# orthogonal complement question

v= R^4 is an Inner product space and u=span{(1,0,-1,0)} subspace. how can I find a base for the vectors which orthogonal to U(the complement of U)?

Thanks!

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– user333870
Jun 3, 2016 at 1:10

If $\textbf{b} = (1,0,-1,0)$ and $U = span\{b\}$, then every vector in $U$ is of the form $\alpha \textbf{b},$ $\alpha \in \mathbb{R}$. It is easy to check that a vector $\textbf{v} \in \mathbb{R}^4$ is orthogonal to every vector in $U$ if and only if it is orthogonal to $\textbf{b}$.

Thus

$$U^\perp = \{\textbf{x}\in\mathbb{R}^4 : \langle \textbf{x}, \textbf{b}\rangle = 0 \text \}.$$

If $\textbf{x} = (x_1,x_2,x_3,x_4)$, then $\textbf{x} \in U^\perp$ if and only if

$$x_1 + 0x_2 - x_3 + 0x_4 = 0.$$

Thus finding a basis for $U^\perp$ is equivalent with finding a basis for the nullspace of the $1\times 4$ matrix

$$\begin{bmatrix}1 & 0 & -1 & 0 \end{bmatrix}.$$

If you don't know how to find a basis for the nullspace of a matrix, I can go into more details about that.

$n = e_1 - e_3$ is the base vector for $U$.

If we had three other vectors, such that all four are linear independent, we could use Gram-Schmidt starting with $n$ to come up with a orthonormal basis for $V$. The other three resulting vectors will form a basis of $U^\top$.

We note that \begin{align} \det(n, e_2, e_3, e_4) &= \det(e_1 - e_3, e_2, e_3, e_4) \\ &= \det(e_1, e_2, e_3, e_4) - \det(e_3, e_2, e_3, e_4) \\ &= \det(e_1, e_2, e_3, e_4) \\ &= 1 \end{align} so they are linear independent.

Calculation:

We enter the four vectors

>> n = [1;0;-1;0]
>> e2 = [0;1;0;0]
>> e3 = [0;0;1;0]
>> e4 = [0;0;0;1]


and start the Gram-Schmidt procedure with $n$:

>> nn = n / norm(n)
nn =

0.70711
0.00000
-0.70711
0.00000


then continue with $e_2$:

>> e2r = e2 - (nn' * e2) * nn
e2r =

0
1
0
0

>> e2n = e2r / norm(e2r)
e2n =

0
1
0
0


which changes nothing, as $e_2$ was orthogonal to $n$, then with $e_3$:

>> e3r = e3 - (nn' * e3) * nn - (e2n' * e3) * e2n
e3r =

0.50000
0.00000
0.50000
0.00000

>> e3n = e3r/norm(e3r)
e3n =

0.70711
0.00000
0.70711
0.00000


and finally $e_4$:

>> e4r = e4 - (nn' * e4) * nn - (e2n' * e4) * e2n - (e3n' * e4) * e3n
e4r =

0
0
0
1

>> e4n = e4r/norm(e4r)
e4n =

0
0
0
1


which stays unchanged as well. This is the new orthonormal basis for $V$:

>> B = [nn, e2n, e3n, e4n]
B =

0.70711   0.00000   0.70711   0.00000
0.00000   1.00000   0.00000   0.00000
-0.70711   0.00000   0.70711   0.00000
0.00000   0.00000   0.00000   1.00000


The first column vector is just the normed $n$, which still spans $U$, the remaining three vectors span $U^\top$.