let $T$ be a linear operator from a Banach space $X$ to Banach space $Y$.and $X=ker(T)\oplus M_1$ where $M_1$ is closed subspace of $X$.let $M$ be a closed subspce of $X$ then I want to prove that there exist a finite dimensional subspace $M_0$ such that $M=M \cap M_1 +M_0$


If $\dim (Ker(T))= +\infty$ the claim is false. Indeed putting $M = Ker(T)$ yeld $M\cap M_1 = \emptyset$ thus $M_0 = M$ must have infinite dimension.

Suppose that $\dim (Ker(T))<+\infty$.

From linear algebra $M = M\cap M_1 \oplus M_0$ for some $M_0$ (here $\oplus$ is only algebraic, it do not implies that $M\cap M_1$ is complemented as a Banach space). Consider the map $M\to X \to \frac X M_1$ where the first map is the inclusion and the second one is the projection. The kernel of this map is $M\cap M_1$ and the image is $M/M_1$. It follows from elementary algebra that $M/(M\cap M_1)\approx M/M_1$ which is finite dimensional since $M/ M_1 \lhd X/M_1 \approx Ker(T)$ and $\dim (Ker(T))<+\infty$. Thus $M_0\approx M/(M\cap M_1)$ is finite dimensional.


It is probably better to post the original problem.

If $X$ is separable Hilbert with $X=H_1\oplus H_2$ and $H_1, H_2$ both infinite-dimensional, and $T$ is the orthogonal projection onto $H_2$ then $M_1=H_2$.

Take $M = H_1$. Then $M\cap M_1=\{0\}$. Therefore $M_0$ would have to be $H_2$.

In your case, do you know something else about $T$?

  • $\begingroup$ In my question I am considering problem in banach space but u are considering hilbert space.@olivia $\endgroup$ – Manu Rohilla Jul 9 '17 at 4:29
  • $\begingroup$ @ManuRohilla Hilbert spaces are, in particular, Banach spaces. $\endgroup$ – Olivia Jul 9 '17 at 11:56

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