# Matrix Theory: Orthogonal Matrix

Consider a vector    $$\begin{array}{l} a=(1,-3,0,1), \\ b=(1,2,0,-1), \\ c=(0,0,1,0) \end{array}$$ Question:

(a) A non-zero vector $u$ orthogonal to all three of $a, b, c$ (there are many, just find one).

(b) Describe the set of all vectors with these properties using a parameter or parameters. Can anyone tell me how to solve this?

(a) $$u=(u_1,u_2,u_3,u_4)$$

So that $u$ is orthogonal to $a$: $u \cdot a=0 \Rightarrow u_1-3u_2+u_4=0 \ \ \ (1)$

So that $u$ is orthogonal to $b$: $u \cdot b=0 \Rightarrow u_1+2u_2-u_4=0 \ \ \ (2)$

So that $u$ is orthogonal to $c$: $u \cdot b=0 \Rightarrow u_3=0$

$$(1)+(2) \Rightarrow 2u_1=u_2$$

For $u_1=1$ we have $u_2=2$.

$(1) \Rightarrow 1-6+u_4=0 \Rightarrow u_4=5$.

Therefore, a non-zero vector $u$ orthogonal to all three of $a, b, c$ is $(1,2,0,5)$.

EDIT:

(b)

From the above relations we have that: $$2u_1=u_2$$ $$u_3=0$$ $$\text{ Replacing } 2u_1=u_2 \text{ at the relation } (2) \text{ we get: } u_1+4u_1-u_4=0 \Rightarrow u_4=5u_1$$

Therefore, $$u=(u_1,u_2,u_3,u_4)=(u_1,2u_1, 0, 5u_1)=u_1(1,2,0,5)$$

So, the set of all vectors with these properties is: $$\{u_1(1,2,0,5):u_1 \in \mathbb{R} ^*\}$$

• Thanks for helping. But what will be (b)? Aug 4, 2014 at 14:13
• I edited my answer.. Aug 4, 2014 at 14:24

Maybe you already realize this since you've used the tag (matrices), but you can solve this problem by computing the kernel of the matrix $$\begin{pmatrix} 1 & -3 & 0 & 1\\ 1 & 2 & 0 & -1\\ 0 & 0 & 1 & 0 \end{pmatrix} \, .$$ According to Wolfram Alpha, this matrix has RREF $$\begin{pmatrix} 1 & 0 & 0 & -1/5\\ 0 & 1 & 0 & -2/5\\ 0 & 0 & 1 & 0 \end{pmatrix}$$ which should allow you to read the answer off.