1. ## Linear functionals

1)Let T:V-->V be any linear map and V* be its dual.Let W be a k-dimensional subspace of V.(V is finite dimensional).Then prove that there exits a Linearly independent subset {f1,f2,...f(n-k)} subset of V* such that W=inetsection of Kernel(fi), where i=1,2,...,n-k.

2)Let T* denote the adjoint of T. If T is surjective then show that T* is injective

2. Originally Posted by math.dj
1)Let T:V-->V be any linear map and V* be its dual.

You mean: $\displaystyle V^{*}$ is the dual space of $\displaystyle V$.

Let W be a k-dimensional subspace of V.(V is finite dimensional).Then prove that there exits a Linearly independent subset {f1,f2,...f(n-k)} subset of V* such that W=inetsection of Kernel(fi), where i=1,2,...,n-k.

Let $\displaystyle \{w_1,\ldots,w_k\}$ be any basis of $\displaystyle W$, and let us extend it to a basis $\displaystyle \{w_1,\ldots,w_k,w_{k+1},\ldots,w_n\}$ of $\displaystyle V$ . Let $\displaystyle \{f_1,\ldots,f_n\}$ be the dual basis in $\displaystyle V^{*}$ of the above basis of $\displaystyle V$ , i.e. $\displaystyle f_iw_j=\delta_{ij}=\left\{\begin{array}{cc}1&\mbox {, if }\, i=j\\0&\mbox{, if }\, i\neq j\end{array}\right.$ .

Well, now prove that $\displaystyle \mbox{ }\,W=\bigcap\limits_{j=k+1}^nKer(f_j)$

2)Let T* denote the adjoint of T. If T is surjective then show that T* is injective
Ok, so we ALSO have an inner product on $\displaystyle V$. Assume $\displaystyle T^{*}x=T^{*}y$ and let be $\displaystyle z\in V\,\,\,s.t.\,\,\,Tz=x-y$ , so:

$\displaystyle <x-y,x-y>=<Tz,x-y>=<z,T^{*}(x-y)>=<z,0> =$ ...

Tonio