# Linear functionals

• Dec 20th 2009, 03:44 AM
math.dj
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
• Dec 20th 2009, 04:09 AM
tonio
Quote:

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

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

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

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

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 $V$. Assume $T^{*}x=T^{*}y$ and let be $z\in V\,\,\,s.t.\,\,\,Tz=x-y$ , so:

$=== =$ ...

Tonio