# september 4

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}} \def\Fun{\mathbf{Fun}} \def\e{\mathbf{e}} \def\f{\mathbf{f}}$$

#### Linear operators

Their range=image, null-space=kernel (both linear subspaces).

Rank=dimension of the image (if finite-dimensional).

Important: dimension of the kernel + rank=dimension of the domain…

##### Example:

$$V=\Fun(S,\Field)$$, where, as before $$S=\{0,1,2,3,4,5\}\subset\Comp$$. Let $$Z$$ is the operator of multiplication by $$z$$. What is the kernel? Image? What about differentiation?

#### Matrices

A linear operator
$A:V\to W,$
and bases $$\f,\e$$ in $$V,W$$ yield the matrix of linear operator $$A$$:if
$A(f_l)=\sum_l A(k,l)e_k,$
then the matrix is
$A= \left( \begin{array}{ccc} A(1,1)&\ldots &A(1,n)\\ \vdots &\ddots &\vdots \\ A(m,1)& \ldots&A(m,n)\\ \end{array} \right).$

What happens if one changes the bases in $$V$$ and $$W$$, replacing $$\e’\mapsto \e$$ and $$\f\mapsto \f’$$? Nothing unexpected.

What happens if $$V=W$$, $$\e’=\f’$$?
Similarity transformation: if in the basis $$e$$ of $$V$$ the matrix of a linear operator is $$A$$, and $$P$$ is the matrix of basis change from $$\f$$ to $$\e$$ (i.e. $$\e_k=\sum_lP(k,l)\f_l$$), then the matrix of the same transformation in the basis $$\f$$ is
$P^{-1}AP: (AP)_{kn}=\sum_l (P^{-1})(k,l)A(l,m)P(m,n).$

Similarity transformations do not change the operator, but its matrix. Matrix can be made simpler, in some sense.

##### Example

multiplication by $$z$$ on $$S\subset\Comp$$ as above.
Hint:
$z(z-1)(z-2)(z-3)(z-4)(z-5)=-120 z + 274 z^2 – 225 z^3 + 85 z^4 – 15 z^5 + z^6.$

##### Example

Laplace operator: $$n$$ altruists stand in a circle, and at each step each of them splits what she has in two and gives to her neighbors, left and right.

Eigenvectors (of a linear operator $$A$$) are the elements, on which the action of the operator is especially easy:
$Av=\lambda v.$

Eigenvalues: roots of the characteristic polynomial
$\det(A-\lambda E).$
If all roots are simple (multiplicity one), then the linear operator is isomorphic to multiplication by $$z$$ on $$\Fun(\sigma(A))$$.

If not – Jordan normal form