# Exercise 2.2

Suppose $V$​ is a vector space with basis vector $|v_1\rangle = |0\rangle$​ and $|v_2\rangle = |1\rangle $​, and $A: V\to V$​ is a linear operator. Since we have $A|0\rangle = |1\rangle$​ and $A|1\rangle = |0\rangle$​, according to eq. (2.12) in the book, we could find the matrix entries from 

$$
\begin{align}
A|v_1\rangle &= A|0\rangle = A_{11}|0\rangle + A_{21}|1\rangle = |1\rangle \\
A|v_2\rangle &= A|1\rangle = A_{12}|0\rangle + A_{22}|1\rangle = |0\rangle 
\end{align}
$$(eqn:2.2.1)

From eq. {eq}`eqn:2.2.1` we could conclude that 

$$
A = \begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
1 & 0 \\ 
\end{pmatrix} 
$$(eqn:2.2.2)

Suppose we have a different basis of $V$ with 

$$
|+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}}, |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}}
$$ (eqn:2.2.3)

To check whether they are basis, we could do

$$
\begin{align}
\langle +|+\rangle &= \frac{\langle 0 | + \langle 1|}{\sqrt{2}}\cdot\frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{\langle 0 |0\rangle + \langle 1|1\rangle}{2} = 1 \\
\langle +|-\rangle &= \frac{\langle 0 | + \langle 1|}{\sqrt{2}}\cdot\frac{|0\rangle - |1\rangle}{\sqrt{2}} = \frac{\langle 0 |0\rangle - \langle 1|1\rangle}{2} = 0 \\
\langle -|+\rangle &= \frac{\langle 0 | - \langle 1|}{\sqrt{2}}\cdot\frac{|0\rangle + |1\rangle}{\sqrt{2}} = \frac{\langle 0 |0\rangle - \langle 1|1\rangle}{2} = 0 \\
\end{align}
$$(eqn:2.2.4)

Then we could conclude from eq. {eq}`eqn:2.2.4` that the basis defined in eq. {eq}`eqn:2.2.3` is a valid orthonormal basis. Under the basis of eq. {eq}`eqn:2.2.3`, basis vectors $|0\rangle$ and $|1\rangle$ becomes

$$
|0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}}, |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}}
$$(eqn:2.2.5)

From eq. {eq}`eqn:2.2.5`, eq. {eq}`eqn:2.2.3` becomes

$$
\begin{align}
A|0\rangle = A\left(\frac{|+\rangle + |-\rangle }{\sqrt{2}}\right) = \frac{A|+\rangle + A|-\rangle }{\sqrt{2}} = |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}}\\
A|1\rangle = A\left(\frac{|+\rangle - |-\rangle }{\sqrt{2}}\right) = \frac{A|+\rangle - A|-\rangle }{\sqrt{2}} = |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\
\end{align} 
$$(eqn:2.2.6)

From eq. {eq}`eqn:2.2.6` we could find that 

$$
A|+\rangle = |+\rangle, A|-\rangle = -|-\rangle 
$$(eqn:2.2.7)

Using similar method with eq. {eq}`eqn:2.2.1`, we obtain

$$
\begin{align}
A|v_1\rangle = A|+\rangle = A_{11}|+\rangle + A_{21}|-\rangle = |+\rangle\\
A|v_2\rangle = A|-\rangle = A_{12}|+\rangle + A_{22}|-\rangle = -|-\rangle\\
\end{align}
$$(eqn:2.2.8)

Thus, the matrix representation of $A$ in basis $|+\rangle$ and $|-\rangle$ becomes

$$
A = \begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{pmatrix} = \begin{pmatrix}
1 & 0 \\
0 & -1 \\
\end{pmatrix}
$$(eqn:2.2.9)

Then under basis $|+\rangle$ and $|-\rangle$, the matrix representation of $A$ is different.