A question on linear transformation(Projection) Let $V$ be a finite dimensional vector space over the real field $\mathbb{R}$. Let $I$ denote the identity transformation of $V$. For which $c\in \mathbb{R}$ is it true that for every projection $P$ the transformation $P+cI$ is invertible?
 A: As $P$ is a projection, its eigenvalues are $0$ and $1$. So the eigenvalues of $P+cI$ are $c$ and $1+c$. Then the statement is true for any $c$ not equal to $0$ or $-1$.
A: If $\dim V=0$, then any transformation (there is only one) is invertible, hence in that case any $c$ is allowed. Therefore, assume $\dim V>0$ from now on.
A projection is by definition a transformation $P$ with $P^2=P$.
If $c=0$, then $P=0$ is a projection and if $c=-1$ then $P=I$ is a projection with $P+cI=0$, which is not invertible.
Now consider $c\notin\{0,-1\}$ and let $P$ be a projection.
Then $(P+cI)v=0$ implies $0=P(P+cI)v=PPv+cPv=(1+c)Pv$, hence  $Pv=0$ because $1+c\ne0$. But then also $cv=Pv+cv=(P+cI)v=0$ and because of $c\ne 0$ finally $v=0$. We conclude that $P+cI$ has trivial kernel if $c\notin\{0,-1\}$. As $\dim V$ is finite, this implies that $P+cI$ is invertible.
In summary: If $\dim V=0$, then the statement is true for all $c\in\mathbb R$. If $\dim V>0$, the statement is true exactly for $c\notin\{0,-1\}$.
