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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.

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  • $\begingroup$ 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 $W^\perp$ $\endgroup$ – Smilia Sep 30 '18 at 20:35
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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.

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  • $\begingroup$ 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. $\endgroup$ – Picaud Vincent Oct 1 '18 at 5:19
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    $\begingroup$ Thank you, I realize that the biggest problem I was having was computational difficulties myself, but this solution helped me realize my own gaps and correct them. Much appreciated. $\endgroup$ – Raj Oct 1 '18 at 14:49
  • $\begingroup$ fine, I also re-learned something... to double check my answers :) $\endgroup$ – Picaud Vincent Oct 1 '18 at 14:51

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