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.