multivariate-statistics/content/chapters/part1/2.tex

\chapter{Elements of Linear Algebra}
\label{ch:elements-of-linear-algebra}

\begin{remark}[vector]
    Let $u$ a vector, we will use interchangeably the following notations: $u$ and $\vec{u}$
\end{remark}

Let $u = \begin{pmatrix}
        u_1    \\
        \vdots \\
        u_n
    \end{pmatrix}$ and $v = \begin{pmatrix}
        v_1    \\
        \vdots \\
        v_n
    \end{pmatrix}$

\begin{definition}[Scalar Product (Dot Product)]
    \begin{align*}
        \scalar{u, v} & = \begin{pmatrix}
            u_1, \ldots, u_v
        \end{pmatrix}
        \begin{pmatrix}
            v_1    \\
            \vdots \\
            v_n
        \end{pmatrix} \\
    & = u_1 v_1 + u_2 v_2 + \ldots + u_n v_n
    \end{align*}

    We may use $\scalar{u, v}$ or $u \cdot v$ notations.
\end{definition}
\paragraph{Dot product properties}
\begin{itemize}
    \item $\scalar{u, v} = \scalar{v, u}$
    \item $\scalar{(u+v), w} = \scalar{u, w} + \scalar{v, w}$
    \item $\scalar{u, v}$
    \item $\scalar{\vec{u}, \vec{v}} = \norm{\vec{u}} \times \norm{\vec{v}} \times \cos(\widehat{\vec{u}, \vec{v}})$
\end{itemize}

\begin{definition}[Norm]
    Length of the vector.
    \[
        \norm{u} = \sqrt{\scalar{u, v}}
    \]

    $\norm{u, v} > 0$
\end{definition}

\begin{definition}[Distance]
    \[
        dist(u, v) = \norm{u-v}
    \]
\end{definition}

\begin{definition}[Orthogonality]

\end{definition}

\begin{remark}
    \[
        (dist(u, v))^2 = \norm{u - v}^2,
    \] and
    \[
        \scalar{v-u, v-u}
    \]
\end{remark}

\begin{figure}
    \centering
    \includestandalone{figures/schemes/vector_orthogonality}
    \caption{Scalar product of two orthogonal vectors.}
    \label{fig:scheme-orthogonal-scalar-product}
\end{figure}

\begin{align*}
    \scalar{v-u, v-u} & = \scalar{v, v} + \scalar{u, u} - 2 \scalar{u, v} \\
                    & = \norm{v}^2 + \norm{u}^2                         \\
                    & = -2 \scalar{u, v}
\end{align*}

\begin{align*}
    \norm{u - v}^2 & = \norm{u}^2 + \norm{v}^2 - 2 \scalar{u,v} \\
    \norm{u + v}^2 & = \norm{u}^2 + \norm{v}^2 + 2 \scalar{u,v}
\end{align*}

\begin{proposition}[Scalar product of orthogonal vectors]
\[
    u \perp v \Leftrightarrow \scalar{u, v} = 0
\]
\end{proposition}

\begin{proof}[Indeed]
    $\norm{u-v}^2 = \norm{u+v}^2$, as illustrated in \autoref{fig:scheme-orthogonal-scalar-product}.
    \begin{align*}
        \Leftrightarrow & -2 \scalar{u, v} = 2 \scalar{u, v} \\
        \Leftrightarrow & 4 \scalar{u, v} = 0                \\
        \Leftrightarrow & \scalar{u, v} = 0
    \end{align*}
\end{proof}

\begin{theorem}{Pythagorean theorem}
    If $u \perp v$, then $\norm{u+v}^2 = \norm{u}^2 + \norm{v}^2$ .
\end{theorem}

\begin{definition}[Orthogonal Projection]

\end{definition}
Let $y = \begin{pmatrix}
        y_1 \\
        .   \\
        y_n
    \end{pmatrix} \in \RR[n]$ and $w$ a subspace of $\RR[n]$
$\mathcal{Y}$ can be written as the orthogonal projection of $y$ on $w$:
\[
    \mathcal{Y} = proj^w(y) + z,
\]
where
\[
    \begin{cases}
        z \in w^\perp \\
        proj^w(y) \in w
    \end{cases}
\]
There is only one vector $\mathcal{Y}$ that ?

The scalar product between $z$ and (?) is zero.

\begin{property}
    $proj^w(y)$ is the closest vector to $y$ that belongs to $w$.
\end{property}

\begin{definition}[Matrix]
    A matrix is an application, that is, a function that transform a thing into another, it is a linear function.
\end{definition}

\begin{example}[Matrix application]

    Let $A$ be a matrix:
    \[
        A = \begin{pmatrix}
            a & b \\
            c & d
        \end{pmatrix}
    \] and
    \[
        x = \begin{pmatrix}
            x_1 \\
            x_2
        \end{pmatrix}
    \]
    Then,
    \begin{align*}
        Ax & = \begin{pmatrix}
                a & b \\
                c & d
            \end{pmatrix}
        \begin{pmatrix}
            x_1 \\
            x_2
        \end{pmatrix}        \\
        & = \begin{pmatrix}
                a x_1 + b x_2 \\
                c x_1 + d x_2
            \end{pmatrix}
    \end{align*}

    Similarly,
    \begin{align*}
        \begin{pmatrix}
            a & b & c & d \\
            e & f & g & h \\
            i & j & k & l
        \end{pmatrix}
        \begin{pmatrix}
            x_1 \\
            x_2 \\
            x_3 \\
            x_4
        \end{pmatrix}
        & = \begin{pmatrix}
                a x_1 + b x_2 + c x_3 \ldots
            \end{pmatrix}
    \end{align*}
\end{example}

The number of columns has to be the same as the dimension of the vector to which the matrix is applied.

\begin{definition}[Tranpose of a Matrix]
    Let $A = \begin{pmatrix}
        a & b \\
        c & d
    \end{pmatrix}$, then $A^\T =  \begin{pmatrix}
        a & c \\
        b & d
    \end{pmatrix}$
\end{definition}

\begin{figure}
    \centering
    \includestandalone{figures/schemes/coordinates_systems}
    \caption{Coordinate systems}
\end{figure}