# Find the orthogonal complement with respect to $\phi$

Given is a symmetric bilinear form $$\phi:\mathbb R^4\times\mathbb R^4\to \mathbb R$$ with $$\phi(v,w)=v^{T}Aw$$ and $$A=\begin{bmatrix}1&1&1&1\\1&2&2&2\\1&2&-2&0\\1&2&0&-3\end{bmatrix}$$

How can I compute the orthogonal complement $$U$$ of $$W=(e_1,e_1+e_2+e_3)$$ with respect to $$\phi$$ and second question; Is $$U\cap W$$ not the empty set ?

$$U=\{v\in\mathbb R^4:\phi(e_1,v)=0,\phi(e_1+e_2+e_3,v)=0\}$$ but $$e_1+e_2+e_3$$ is not in $$\mathbb R^4$$, it is a matrix ?

can I say $$\phi(e_1+e_2+e_3,v)=\phi(e_1,v)+\phi(e_2,v)+\phi(e_3,v)$$ and then ?

The set $W$ can clearly be expressed as the set of vectors $(a+b,b,b,0)^T$.

Expanding orthogonality relationship

$$\begin{bmatrix}x&y& z& t\end{bmatrix}\begin{bmatrix}1&1&1&1\\1&2&2&2\\1&2& -2& 0&\\1&2&0&-3\end{bmatrix}\begin{bmatrix}a+b\\b\\ b\\ 0\end{bmatrix}=0$$

one gets:

$$a(x+y+z+t)+b(3x+5y+z+3t)=0$$

that must be verified for all $a,b$, meaning equivalently that $x, y, z, t$ are solutions of the underdetermined system:

$$\begin{cases} x+y+z+t&=&0\\3x+5y+z+3t&=&0\end{cases}$$

That we can for example express in function of the two first variables, giving:

$$\begin{cases} z=y\\t=-x-2y\end{cases}$$

giving the 2 dimensional vector set:

$$\begin{bmatrix}x\\y\\ z\\ t\end{bmatrix}=\begin{bmatrix}x\\y\\y\\ -x-2y\end{bmatrix}=x\begin{bmatrix}1\\0\\0\\-2\end{bmatrix}+y\begin{bmatrix}0\\1\\1\\-2\end{bmatrix}$$

Regarding your second question, a vector of $U \cap W$ could be written, for certain $a,b,x,y$:

$$\begin{cases}x&=&a+b\\y&=&b\\y&=&b\\-x-2y&=&0\end{cases}$$

from which you will deduce easily that : $a=b=x=y$ (begin by adding the 4 equations): thus we get the null vector; said otherwise: $U \cap W=\{0\}$ (but not "void": beware).

A remark: Matrix $A$ is not positive definite: it has 2 positive eigenvalues and 2 negative eigenvalues.

• I think that you were blocked by the fact you didn't use coordinates. Apr 7, 2016 at 16:48
• Now it's clear thanks a lot Apr 7, 2016 at 17:03