Can we conclude that $\text {rank}\ P = k\ $? [closed]

Let $V$ be a normed vector space of dimension $n.$ Let $\{P_n \}_{n \geq 1}$ be a sequence of projections on $V$ with $\text {rank}\ P_n = k$ for all $n \geq 1.$ If $P_n \to P$ for some projection $P$ on $V$ in norm then $\text {rank}\ P = k.$

I can't conclude this. Any help would be appreciated.

Solution 1:

For a projection $P$ on $V$ the rank equals the trace: For $l=\mathrm{rank}\,P$ choose a basis $(v_1,..,v_l)$ of $\mathrm{im}(P)$ and a basis $(v_{l+1},...,v_n)$ of $\ker P$. Then the representation matrix of $P$ with respect to the basis $(v_1,...,v_n)$ is $A_P=\mathrm{diag}(1,...,1,0,...,0)$, with $l$ ones. So $\mathrm{rank}\,A=\mathrm{tr}(A)=l$.

Now just use that the trace is continous with respect to any norm on $\mathrm{End}(V)$.