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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.