Question

3. Let R be equipped with the inner product (x,y) = AX Ay, where A is the matrix shown below: TO-4 21 A = 3 2 LO 0 5) a.) (5 points) Let v = (1,-1,3). Find || V || 1 b.) (5 points) Let x = (2,3,0) and y = (-3,2,1). Are x and y orthogonal in this inner product space? Justify your answer

Answer 1

Answer:

Given that
R3
be equipped with the inner product $\left \langle x,y \right \rangle=Ax.Ay$ where  
0 -4 2 A= 3 2 0 0 5 1
.

a). Let  $v=\begin{pmatrix} 1 & -1 & 3 \end{pmatrix}$

now $\left \| v \right \|=\sqrt \left \langle v,v \right \rangle$

$\Rightarrow \left \| v \right \|=\sqrt {Av.Av }$

  $\Rightarrow \left \| v \right \|=\sqrt {\begin{bmatrix} 0 & -4 &2 \\3 &2 &1 \\0 &0 &5 \end{bmatrix}\begin{bmatrix} 1\\-1 \\3 \end{bmatrix}. \begin{bmatrix} 0 & -4 &2 \\3 &2 &1 \\0 &0 &5 \end{bmatrix}\begin{bmatrix} 1\\-1 \\3 \end{bmatrix}}$

  $\Rightarrow \left \| v \right \|=\sqrt {\begin{bmatrix} 10\\4 \\15 \end{bmatrix}. \begin{bmatrix} 10\\4 \\15 \end{bmatrix}}$

$\Rightarrow \left \| v \right \|=\sqrt { 10^2+4^2+15^2}$

$\Rightarrow \left \| v \right \|=\sqrt { 100+16+225}$

  $\Rightarrow \left \| v \right \|=\sqrt {341}$

  

b). Let $x=\begin{pmatrix} 2 &3 &0 \end{pmatrix}$ and  $y=\begin{pmatrix} -3 &2 &1 \end{pmatrix}$.

   will be orthogonal if  $\left \langle x,y \right \rangle=0$.

Now

  $\left \langle x,y \right \rangle=Ax.Ay$

  $={\begin{bmatrix} 0 & -4 &2 \\3 &2 &1 \\0 &0 &5 \end{bmatrix}\begin{bmatrix} 2\\3 \\0\end{bmatrix}. \begin{bmatrix} 0 & -4 &2 \\3 &2 &1 \\0 &0 &5 \end{bmatrix}\begin{bmatrix} -3\\2 \\1 \end{bmatrix}}$

  $= \begin{bmatrix} -12\\12\\0 \end{bmatrix}. \begin{bmatrix} -6\\-4 \\5\end{bmatrix}$

$=(-12).(-6)+12.(-4)+0.5$

$=24$ which is not equal to 0.

Therefore are not orthogonal.