# Extending a bounded linear operator of finite rank

Let $X$ and $Y$ be normed spaces and let $W$ be a subspace of $X$. Assume that $T$ is a bounded linear operator from $W$ to $Y$, that is of finite rank. Show that $T$ can be extended to a bounded linear operator $T'$ from $X$ to $Y$ such that $T'(X) = T(W)$.

I think the Hahn-Banach extension theorem is needed somewhere, but since that theorem deals with extensions of functionals, i have no idea where to go..

Hint: Say $b_1,\dots,b_n$ is a basis for the finite-dimensional space $T(W)$. Then for every $w\in W$ there is a unique expansion $$Tw=\sum_{j=1}^n a_jb_j.$$Say $$a_j=\Lambda_jw\quad(w\in W).$$Since every linear functional on a finite-dimensional space is bounded, there exist $c$ so that $$|\Lambda_j w|\le c||Tw||.$$
• So $a_j = \Lambda_jw$. This is a bounded linear function to $\mathbb{F}$, so there exists an extension $\Lambda'_j : X \to \mathbb{F}$ by Hahn-Banach for every $j$. Then define $$T' \ : \ X \to T(W) \ : \ x \mapsto \sum_{j=1}^n\Lambda'_j(x)v_j.$$ This is a bounded linear operator from $X$ to $Y$, with $T'(X) = T(W)$. – ronalddb89 May 27 '16 at 16:12
• Right. Note that $\Lambda_j$ is bounded by one of the inequalities in my answer, plus the fact that $T$ is bounded... – David C. Ullrich May 27 '16 at 16:22