 $\dim(V^\bot)=1$.$\ \ \ \ \square$

Answer 2

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 $\mathbb R^4$.$ $

Answer 3

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 $$\dim(V^\perp)=\dim\ker(A)={\rm nullity}(A)=1\ .$$