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Posts Tagged ‘linear’

Max Dimension for a linear space of singular matrices

September 27, 2011 Leave a comment

Denote by V a vector space of singular (\det =0) n \times n matrices. What is the maximal dimension of such a space.

Denote by W a vector space of n \times n matrices with rank smaller or equal to p < n. What is the maximal dimension of such a space?

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Categories: Higher Algebra, Linear Algebra Tags:

Hahn-Banach Theorem (real version)

January 18, 2010 Leave a comment

Suppose X is a vector space over \mathbb{R}, p:\mathbb{X} \to \mathbb{R} has the following properties: p(\lambda x)=\lambda p(x),\ \forall x \in X, \lambda \in\mathbb{R}_+ and p(x+y)\leq p(x)+p(y),\ \forall x,y \in X.
Let X_0 be a subspace of X and u: X_0\to \mathbb{R} a linear functional such that u(x) \leq p(x),\ \forall x \in X_0.
Then we can find f:X \to \mathbb{R} a linear functional such that f| _{X_0} =u and f(x) \leq u(x),\ \forall x \in X. Read more…

Categories: Uncategorized Tags: , ,

Fuglede & Putnam Theorem

January 15, 2010 Leave a comment

Fuglede’s Theorem Let T,N be bounded linear operators on a Hilbert space X with N being normal ( NN^*=N^*N ). Then TN=NT implies TN^*=N^*T.
Putnam’s Theorem Let T,N,M be bounded linear operators on a Hilbert space X with N,\ M being normal. Then MT=TN implies M^*T=TN^*.

Categories: Functional Analysis Tags: ,

Fixed point for an operator

January 12, 2010 Leave a comment

Suppose X is a Hilbert space and T is a linear operator on X with \| T\| \leq 1. Prove that Tx=x if and only if T^* x =x.
Solution: Using the Cauchy-Buniakovski inequality, we get
\| x \| ^2 = \langle  x,  x \rangle  =  \langle Tx,x \rangle= \langle x,T^*x \rangle \leq \|x \| \| T^* x\| \leq \|x\|^2. Since this implies equality in the C-B inequality, we must have T^* x= \lambda x,\ \lambda\geq 0. We easily find that \lambda=1. The converse is equivalent to the implication above.

Impossible relation

January 11, 2010 Leave a comment

Prove that we cannot find any two linear continuous maps f,g : X \to X such that (f\circ g -g\circ f)x=x,\ \forall x \in X. Read more…

Disjoint convex sets

October 30, 2009 Leave a comment

Prove that for any disjoint non-void convex sets A,B from a linear space X we can find C,D \subset X convex and disjoint, such that A\subset C,\ B\subset D and C\cup D=X.
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Linear maximal subspaces & Linear functionals

October 26, 2009 1 comment

Take X to be a vector space over the field \Gamma.
1. Let f:X \to \Gamma be a linear functional, which is not identically zero. Then it’s kernel, \ker f=\{x \in X : f(x)=0\}, is a linear maximal subspace of X.

2. Conversely, given a maximal subspace X_0 of X, there exists a linear functional f such that \ker f=X_0.

3. As a consequence of the above prove that if we have two linear functionals (different from zero identity) f_1,f_2 such that \ker f_1=\ker f_2 then there exists \lambda \in \Gamma such that f_1=\lambda f_2.

As application to the above solve the following:
4. Given f,g_1,...,g_n linear functionals on X such that \ker f \supset \bigcap\limits_{i=1}^n \ker g_i prove that we can find scalars a_1,...,a_n such that f(x)=a_1g_1(x)+...+a_ng_n(x),\ \forall x \in X.

Prove that if f_i,\ i=1..n are linearly independent linear functionals on X then there exist elements x_1,...,x_n \in X such that f_i(x_j)=\delta_{ij}, where \delta_{ij}=\begin{cases} 0, & i \neq j \\ 1, & i=j \end{cases}.

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