While reading Theoretical Minimum by Leonard Susskind, I came across the exercise 3.4 where he asked to find the eigenvalues and the eigenvectors of the matrix that represents the $\sigma_{n}$ component of the operator $\sigma$ where: $$\sigma_{n}=\vec\sigma.\hat n$$ Which leads to: $$\sigma_n=\begin{pmatrix}n_{z}&(n_{x}-i.n_{y})\\(n_{x}+in_{y})&-n_{z}\end{pmatrix}$$ Given that: $$n_{z}=\cos(\theta)$$ $$n_x=\sin(\theta) \cdot \cos(\phi)$$ $$n_y=\sin(\theta) \cdot \sin(\phi)$$ How can I approach the question as I am a little bit confused, and I searched online for the answer and found that we need to find the determinant of the $\sigma_n$ where is equals: $$\begin{pmatrix}\cos(\theta)-\lambda&\sin(\theta)\cdot\cos(\phi)-i\cdot \sin(\theta)\cdot\sin(\phi)\\ \sin(\theta)\cdot \cos(\phi)+i \cdot \sin(\theta)\cdot \sin(\phi)&-\cos(\theta)-\lambda\end{pmatrix}$$ How does finding the determinant of $\sigma_n$ help solve the question and where did the $\lambda$ come from in the first place?

  • $\begingroup$ Can you link the online answer you are talking about? And also, how familiar are you with linear algebra? So that people can answer your question accordingly. $\endgroup$
    – FDGod
    Dec 22, 2023 at 22:01
  • $\begingroup$ Here is where I got the answer: quantum-abc.de/Exercises%20complete1.pdf. My mathematical knowledge is mostly from high school. $\endgroup$
    – zizaaooo
    Dec 22, 2023 at 22:18

1 Answer 1


From your post and the comments therein, it seems that you would benefit in reading a little bit more about linear algebra. There are plenty of resources for that, including some excellent posts on the Mathematics StackExchange.

In this post, I will only answer your question about why finding this determinant allows to compute the eigenvalues of a matrix. However, it's quite hard to answer such a question with high school knowledge only, so I'll assume just a tad bit of knowledge about matrices (hopefully not much, but feel free to ask questions in the comments).

We will need several facts:

  1. If a matrix is not invertible, its determinant is nil. Reciprocally, it is invertible if its determinant isn't nil.
  2. If there is a non-zero vector $x$ such that $Mx$ is the zero vector, then $M$ isn't invertible. Reciprocally, if the only $x$ such that $Mx=0$ is $x=0$, then $M$ is invertible.
  3. If there is a non-zero vector $x$ such that $Mx=\lambda x$ for a certain $\lambda$, then $\lambda$ is an eigenvalue of $M$.

With that being said, let us consider an eigenvalue $\lambda$ of a matrix $M$. From Fact 3. we have: $$Mx=\lambda x$$ Which we can rewrite as: $$Mx=\lambda Ix$$ with $I$ being the identity matrix. This is equivalent to: $$(M-\lambda I)x=0$$ But now from Fact 2. we have that $M-\lambda I$ is not invertible, which means from Fact 1. that the determinant of $M-\lambda I$ is nil.

Thus, we have shown that if $\lambda$ is an eigenvalue of $M$, then the determinant of $M-\lambda I$ is nil.

Reciprocally, if $\lambda$ is not n eigenvalue of $M$, then we can show that $M-\lambda I$ is invertible. Suppose that: $$(M-\lambda I)x=0$$ Then it means that: $$Mx=\lambda x$$ But we have assumed that $\lambda$ is not an eigenvalue of $M$, so it necessarily means that $x=0$. Thus, from Fact 2. $M-\lambda I$ is invertible, which means from Fact 1. that its determinant is not nil.

All in all, we have shown that the eigenvalues of $M$ are exactly the solutions to the equation $$\text{det}(M-\lambda I)=0$$ Where we want to solve for $\lambda$.

If you want to learn more on this topic, you should read about the characteristic polynomial of a matrix.

In your case, you will find that the determinant is actually a degree 2 polynomial in $\lambda$. You should thus be able to quickly find the eigenvalues.

Once you know an eigenvalue $\lambda$, there are several ways to compute the eigenvectors. The simplest is simply to solve for $x$ in $Mx=\lambda x$. Such an $x$ isn't unique, but you're interested in linearly independent solutions here.

This might be a bit too much for an high-school level and I apologize for that. But to put it in a nutshell:

  • The determinant of $M-\lambda I$ is called the characteristic polynomial of $M$, and its roots are exactly the eigenvalues of $M$.
  • Using the definition of an eigenvalue, it is possible to find an eigenvector associated to a known eigenvalue.
  • You should definitely get yourself an introductory course on linear algebra if you're interested in that topic, you'll learn a lot of things!
  • 2
    $\begingroup$ As for learning linear algebra, I would highly recommend Greg Sanderson's Essence of Linear Algebra series to get an excellent intuitive understanding. If you are looking for something slightly advanced, then I would recommend Gilbert Strang's Linear Algebra Course at MIT. $\endgroup$
    – FDGod
    Dec 23, 2023 at 5:14
  • $\begingroup$ Thanks a lot, but I why did you make a point about $x$ being 0 if $\lambda$ is not an eigenvalue of $M$? $\endgroup$
    – zizaaooo
    Dec 24, 2023 at 8:49
  • $\begingroup$ @zizaaooo In this step, our goal is to show that if $\lambda$ is not an eigenvalue of $M$, then $M-\lambda I$ isn't invertible. In order to do this, we use the second part of Fact 2. : we show that the only $x$ satisfying $(M-\lambda I)x=0$ is $x=0$. $\endgroup$
    – Tristan Nemoz
    Dec 24, 2023 at 9:27
  • $\begingroup$ Do you mean that $M-\lambda$ is invertible because the only solution is $x=0$ if the $\lambda$ is not an eigenvalue of $M$? Because you wrote in the first line of your comment that $M-\lambda$ isn't invertible. I might be repeating what you just said but I want to make sure that I understood your answer. $\endgroup$
    – zizaaooo
    Dec 24, 2023 at 10:15
  • 1
    $\begingroup$ @zizaaooo Oops, you're right! Its determinant isn't nil, so $M-\lambda I$ is invertible indeed, sorry for that! $\endgroup$
    – Tristan Nemoz
    Dec 24, 2023 at 10:26

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