Finding a basis for a subspace that is orthogonal to a a set of 2 vectors

Question

I've been plugging away at homework but this question has left me stuck and frustrated to the point where I am going to lose my marbles. I'm having trouble actually going about the steps to do it and hope that someone here could give me some pointers.

Problem 3: Consider the subspace consisting of all vectors in $\Bbb R^4$ that are orthogonal to both of the vectors $$\vec n_1=\begin{bmatrix}1 \\ 0 \\ -1 \\ 0\end{bmatrix}, \ \ \vec n_2=\begin{bmatrix}1 \\ 0 \\ 1 \\ 0\end{bmatrix}.$$ Find a basis for this subspace.

What I feel like I know about this question:

Firstly I know we want to find a vector $\vec{x} = (x_{1}, x_{2}, x_{3}, x_{4})$. Because we know it is orthogonal to both $\vec{n_{1}}$ and $\vec{n_{2}}$, we want the dot product of $\vec{x} \cdot \vec{n_{1}}$ = $\vec{x} \cdot \vec{n_{2}}$ $= 0$.

Since we want to prove a basis for the subspace we will have to show linear independence and that it is within the spans $\vec{x}$.

I believe I have done the linear independence part, as I did a system of equations on the 2 non trivial equations from the vector sets and ended up with $2x_{2} = 0$ which I input into an equation and get the expected answer of $0$.

As of now I am stumped on how to even find if it is in the span. A lot of the questions asked online use matrix row reduction and such but we have not covered that in our class yet so I am not familiar with how to use it :(. Any pointers are appreciated.

Answer 1

Let $\vec x\in\Bbb R^4$. Then $$(*) \ \ \ \ \ \ \vec x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}, \ \text{with} \ x_1,x_2,x_3,x_4\in\Bbb R.$$ Now, in order to find the subspace consisting of all vectors in $\Bbb R^4$ such that $$(**) \ \ \ \ \ \ \vec x\cdot\vec n_1=0 \ \ \text{and} \ \ \vec x\cdot\vec n_2=0,$$ we have to substitute $(*)$ and $\vec n_1,\vec n_2$ (which are given in the problem) into the equations $(**)$. Thus, we get $\begin{cases}x_1-x_3=0\\x_1+x_3=0\end{cases}\implies x_1=x_3=0.$ Substituting $x_1=x_3=0$ into $(*)$, we get $$\vec x=\begin{bmatrix}0\\x_2\\0\\x_4\end{bmatrix}.$$

Thus, the subspace consisting of all vectors in $\Bbb R^4$ that are orthogonal to both $\vec n_1$ and $\vec n_2$ is given by $$U=\left\{\left.\begin{bmatrix}0\\x_2\\0\\x_4\end{bmatrix}\right|x_2,x_4\in\Bbb R\right\}.$$ Note that if $u\in U$, then $$u=\begin{bmatrix}0\\x_2\\0\\x_4\end{bmatrix}=x_2\begin{bmatrix}0\\1\\0\\0\end{bmatrix}+x_4\begin{bmatrix}0\\0\\0\\1\end{bmatrix}:=x_2 v+x_4 w.$$ This means that $v$ and $w$ spans $U$. Moreover, since $v$ and $w$ is linearly independent, we can conclude that the set $$\left\{\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\right\}$$ is a basis for $U$.

Answer 2

Another way, we have

$$\vec n_1+\vec n_2=\begin{bmatrix}2\\0\\0\\0\end{bmatrix}=2\hat e_1 \;\;\;\;\;\;\vec n_2-\vec n_1=\begin{bmatrix}0\\0\\2\\0\end{bmatrix}=2\hat e_3$$

therefore $\hat e_2$ and $\hat e_4$ are a basis of the orthogonal subspace.