\(\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&0\\0&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}}

\)

**1.** Linear mappings, or linear operators are just linear functions taking values in a linear space:

\[

A:U\to V,\ \mathrm{where\ both\ } U\ \mathrm{and\ } V\ \mathrm{ are\ linear\ spaces.}

\]

The same properties (with respect to the additivity and multiplication by a constant) are assumed:

\[

A(\lambda_1u_1+\lambda_2u_2)=\lambda_1A(u_1)+\lambda_2A(u_2), \mathrm{etc.}

\]

**2.** We will denote the space of linear operators from \(U\) to \(V\) as \(L(U,V)\). It is again a linear space (this is a routine exercise to verify).

If the bases \(e_j, j=1,\ldots,\dim U, f_i, i=1,\ldots,\dim V\) in both linear spaces \(U\) and \(V\) are fixed, then one can represent the operator as a matrix whose coefficient \(A_{ij}\) comes from expanding \(Ae_j\) in terms of \(f\)’s. Easiest mnemonic rule uses bra-ket representations of unity in \(U,V\):

\[

A=E_VAE_U=\sum_i\ket{f_i}\bra{f_i}A\sum_j\ket{e_j}\bra{e_j}=\sum_{ij}A_{ij}\ket{f_i}\bra{e_j}.

\]

If \(U=V\) is the same space (i.e. \(A\) is a self-mapping), one typically takes the same bases.

**3.** So, given the bases \(e_j, j=1,\ldots,\dim U, f_i, i=1,\ldots,\dim V\) one obtains a basis in \(L(U,V\) given by \(\ket{f_i}\bra{e_j}, j=1,\ldots,\dim U, i=1,\ldots,\dim V\) (this is, again, an easy theorem). Hence

\(\dim L(U,V)=\dim(U)\dim(V)\).

*Exercise*: Let \(U\) be the space of polynomials in \(x,y\) of degree at most 2. Find a basis for this space. Consider the operator

\[

AP=\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\right)P.

\]

Check that it maps \(U\) into \(U\). Find its matrix in your basis.

**4.** If one is given operators \(B:U\to V,A:V\to W\), one can define their composition \(AB\) (bizarrely, we first apply \(B\)…) – the matrix of this composition can be easily recovered from the matrices of \(A, B\).

**5.** Null-space and range of a linear operator are defined as follows:

\[

\ker A=\{u\in U: Au=0\}. \im A=\{v: \mathrm{for\ some\ } u\in U, v=Au\}.

\]

An easy check: both kernel and image of an operator \(A\in L(U,V)\) are linear subspaces (of \(U\) and \(V\), respectively).

Exercise: Let \(U=\{a_0z^4+a_1z^3+\ldots+a_4\}\) be the space of all polynomials in one variable of degree 4. Let \(A:U\to U\) is given by

\[

AP=Q \Leftrightarrow Q(z)=P(z)+P(-z).

\]

Find \(\ker A\) and \(\im A\).

One often says that \(im A\) is *isomorphic* to the factorspace \(U/\ker A\). One can see this as the definition of the factorspace.

**6.** There exists a fundamental relation

\[

\dim U=\dim\ker A+\dim\im A

\]

for any \(A\in L(U,V)\).

**7.** An important class of sub- and factorspaces of a linear space \(U\) comes from duality: if \(V\subset U\) is a linear subspace, then any linear function on \(U\) is also a linear function on \(V\). Hence we get automatically a mapping

\[

U^*\to V^*.

\]

The image is all of \(V^*\) (why?).

The kernel of this mapping consists of exactly the functionals that vanish on \(V\) (there is a special name for that, the annulator of \(V\), denoted as \(V^\perp\)). So,

\[

V^*=U^*/V^\perp.

\]

**8.** More generally, if we have a linear operator

\[

A:U\to V,

\]

we automatically obtain a linear operator

\[

A^*:V^*\to U^*,

\]

The matrix of \(A^*\) is the transposition of the matrix of \(A\) (if one uses the dual bases). In terms of bra-ket notations, it just boils down to multiplying from the left.

Generalizing the relation between the annulator and the image, we have

\[

(\im A)^*=V^*/(\ker A^*).

\]

In particular, \(\dim U-dim\ker A=\dim \im A=\dim(\im A)^*=\dim V^*-\dim\ker A^*\).

Exercise: Let \(U\) be the space of polynomials in one variable of degree 4. Consider the subspace of polynomials divisible by \(z^2-1\). Describe its annulator.

**9.** Rank of an operator (or of a matrix) is the dimension of its image. Equivalently, it is the maximal number of linearly independent columns.

**10.** Finding preimages of linear operators is the same as solving systems of linear equations. Here we have a simple but important

*Theorem* (Fredholm alternative):

If \(\dim U=\dim V\), and \(A:U\to V\) is linear, then either both systems \(Ax=b, A^*y=c\) have unique solutions for any \(b\in V, c\in U^*\) or the null-spaces of both \(A\) and \(A^*\) are nontrivial (and have the same dimension, \(\dim\ker A=\dim\ker A^*>0\)).

One remarkable property of the linear mapping is that if \(A:U\to V\) is one-to-one, then the inverse mapping is also linear. It means that in this case there exists a linear operator \(B:V\to U\) such that \(BA=E_U\) (and, therefore, \(AB=E_V\)).

This leads immediately to determinants.

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