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:

  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?


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.

  • $\begingroup$ Isn't the algebra pretty easy in this case, though? Just look at $\varphi_1$: it's got an $x$ term and a constant term. What do you need to multiply $\psi_1$ by to get the constant term in $\varphi_1$? Oh $\frac 1{\sqrt{2}}$. What about the $x$ term? Oh, also $\frac 1{\sqrt{2}}$. So $\varphi_1 = \frac{1}{\sqrt{2}}\psi_1 + \frac{1}{\sqrt{2}}\psi_2$. Find $\varphi_2$ similarly. $\endgroup$
    – user137731
    Oct 23, 2015 at 13:52
  • $\begingroup$ @Bye_World Yes the algebra is easy. But this is different than my question.. I'm asking (in question #1) when he needs to find $\varphi$ as a linear combination of $\psi$ ? Why not the opposite? They make it look so easy until I try this myself and gets stumped. $\endgroup$
    – Dor
    Oct 23, 2015 at 13:57
  • $\begingroup$ Sometimes the algebra is simple enough to do in your head (like the example your professor gave) and sometimes it's not. But it's always a very methodical process. If you're having trouble with change of basis, crack open your old linear algebra text and start working through the exercises. $\endgroup$
    – user137731
    Oct 23, 2015 at 14:00
  • $\begingroup$ @amd Why I did it backwards? Which phase in the process I had it wrong? The matrix $P_\varphi^\psi$ is defined to transfer a vector in base $\psi$ to a vector in base $\varphi$ - which is exactly what we're looking for. That matrix is also defined as follows: $ P_\varphi^\psi = \begin{pmatrix} [\psi_1]_\varphi & [\psi_2]_\varphi \end{pmatrix}$ $\endgroup$
    – Dor
    Oct 28, 2015 at 21:42
  • $\begingroup$ @amd please refer to my last comment :-) $\endgroup$
    – Dor
    Oct 30, 2015 at 2:48

2 Answers 2


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.

  • $\begingroup$ Indeed perhaps the instructor thought it doesn't matter whether to find $\varphi_i$ as a linear combination of $\psi_i$ or otherwise, because that the matrix is a Unitary operator. I understand now, thank you! $\endgroup$
    – Dor
    Nov 6, 2015 at 10:23

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

  • $\begingroup$ In the middle row you actually wrote $M_E^\phi = \pmatrix{[\varphi_1]_E & [\varphi_2]_E}^{-1} = \pmatrix{[e_1]_\varphi & [e_2]_\varphi}$ which IMHO should be wrong :| $\endgroup$
    – Dor
    Oct 28, 2015 at 22:03
  • $\begingroup$ Good catch (although you don’t mention the same error for the other matrix). I had subscripts and superscripts swapped, which I’ve fixed, but the matrices themselves were correct all along. We’re given $[\varphi_1]_E$ and $[\varphi_2]_E$, which lets us build $M_E^\varphi$ directly just as we did $M_E^\psi$, but we want $M_\varphi^E=(M_E^\varphi)^{-1}$. $\endgroup$
    – amd
    Oct 29, 2015 at 0:08
  • $\begingroup$ That is exactly what I did, but I was told by the instructor that I mixed the position of the base items. I don't understand the bug in my solution :s $\endgroup$
    – Dor
    Oct 30, 2015 at 2:50
  • $\begingroup$ Perhaps he meant that the standard basis should’ve been $\{x,1\}$ instead of $\{1,x\}$, as both of us took it to be. That doesn’t make any difference to the final answer as long as you’re consistent, though, since any permutation of $E$ gets undone when you multiply the two matrices together. You can see that yourself by exchanging the rows of $M_E^\psi$ and $M_\varphi^E$. $\endgroup$
    – amd
    Nov 4, 2015 at 19:36

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