Author Archive | yuliy

solutions for problems on vector analysis and differential forms

Solutions

Exercise:
Assume that 4 vectors in $$\mathbb{R}^3$$ satisfy $$A+B+C+D$$=0.
Simplify
$$A\times B-B\times C+C \times D-D\times A$$

Solution: By the anti-symmetry $$-B\times C = C\times B$$
and $$-D\times A = A\times D$$ we can rewrite the expression as

2.28 operators in Hermitian spaces and their spectra.

$$\def\Real{\mathbb{R}} \def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}} \def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}} \def\eye{\left(\begin{array}{cc}1&amp;0\\0&amp;1\end{array}\right)} \def\bra#1{\langle #1|}\def\ket#1{|#1\rangle}\def\j{\mathbf{j}}\def\dim{\mathrm{dim}} \def\ker{\mathbf{ker}}\def\im{\mathbf{im}} \def\tr{\mathrm{tr\,}} \def\braket#1#2{\langle #1|#2\rangle}$$

1.
Given a bilinear form $$Q(\cdot,\cdot)$$, or, equivalently, a mapping $$Q:U\to U^*$$, one can bake easily new bilinear forms from linear operators $$U\to U$$: just take $$Q_A(u,v):=Q(u,Av)$$.

This opens …

midterm

problems and solutions here.…

solutions to 2.14

Exercise:
Let $$V$$ be the space of real polynomial functions of degree at most $$3$$. Consider the quadratic form $$q_1:V \to \mathbb{R}$$ given by $$q_1(f):=|f(-1)|^2+|f(0)|^2+|f(1)|^2.$$ Is this form positive definite?
Consider another form $$q:V\to \mathbb{R}$$ given by  …

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}} \def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}} \def\eye{\left(\begin{array}{cc}1&amp;0\\0&amp;1\end{array}\right)} \def\bra#1{\langle #1|}\def\ket#1{|#1\rangle}\def\j{\mathbf{j}}\def\dim{\mathrm{dim}} \def\ker{\mathbf{ker}}\def\im{\mathbf{im}} \def\tr{\mathrm{tr\,}} \def\braket#1#2{\langle #1|#2\rangle}$$
1. Bilinear forms are functions $$Q:U\times U\to k$$ that depend on each of the arguments linearly.