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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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  • $\begingroup$ I notice that you have asked a number of questions over the past week, but you have not accepted any of the answers. If you have gotten the help that you needed from the answers that people took the time to write for you, it is customary to accept the answer as a thank you. $\endgroup$ – user333870 Jun 3 '16 at 1:10
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$n = e_1 - e_3$ 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 $U^\top$.

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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.

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