# about djoint operator

I'm trying to prove that if $X,Y$ are normed spaces and $T:X\to Y$ is linear operator (assume bounded, if needed), then $T^*$ linear.

What i've been trying so far is taking $Tx_1,...,Tx_n$ a basis to $ImT$ and try to construct linear functionals $f_1,...f_n\in X^*$ corresponding to $x_1,...,x_n$ which means $f_i(x_j)=\delta_{ij}$ or $|f_i(x_i)|=\|x_i\|$, etc... Using some consequences of hahn-banach, and similar theorems..

Thanks for your help

• Can you do the case of rank $1$? And then can you do the general case using that? – GEdgar Jan 4 '15 at 1:32
• @AlwaysLearning: You do not need to use $T,T^{*}$ are compact. In fact you can show any compact operator in Hilbert space is a limit of finite rank operator. It is easier to use the fact $T$ is finite rank and you do not need to appeal to big theorems. The fact $|T|=|T^{*}|$ should be automatic once you finish the proof. – Bombyx mori Jan 4 '15 at 2:09
• "P.S" should be a separate question. – user147263 Jan 4 '15 at 2:37
• It's already been asked. I will go compile an answer there. – user147263 Jan 4 '15 at 2:41
• Review the definition of adjoint (between Hilbert spaces) to see why $T$ satisfies the definition of the adjoint of $T^*$. – user147263 Jan 4 '15 at 2:57

Without working with a basis explicitly, one can argue as follows. Let $V=T(X)$. Let $R:Y^*\to V^*$ be the restriction operator, namely $R\phi = \phi_{|V}$. The range of $R$ is finite-dimensional, since $V^*$ is finite-dimensional. By definition of $T^*$, we have $T^*\phi = \phi\circ T = (R\phi)\circ T$. Thus, $T^*$ is the composition of finite-rank operator $R$ with another linear operator.