Linear Algebra - change of basis matrix - quick method In the class, the instructor supplied the following exercise:
Given the following vectors:
$$
\begin{align}
\psi_1(x) &= \frac{1}{\sqrt 2} \\
\psi_2(x) &= \sqrt{\frac{3}{2}} \cdot x \\
\varphi_1(x) &= \frac{\sqrt 3}{2} \cdot x + \frac{1}{2} \\
\varphi_2(x) &= \frac{\sqrt 3}{2} \cdot x - \frac{1}{2} 
\end{align}
$$
Then $(\psi_1, \psi_2)$ and $(\varphi_1, \varphi_2)$ form a basis.
Find a matrix which converts a vector $\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}_{\psi}$ in base $\psi$ to a vector $\begin{pmatrix} d_1 \\ d_2 \end{pmatrix}_{\varphi}$ in base $\varphi$.

The solution according to the instructor:
The instructor guessed the following:
$$
\begin{align}
\varphi_1(x) &= \frac{1}{\sqrt 2} \cdot \big(\; \psi_1(x) + \psi_2(x) \;\big) \\
\varphi_1(x) &= \frac{1}{\sqrt 2} \cdot \big( -\psi_1(x) + \psi_2(x) \;\big)
\end{align}
$$
So the required matrix $P$ is:
$$
P = \begin{pmatrix}
1 & 1\\
-1 & 1\\
\end{pmatrix} \cdot \frac{1}{\sqrt 2}
$$
and the conversion is done as follows: $\vec d = P \cdot \vec c$.
Or, when elaborating:
$$
\begin{pmatrix} d_1 \\ d_2 \end{pmatrix} = 
\begin{pmatrix}
\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\
-\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\
\end{pmatrix}
\cdot
\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}
$$
Now, this guess seems easy. But I'm having difficulties with:


*

*How did he know that he needs to find $\varphi_i$ as a linear combination of $\psi_i$ in order to find the matrix in question?

*What should I do if guessing isn't as easy as in the example?
Solve it using the standard method of $P = M_\varphi^E M_E^\psi $ ? ($E$ is standard basis ; $E = (1, x)$)
It takes precious time. Isn't there a quicker method?



Edit:
I solved it as follows:  
$$
P = M_\varphi^E \cdot M_E^\psi \equiv P_\varphi^\psi
$$
when:


*

*The basis in the subscript is the range, and the basis in the superscript is the domain

*Matrix M is a also a change of basis matrix (identity map $Id$)

*E is standard basis


Therefore, according to the definition of change-of-basis matrix:
$$
P_\varphi^\psi = \begin{pmatrix}
 | & | \\
 [\psi_1]_\varphi & [\psi_2]_\varphi \\
 | & | \\
\end{pmatrix} = \begin{pmatrix}
\lambda_1 & \lambda_3 \\
\lambda_2 & \lambda_4 \\
\end{pmatrix}
$$
Now, according to the definition of coordinate vector:
$$
\begin{align*}
& [\psi_1]_\varphi = 
\begin{pmatrix} \lambda_1 \\ \lambda_2 \end{pmatrix} =
\lambda_1 \cdot \varphi_1 + \lambda_2 \cdot \varphi_2 \\
& 
[\psi_2]_\varphi = 
\begin{pmatrix} \lambda_3 \\ \lambda_4 \end{pmatrix} =
\lambda_3 \cdot \varphi_1 + \lambda_4 \cdot \varphi_2
\\
\end{align*}
$$
According to all the actions that I did until now - we know that we need to find $\psi$ as a function of $\phi$ - which is the opposite of what the instructor did.
Let's solve the above equations:
$$
\begin{align*}
 \left[ \frac{1}{\sqrt 2} \right]_\varphi &=  \frac{1}{\sqrt 2} \cdot (\varphi_1 - \varphi_2) \\
 \left[ \sqrt{\frac{3}{2}} \cdot x \right]_\varphi &=  \frac{1}{\sqrt 2} \cdot (\varphi_1 + \varphi_2)
\end{align*}
$$
$$
\Longrightarrow P_\varphi^\psi = \begin{pmatrix}
 1 & 1 \\
 -1 & 1 \\
\end{pmatrix} \cdot \frac{1}{\sqrt 2}
$$
Which is exactly the same matrix that the instructor received. He told me that there's symmetry because that the matrix is a Unitary operator. When I showed this (my solution) to the instructor, he said that I mixed the positions of the bases vectors, so I got a wrong result. But again I don't understand what's wrong - everything is done that same as I did in Linear Algebra course.

Using the “standard method:” $$
M_E^\psi=\pmatrix{\frac1{\sqrt2} & 0 \\ 0 & \sqrt\frac32} \\
M_\varphi^E=\pmatrix{\frac12 & -\frac12 \\ \frac{\sqrt3}2 & \frac{\sqrt3}2}^{-1}=\pmatrix{1 & \frac1{\sqrt3} \\ -1 & \frac1{\sqrt3}} \\
M_\varphi^E M_E^\psi = \pmatrix{1 & \frac1{\sqrt3} \\ -1 & \frac1{\sqrt3}}\pmatrix{\frac1{\sqrt2} & 0 \\ 0 & \sqrt\frac32}=\pmatrix{\frac1{\sqrt2} & \frac1{\sqrt2} \\ -\frac1{\sqrt2} & \frac1{\sqrt2}}.
$$
