Let $S \subset \mathbb{R}^4$ be the vectors whose components satisy $x_1 + x_2 – x_3 + x_4 = 0$
Find the dimension of S and then find a basis for the orthogonal complement of S
So to find the dimension, i understand I am looking for the nullity as the equation equals 0, so setting $x_4=r$ $x_3=t$ and $x_2=s$ and having $x_1=-x_2+x_3-x_4$ so you get
$\begin{bmatrix}
-2+t-r \\
s\\
t\\
r
\end{bmatrix} = s\begin{bmatrix}
-1 \\
1\\
0\\
0
\end{bmatrix}+t\begin{bmatrix}
1 \\
0\\
1\\
0
\end{bmatrix}+r\begin{bmatrix}
-1 \\
0\\
0\\
1
\end{bmatrix}$ therefore giving dim(S)=3
However I am now stuck on how to find the orthogonal complement?
Best Answer
The subspace $S$ is the null space of the matrix $$ A=\begin{bmatrix}1 & 1 & -1 & 1\end{bmatrix} $$ so the orthogonal complement is the column space of $A^T$. Thus $S^\perp$ is generated by $$ \begin{bmatrix}1 \\ 1 \\ -1 \\ 1\end{bmatrix} $$
It is a general theorem that, for any matrix $A$, the column space of $A^T$ and the null space of $A$ are orthogonal complements of each other (with respect to the standard inner product).
To wit, consider $x\in N(A)$ (that is $Ax=0$) and $y\in C(A^T)$ (the column space of $A^T$). Then $y=A^Tz$, for some $z$, and $$ y^Tx=(A^Tz)^Tx=z^TAx=0 $$ so $x$ and $y$ are orthogonal. In particular, $C(A^T)\cap N(A)=\{0\}$. Let $A$ be $m\times n$ and let $k$ be the rank of $A$. Then $$ \dim C(A^T)+\dim N(A)=k+(n-k)=n $$ and so $C(A^T)\oplus N(A)=\mathbb{R}^n$, thereby proving the claim.