Is this in general true (about projection matrices)?

[td21]

$$A$$ is a hermitian matrix with eigenvalues +1 and -1. Let $$\left|+\right>$$ and $$\left|-\right>$$ be the eigenvector of $$A$$ with respect to eigenvalue +1 and eigenvalue -1 respectively.
Therefore, $$P_{+} = \left|+\right>\left<+\right|$$ is the projection matrix with respect to eigenvalue +1. $$P_{-} = \left|-\right>\left<-\right|$$ is the projection matrix with respect to eigenvalue -1.

We all know that $$A = P_{+} + (-1)P_{-}$$. But is $$I = P_{+} + P_{-}$$ true? $$I$$ is the identity matrix.

 

[DEvens]

It is the identity matrix in context of the vector space spanned by $\left|+\right>$ and $\left|-\right>$.

Why do I weasel? Because you can probably find vectors that are orthogonal to $\left|+\right>$ and $\left|-\right>$, and it would be surprising if you applied the identity matrix to a vector and got zero. So $P_+ + P_-$ is only the identity matrix with respect to the appropriate vector space.

How do you prove it? You apply $I$ to an arbitrary combination of $x\left|+\right> + y\left|-\right>$ and see what you get. And you determine what is the square of $I$.

 

[Geofleur]

If the eigenvectors $| + \rangle$ and $| - \rangle$ form a basis of the underlying vector space, then yes. Because then, for any $| v \rangle$,

$(P_+ + P_-) | v \rangle = | + \rangle \langle + | v \rangle + | - \rangle \langle - | v \rangle = I | v \rangle$.

That is, if we can expand any ket in terms of the eigenvectors, then we can write the identity as a sum of the corresponding projection operators.

Oops, DEvens beat me to it!