Linear functionals

**math.dj** · December 20th 2009, 02:44 AM

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

**tonio** · December 20th 2009, 03:09 AM

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:

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

Tonio

Thread: Linear functionals

LinkBack

Thread Tools

Display

Linear functionals

Similar Math Help Forum Discussions

Linear Functionals

Bdd linear functionals on L-inf

multiplicative linear functionals

Linear functionals

dot products and linear functionals

Search Tags