Explanation for Eigenvalues or Characteristic values of Projection Operator Question: I need to give what are possible eigenvalues/characteristic value of projection operator with an explanation.
My Attempt: 
Let E be any projection on vector space V. Assume R be the range of E and N be null space of E

My Doubt:


*

*I have got c=1 and c=0;

*there is and in between by the condition of independence

*so how does this imply c is either 1 or 0
Please help me in understanding how the above equation gives that c has possible eigenvalue either 0 or 1.
 A: Ah no, you are forgetting something.
You wrote that $\alpha \in R, \beta \in N$ so they must be independent. This is not true unless both are non-zero vectors!
So if $\alpha  = 0$, then of course $\alpha \in R$ but $\alpha$ won't be independent of $\beta$, even if $\beta \in N$.
Therefore, while the equation $(1-c)\alpha + c\beta = 0$ is correct, you need to now make exceptions for potentially one of $\alpha,\beta$ being zero.
So if $\alpha = \beta = 0$ then $v = 0$ : but $v$ is an eigenvector, so by definition $v \neq 0$.
If $\alpha = 0$ then $\beta \neq 0$ so the equation becomes $c \beta = 0$ which gives $c = 0$.
If $\beta = 0$ then $\alpha \neq 0$ so the equation becomes $(1-c)\alpha = 0$ which gives $c = 1$.
Both $\alpha , \beta \neq 0$ is not possible, since this would lead to $c=0$ and $c=1$  which is not possible.
Thus, the accurate conclusion would be the following : given a projection operator $E$ with decomposition $R + N$, any eigenvector $v$ either :


*

*belongs to $R$ with eigenvalue $1$ , or

*belongs to $N$ with eigenvalue $0$.
