# Orthogonal complement of linear span with respect to matrix of bilinear form

Now I really feel that this problem should be straightforward, but I just don't feel right about what am I doing. I am given:

$$A = \begin{pmatrix} 1 & -1 & -2 \\ -1 & 0 & -3 \\ -2 & -3 & 7 \end{pmatrix}$$ as the matrix of the bilinear form, and the goal is to find find the orthogonal complement of $$\text{span}(a_1,a_2)$$ where $$a_1 = (1,2,3), a_2 = (3,4,5)$$.

This is all seems well enough, I consider essentially that I have $$\alpha : V \times V \to K$$ (field $$K$$, I think I could just say $$\mathbb{R}$$, but this is unspecified) as my bilinear map associated with $$A$$. I consider $$W\subset V$$ as $$W =$$ $$\text{span}(a_1,a_2)$$, then by definition $$W^\perp = \{ y\in V : \alpha(x,y) = 0, \forall x \in W\}$$ (this being my goal to find).

I can express my linear span $$W = \text{span}(a_1,a_2) = \{(x_1+3x_2, 2x_1+4x_2,3x_1+5x_2, x_i \in K\}$$. Then using the definition of what it means for a vector $$y \in V$$ to be in the orthogonal complement I say:

$$\begin{pmatrix} x_1 + 3x_2 \\ 2x_1+4x_2 \\ 3x_1+5x_2 \end{pmatrix}^T \begin{pmatrix} 1 & -1 & -2 \\ -1 & 0 & -3 \\ -2 & -3 & 7 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}=0$$ which gives me $$−11x_2y_1−18x_2y_2+17x_2y_3+18x_1=0$$. But I'm just not certain about this, I feel it makes sense at every step, but somehow I'm left feeling I could do more?

So essentially my question is have I gone about this correctly? Because the math seems fine by my intuition isn't very satisfied. Thanks very much for any help.

• You equation can be written as $x_1 (-7 y_1 - 10 y_2 + 13 y_3) + x_2 (-11 y_1 - 18 y_2 + 17 y_2)=0$ which sould be satisfied for every $x_1,x_2$ This implies that $y$ should belong to the intersection of the planes whose equations are: $-7 y_1 - 10 y_2 + 13 y_3=0$ and $-11 y_1 - 18 y_2 + 17 y_2=0$. The wedge product of the normal vectors will give you a basis of $W^\perp$ – Smilia Sep 30 '18 at 20:35

Your space $$W$$ is generated by the two vectors $$a_1$$ and $$a_2$$, hence $$b\in W^\perp$$ is equivalent to the system: \begin{align*} \langle b,a_1 \rangle_A &=0 \\ \langle b,a_2 \rangle_A &=0 \\ \end{align*} where you scalar product is defined by $$\langle v,w \rangle_A = v^t.A.w$$
With your $$a_1=(1,2,3)$$ and $$a_2=(3,4,5)$$ and $$b=(x,y,z)$$ you get: \begin{align*} \langle b,a_1 \rangle_A &= -7x-10y+13z &=0 \\ \langle b,a_2 \rangle_A &=-11x-18y+17z &=0 \\ \end{align*} The solution is $$b=(x,-\frac{3}{8}x,\frac{1}{4}x)$$ hence $$W^\perp$$ is a one dimensional space generated by the $$(1,-\frac{3}{8},\frac{1}{4})$$ vector.
• Raj and @amd sorry the the delay I just woke up. I had an error in matrix A (wrong value $A_{2,3}=3$ instead of $A_{2,3}=-3$ ). I have updated the post with the corrected value I hope it is ok now. Thanks for your vigilance and sorry for this error. – Picaud Vincent Oct 1 '18 at 5:19