# How do you find the projection operator onto an eigenspace if you don't know the eigenvector?

I was working on exercise 2.60 of Nielsen-Chuang which is as follows:

Show that $\vec{v}\cdot\vec{\sigma}$ has eigenvalues $\pm 1$, and that the projectors onto the corresponding eigenspaces are given by $P_{\pm}=I\pm (\vec{v}\cdot\vec{\sigma})/2$.

I wrote $v=a\vec{x}+b\vec{y}+c\vec{z}$, so $\vec{v}\cdot\vec{\sigma}$ is a matrix with elements $c,a-bi,a+bi,-c$. I then was able to solve for the eigenvalues $\pm 1$ by finding $\lambda$ for which the determinant of $\vec{v}\cdot\vec{\sigma}-\lambda I$ is $0$. But I'm having more trouble with the second part of the problem.

Normally I would simply solve for the eigenvector for each eigenvalue, and use that to find the projection operator but whenever I try to solve for the eigenvector I get $0=0$. For example, for the eigenvalue of $1$ I get the following two equations: $$(c-1)x+(a-bi)y=0$$ $$(a+bi)x+(-c-1)y=0$$ and when I try to cancel the $y$ terms I get $(a^2+b^2+c^2-1)x=0$ which is just $0=0$.

That said, the form of the answer makes me think there's an easier way.

$I+(\vec{v}\cdot \vec{\sigma})$ is just $\vec{v}\cdot\vec{\sigma}-\lambda I$ for $\lambda=-1$. and $I-(\vec{v}\cdot \vec{\sigma})$ is the negative of $\vec{v}\cdot\vec{\sigma}-\lambda I$ for $\lambda=1$. And then there's a $\frac{1}{2}$ factor so the sum of the projection operators is $I$. Can anyone explain why this is true and how one can find the projection operators from scratch if you don't know the eigenvectors? (I can show these are projection operators but don't know how I would find them without the question explicitly telling me what they are.)

Also, can anyone could tell me how to solve the above system of equations?

• I will vote to close as this is a purely mathematical question. That said, there is a nearly trivial way of proving this assertion, because you already have the answer: a) You prove that $P_{\pm}$ are projections with $P_+P_-=0$. b) Have a look at $P_+-P_-$ and think about the uniqueness of the spectral decomposition. May 4, 2015 at 21:27
• Is there any easy way to find the projections from scratch without finding the eigenvectors first? I know how to verify the given operators are projections. May 4, 2015 at 23:25

There is indeed a way to construct the projection-operators, when you only know the operator itself and its eigenvalues. The derivation can be found in Julian Schwingers "Quantum Mechanics: Symbolism of Atomic Measurement" and leads to

$$P_j = \prod_{i\neq j} \frac{A-a_i}{a_j - a_i},$$

where the product goes over all distinct eigenvalues $a_i$ of $A$. It is quite easy to see, that it indeed fulfills

$$P_i |\psi_j\rangle = \delta_{ij}|\psi_j\rangle,$$ where no summation is implied.

It you apply this to your problem, you get the (correct) projectors

$$P_{+} = \frac{\vec v \cdot \vec \sigma -(-1)}{1-(-1)} = \frac{I+\vec v \cdot\vec \sigma}2\\ P_- = \frac{\vec v \cdot \vec \sigma - 1}{(-1)-1} = \frac{I-\vec v \cdot \vec \sigma}{2}$$

• One what page can one find the derivation? Sep 8, 2016 at 19:48
• @AlexProvost : Frobenius covariant. Oct 31, 2017 at 18:48

Eigenvectors are undetermined up to a scalar multiple. So for instance if c=1 then the first equation is already 0=0 (no work needed) and the second requires that y=0 which tells us that x can be anything whatsoever. This is fine, and correct, as $x=re^{i\theta}, y=0$ is a fine solution when c=1.

If $c\neq 1$ then none of the equations are already 0=0 so we can pick any one we want and use it to solve for x in terms of y or for y in terms of x. That solution will actually work in the other equation too.

$$(c-1)x+(a-bi)y=0,$$ $$(a+bi)x+(-c-1)y=0.$$
I can solve the second equation and note that $$x=\frac{a-bi}{a^2+b^2}(c+1)y=0$$ solves $$(a+bi)x+(-c-1)y=0.$$
But $$x=\frac{a-bi}{a^2+b^2}(c+1)y$$ also solves $$(c-1)x+(a-bi)y=0$$ because
$$(c-1)\frac{a-bi}{a^2+b^2}(c+1)y+(a-bi)y$$ equals
$$(a-bi)y\left(\frac{c^2-1}{a^2+b^2}+\frac{a^2+b^2}{a^2+b^2}\right)=0.$$
• If c=1 the solution is x=whatever, y=0. If c is not 1 then the solution is y=whatever, $x=(a-bi)(c+1)y/(a^2+b^2)$. Everything after the word "solves" is just explicit verification that the solution (already found) is indeed a solution. So getting 0 for the LHS is exactly what we want since the RHS is 0. Our equations are redundant, we don't want to use them all, and I'm just showing that for $c \neq 1$ we could use either equation. May 4, 2015 at 23:34