Question

a)Find the values of a,b∈C for which the matrix (1. i. 1+i)
( a. 0. b. )
(1−i. 2+i. 1) is Hermitian.(2)b)
Are there values of a∈C for which the matrix (1. 0. 0 )
(0. −1/√2. 1/√2)
(0. 1/√2. a. )
is unitary? Justify your answer.(3)c)Let (x1,x2,x3) and (y1,y2,y3) represent the coordinates with respect to the bases B1={(1,0,0),(0,1,0),(0,0,1)},B2={(1,0,0),(0,1,2),(0,2,1)}.If Q(X)=x21+2x1x2+2x2x3+x22+x23,find the representation of Q in terms of (y1,y2,y3).

Answer 1

Answer on Question #52681 – Math – Linear Algebra

a) Find the values of $a, b \in \mathbb{C}$ for which the matrix

$$A = \left( \begin{array}{ccc} 1 & i & 1 + i \\ a & 0 & b \\ 1 - i & 2 + i & 1 \end{array} \right)$$

is Hermitian.

**Solution**

$$A^{\dagger} = \left( \begin{array}{ccc} 1 & a^{*} & 1 + i \\ -i & 0 & 2 - i \\ 1 - i & b^{*} & 1 \end{array} \right).$$

If A is Hermitian

$$A = A^{\dagger}.$$

Thus

$$a = -i; b = 2 - i.$$

b) Are there values of $a \in \mathbb{C}$ for which the matrix

$$A = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & a \end{array} \right)$$

is unitary? Justify your answer.

**Solution**

If A is unitary then

$$\det A = \pm 1.$$ $$\det A = 1 \left( -\frac{1}{\sqrt{2}} \cdot a - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \right) = -\frac{1}{\sqrt{2}} \left( a + \frac{1}{\sqrt{2}} \right).$$

If $\det A = -1$

$$\left( a + \frac{1}{\sqrt{2}} \right) = \sqrt{2} \rightarrow a = \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$$

If $\det A = 1$

$$\left( a + \frac{1}{\sqrt{2}} \right) = -\sqrt{2} \rightarrow a = -\sqrt{2} - \frac{1}{\sqrt{2}} = -\frac{3}{\sqrt{2}}$$

A is unitary if $a = \pm \sqrt{2} - \frac{1}{\sqrt{2}}$.

But the matrix A is real and therefore it is orthogonal.

But orthogonal matrix should look like

$$A = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \varphi & \sin \varphi \\ 0 & \sin \varphi & - \cos \varphi \end{array} \right).$$

But in our case

$$- \sqrt {2} - \frac {1}{\sqrt {2}} \neq - \left(- \frac {1}{\sqrt {2}}\right) = \frac {1}{\sqrt {2}}.$$

Thus, there is $a = \frac{1}{\sqrt{2}} \in \mathbb{C}$ for which the matrix $A$ is unitary.

c) Let $(x1, x2, x3)$ and $(y1, y2, y3)$ represent the coordinates with respect to the bases

B1 = {(1,0,0), (0,1,0), (0,0,1)}, B2 = {(1,0,0), (0,1,2), (0,2,1)}. If Q(X) = x21 + 2x1x2 + 2x2x3 + x22 + x23, find the representation of Q in terms of (y1,y2,y3).

Solution

We have a matrix $Q$

$$Q = \left( \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right).$$

We need to find the transition matrix.

$$\overline {{b _ {2 1}}} = \overline {{b _ {1 1}}}, \overline {{b _ {2 2}}} = \overline {{b _ {1 2}}} + 2 \overline {{b _ {1 3}}}, \overline {{b _ {2 3}}} = 2 \overline {{b _ {1 2}}} + \overline {{b _ {1 3}}}.$$ $$P = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{array} \right).$$

Inverting, $\overline{b_{11}} = \overline{b_{21}},\overline{b_{12}} = -\frac{1}{3}\overline{b_{22}} +\frac{2}{3}\overline{b_{23}},\overline{b_{23}} = \frac{2}{3}\overline{b_{22}} -\frac{1}{3}\overline{b_{23}}$

$$P ^ {- 1} = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & - \frac {1}{3} & \frac {2}{3} \\ 0 & \frac {2}{3} & - \frac {1}{3} \end{array} \right).$$

Thus

$$Q _ {B _ {2}} = P ^ {- 1} Q P = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & - \frac {1}{3} & \frac {2}{3} \\ 0 & \frac {2}{3} & - \frac {1}{3} \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & - \frac {1}{3} & \frac {2}{3} \\ 0 & \frac {2}{3} & - \frac {1}{3} \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 2 \\ 1 & 3 & 3 \\ 0 & 3 & 3 \end{array} \right) =$$ $$\left( \begin{array}{ccc}1 & 1 & 2\\ -\frac{1}{3} & 1 & 1\\ \frac{2}{3} & 1 & 1 \end{array} \right).$$

Thus the representation of $Q$ in terms of $(y1, y2, y3)$ is

$$Q ^ {\prime} = y _ {1} ^ {2} + y _ {2} ^ {2} + y _ {3} ^ {2} + \frac {2}{3} y _ {1} y _ {2} + \frac {8}{3} y _ {1} y _ {3} + 2 y _ {2} y _ {3}.$$

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