### Archive

Posts Tagged ‘linear’

## Max Dimension for a linear space of singular matrices

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?

Categories: Higher Algebra, Linear Algebra Tags:

## Hahn-Banach Theorem (real version)

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

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

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

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…

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## Disjoint convex sets

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 $\delta_{ij}=\begin{cases} 0, & i \neq j \\ 1, & i=j \end{cases}$.