Linear Algebra - change of basis matrix - quick method

Question

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 $\varphi$.

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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:

1. 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?
2. 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?

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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.

Answer 1

For your first question, it looks like the instructor worked this problem “backwards,” but got off easy because of the properties of the resulting transformation. He found $[\phi_1]_\psi$ and $[\phi_2]_\psi$ (whether by experience, guesswork, or having done this example dozens of times before) instead of $[\psi_1]_\phi$ and $[\psi_2]_\phi$ as one might have expected. The matrix $P_\psi^\phi=\left[[\phi_1]_\psi\;[\phi_2]_\psi\right]$ needs to be inverted to get the required change-of-basis matrix, but because $P_\psi^\phi$ is both unitary and conformal, $P_\phi^\psi=(P_\psi^\phi)^{-1}=(P_\psi^\phi)^T$, so he could simply write $[\phi_1]_\psi$ and $[\phi_2]_\psi$ as the rows of the change-of basis matrix.

As for your second question, I have to echo Bye_World’s comment. Sometimes, you can just eyeball a solution, sometimes there are other shortcuts you can take, but sometimes you just have to grind through the algebra. Experience in working through such problems will let you develop your own shortcuts.

It shouldn’t be all that time-consuming for problems like this, though. You’re simply solving the system $$\begin{align} \psi_1 &= a_{11}\phi_1+a_{21}\phi_2 \\ \psi_2 &= a_{11}\phi_1+a_{22}\phi_2 \end{align}$$ whatever those two bases might be. For that matter, using the “standard method” here doesn’t require all that much work and pretty much no thought: plug in the vectors, invert one of matrices, then multiply them together.

Answer 2

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