Determine a basis for $U^\perp.$

Question

Let $U\subset\mathbb{R}^3$ be the space spanned by the vectors $v_1=[1,-1,0]^T$ and $v_2=[1,0,-1]^T$. Determine a basis for $U^\perp.$

In the solution they say that:

$U^{\perp}$ is equal to the nullspace of the matrix

$$A^T=\begin{bmatrix}1 & -1 & 0 \\ 1 & 0 & -1\end{bmatrix},$$

where its rows are $v_1^T$ and $v_2^T$. By rowreduction we obtain $U^{\perp}=\text{span}(u)$ where $u=[1,1,1]^T.$

Questions:

1) Does this mean that $U$ is not a space but a plane since two vectors in space can only span a plane?

2) Why is it equal to the nulllspace of $A^T?$

3) Why is it not equal to the nullspace of $A$ instead?

4) What is the answer? They have not given a bssis for $U^{\perp}?$

Answer 1

1) Does this mean that $U$ is not a space but a plane since two vectors in space can only span a plane?

Any plane through the origin is a vector space. So it can be both.

2) Why is it equal to the nulllspace of $A^T?$

For real values matrices, it's because the dot product is defined by the transpose: $u \cdot v= u^Tv$. So since you're looking for all the vectors perpendicular to the column space of $A$ this means finding all $x$ such that $a \cdot x=a^Tx=0$ for all a in $A$. In particular it's enough to check that this works for every vector that is in a basis for $A$ (or a span of the columns). Write out $$ A^Tx=\begin{bmatrix}v_1^Tx\\ v_2^Tx\end{bmatrix}=\begin{bmatrix}v_1\cdot x\\ v_2\cdot x\end{bmatrix} $$ So you can see that settin $A^Tx=0$ gives a solution to all vector perpendictular to these vectors.

3) Why is it not equal to the nullspace of $A$ instead?

The null space of $A$ is the solutions to $Ax=0$ this is looking for solutions to $A^Tx=0$. The nullspace of $A$ is a subspace of $\mathbb{R}^3$ in your example the nullspace of $A^T$ is a subspace of $\mathbb{R}^2$ so they can't even be the same.

4) What is the answer? They have not given a bssis for $U^{\perp}?$

First note they have given you a basis for $U$ then they find the basis using something like the following

Row reduce

$$A^T=\left[\begin{array}{rrr}1 & -1 & 0 \\ 1 & 0 & -1\end{array}\right] \sim \left[\begin{array}{rrr}1 & 0 & -1 \\ 0 & 1 & -1\end{array}\right] $$

Note when you solve $Ax=0$ the 3rd coordinate of $x_3$ variable is free so $x_3=t$. Then multiplying out by $x$ you get equations $$ \begin{align} x_1-x_3&=0\\ x_2-x_3&=0 \end{align} $$ So solving gives $x_1=x_2=x_3=t$ so solutions have the form $x=t\begin{bmatrix}1\\1\\1\end{bmatrix}$. So a basis for $U^\perp$ is $\begin{bmatrix}1\\1\\1\end{bmatrix}$

Answer 2

Yes, U is a plane. Note that U is the span of two vectors. And in $\mathbb{R}^3$ is a plane.

To be in $U^\perp$, the vector needs to be perpendicular to $v_1$ and $v_2$. That is, $v_1\cdot x= 0$ and $v_2\cdot x=0$. Another way to write this is $v_1^T x= 0$ and $v_2^Tx=0$. This can all be written as $A^T x = 0$. Do you see it?

So now you need to find the null space of $A^T$ to get a vector in $U^\perp$. $A^T x = 0$ implies $x_1-x_2=0$ and $x_1-x_3=0$. If we let $x_1=1$ then $x_2=1$ and $x_3=1$. Do a vector in $U^\perp$ is $(1,1,1)$.

Answer 3

1. Yes. The span of two vectors is a plane, not a three dimensional space.
2. I don't know why they call the matrix $A^T$ instead of $A$ nor do I know why they chose this solution. $U^{\perp}$ is one dimensional and is spanned by the vector that is the cross product of $v_1$ and $v_2$.
3. As above, I don't know why they used that notation but the reason the solution is the null space of that matrix is because vectors in $U^{\perp}$ must be perpendicular to both of $v_1$ and $v_2$. That matrix encodes those two dot products and the fact that each must be zero.
4. Take the cross product of $v_1$ and $v_2$.