Linear Algebra - Homogenous Equations

Question

Find a homogenous system of linear equations in five unknowns whose solution space consists of all vectors $\Re^5$ that are orthogonal to the vectors:

$\mathbf v_1$$=\langle3,0,1,2,3\rangle$;

$\mathbf v_2$$=\langle-3,1,0,-1/2,-1\rangle$;

$\mathbf v_3$$=\langle6,0,1,2,-3\rangle$.

What kind of geometric object is the solution set? Find a general solution of the system and confirm that the solution space has the orthogonality and the geometric properties that were stated.

I was told this problem would be really good practice for linear algebra however I have no idea how to approach this problem and would appreciate any guidance.

Answer 1

We are looking for all $\mathbf x=\langle x,y,z,u,v\rangle \in \Re^5$ such that

$\mathbf x \cdot \mathbf v_j=0$ for $j=1,2,3$, where $ \mathbf x \cdot \mathbf v_j$ denotes the usual inner product on $\Re^5$.

Answer 2

A vector $(x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5$ is orthogonal to a vector $(a_1, a_2, a_3, a_4, a_5) \in \mathbb{R}^5$ if and only if $a_1 x_1 + a_2 x_2 + a_3 x_3 + a_4 x_4 + a_5 x_5 = 0$. Here, the set of vectors you are looking for is the set of solutions of the system : \begin{equation} \left\{ \begin{aligned} 3x_1& & +x_3& +2x_4&+3x_5&=&0\\ -3x_1&+x_2& &-\frac{1}{2}x_4&-x_5&=&0\\ 6x_1& &+x_3&+2x_4&-3x_5&=&0\\ \end{aligned} \right. \end{equation}

Now you get a linear system and you can resolve it with your favorite method.

Answer 3

Let $(x,y,z,u,w)$ be such a solution.

You want $$ 3x+z+2u+3w=0$$

and $$-3x+y-1/2u+w=0$$ and $$6x+z+2u-3w =0$$ Obviously you have many solutions for this system.

Answer 4

Others answer cover the linear system quite well. So this one will only focus on the vector space part. And specially on the question "What kind of geometric object is the solution set?"

Let's call

$V = Span(v_1,v_2,v_3)$

Let's call $W$ the set of vectors orthogonals to both $v_1$,$v_2$ and $v_3$.

We can prove the following propositions one after the other (I'll let you prove it. If you need help with the proofs, just ask in comment):

• $W$ is a vector space
• $W$=$V^\bot$
• $Dim(V)=3$
• $Dim(W)=2$

Therefore, $W$ is a sub-space of $\mathbb R^5$ of dimention $2$.