1
$\begingroup$

$\newcommand{\qr}[1]{|#1\rangle}$

Exercise 2.2: (Matrix representations: example) Suppose $V$ is a vector space with basis vectors $\qr0$ and $\qr1$, and $A$ is a linear operator from $V$ to $V$ such that $A\qr{0}$ = $\qr1$ and $A\qr1 = \qr0$. Give a matrix representation for $A$, with respect to the input basis $\qr0,\qr1$, and the output basis $\qr0,\qr1$. Find input and output bases which give rise to a different matrix representation of $A$.

I had never heard of the terms "input basis" and "output basis". Nevertheless, the first part of the exercise is easy enough and I can see that a matrix representing the linear transform is just an $2x2$ identity matrix with the rows swapped. But how can I find input and output bases which give rise to a different matrix representation of $A$? Can you show me with a concrete example?

$\endgroup$
1
$\begingroup$

A matrix is a linear map between two vector spaces. Given a linear map, $\phi:\mathbb R^n\to\mathbb R^m$, there are multiple ways to represent it as an $n\times m$ matrix depending on what basis you use. You can read more about this here, if you want to confirm that you understand this. When it says “unit basis” and “output basis” its talking about the basis for the vector space that the (input/output) is coming from.

You’re correct that the answer with respect to the given basis is the transpose of the identity matrix. Suppose now you were working with $|0\rangle$ and $|\frac{1}{2}\rangle$ for both the input and the output bases instead. Do you see how the matrix would change? Hint: what’s the relationships between the old bases and the new bases.

——-

Let’s say you have a linear transformation, $T$. We can write it as a matrix, $A=[T]_\beta^\gamma$ where $\beta$ is the input basis and $\gamma$ is the output basis. To find $A$, we use the following formula:

$$A=[T]_\beta^\gamma=\big[[T(v_1)]_\gamma\;[T(v_2]_\gamma\;\ldots\;[T(v_k)]_\gamma\big]$$

Here $[T(v_i)]_\gamma$ is the matrix representation in the basis $\gamma$ of the vector that is the linear transformation $T$ applied to the $i^{th}$ basis vector of $\beta$. Therefore $[T(v_i)]_\gamma=\varphi_\gamma(T(v_i))$ where $\varphi_\gamma$ is the transformation that is a change of basis into basis $\gamma$.

So let’s do an example. Say that we want to keep the input basis the same but change the output basis to $\gamma=\{(0,1),(2,0)\}$. Your problem doesn’t name the linear operator (only the matrix) so let’s say that the operator is called $T$ and it’s representation in the standard basis is $A$. We wish to find $A’$, the representation in our new basis. From our formula, we know that the columns of $A’$ are $[T(0,1)]_\gamma$ and $[T(1,0)]_\gamma$. We know that $T$ switches the order, so the columns are $w_1=(1,0)$ and $w_2=(0,1)$. But those vectors are written with respect to the standard basis, not with respect to $\gamma$! Luckily we can easily write them with respect to gamma by seeing that $(0,1)=0\cdot w_1+1\cdot w_2$ and $(2,0)=2\cdot w_1+0\cdot w_2$. Therefore with respect to basis $\gamma$ we have that the columns are $(0,1)$ and $(2,0)$, so our matrix is

$$A’=\begin{bmatrix}0&2\\1&0\end{bmatrix}$$

$\endgroup$
  • $\begingroup$ I believe you meant to write $|0\rangle$ and $|\frac{1}{2}\rangle$ as my new basis? Gonna work on this and get back to you. (I can be very slow at math!) $\endgroup$ – R. Chopin Jan 15 '18 at 13:10
  • $\begingroup$ @R.Chopin Indeed I did! Lemme know if you want a hint. $\endgroup$ – Stella Biderman Jan 15 '18 at 13:43
  • $\begingroup$ $\newcommand{\qr}[1]{|#1\rangle}$Hi, @Stella. My question is now this. If $\qr0$ and $\qr{\frac{1}{2}}$ are now my basis, what am I to expect from the transformation $A$? It maps $\qr0$ to $\qr1$ and $\qr1$ to $\qr0$. I don't know to what A maps $\qr{\frac{1}{2}}$. If I did know, then I get a matrix in the same way I got the first matrix. More than a hint, I'd like a little linear algebra culture. :-) I don't seem to be on the same page as you and the exercise. $\endgroup$ – R. Chopin Jan 15 '18 at 17:26
  • $\begingroup$ @R.Chopin The function is linear, so $\frac{1}{2}\phi(x)=\phi(\frac{1}{2}x)$ $\endgroup$ – Stella Biderman Jan 15 '18 at 17:48
  • $\begingroup$ $\newcommand{\qr}[1]{|#1\rangle}$ @Stella, what is meant by $\qr{\frac{1}{2}}$? I know $\qr0$ is the column vector $(1, 0)$ and that $\qr1$ is the column vector $(0, 1)$. So I think $\qr{\frac{1}{2}}$ must equal $(0, \frac{1}{2})$ because maybe $\qr{\frac{1}{2}} = \frac{1}{2}\qr{1} = (0, \frac{1}{2}).$ Does that make sense? $\endgroup$ – R. Chopin Jan 17 '18 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.