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% 19/10/2022 11AM

\subsection{Change of basis and equivalent
matrices}

Let $\beta : U \to V$, $\alpha : V \to W$ and
$\mathcal{A}, \mathcal{B}, \mathcal{C}$ bases of
$U, V, W$.
\[ \implies [\alpha \circ \beta]_{\mathcal{A},
\mathcal{C}} = [\alpha]_{\mathcal{B},
\mathcal{C}} [\beta]_{\mathcal{A}, \mathcal{B}} \]

\subsubsection*{Change basis}

Let $\alpha : V \to W$ and let $\mathcal{B},
\mathcal{B}'$ and $\mathcal{C}, \mathcal{C}'$ be
bases for $V$ and $W$.

\begin{flashcard}[change-basis-matrix]
\begin{definition*}
    The ``change of basis matrix'' from
    $\mathcal{B}'$ to $\mathcal{B}$ is
    \[ P = (p_{ij}) \]
    given by
    \[ P = \cloze{([v_1']_{\mathcal{B}} \cdots
    [v_n']_{\mathcal{B}}) =
    [\id]_{\mathcal{B}', \mathcal{B}}} \]
\end{definition*}
\end{flashcard}

\begin{lemma*}
    $[v]_{\mathcal{B}} = P[v]_{\mathcal{B}'}$
\end{lemma*}

\begin{proof}
    \begin{itemize}
        \item $[\alpha(v)]_{\mathcal{C}} =
            [\alpha]_{\mathcal{B}, \mathcal{C}}
            [v]_{\mathcal{B}}$
        \item $P = [\id]_{\mathcal{B}',
            \mathcal{B}}$
            \[ \implies [\id(v)]_{\mathcal{B}} =
            [\id]_{\mathcal{B}', \mathcal{B}}
            [v]_{\mathcal{B}'} \]
            \[ \implies [v]_{\mathcal{B}} =
            P[v]_{\mathcal{B}'} \]
    \end{itemize}
\end{proof}

\begin{remark*}
    $P$ is a $n \times n$ invertible matrix, and
    $P^{-1}$ is the change of basis matrix from
    $\mathcal{B}$ to $\mathcal{B}'$. \myskip
    Indeed
    \[ [\alpha \circ \beta]_{\mathcal{A},
    \mathcal{C}} = [\alpha]_{\mathcal{B},
    \mathcal{C}} [\beta]_{\mathcal{A},
    \mathcal{B}} \]
    \[ \implies [\id]_{\mathcal{B}, \mathcal{B}'}
    [\id]_{\mathcal{B}, \mathcal{B}'} =
    [\id]_{\mathcal{B}', \mathcal{B}'} \equiv I_n \]
    \[ \implies [\id]_{\mathcal{B}', \mathcal{B}}
    [\id]_{\mathcal{B}, \mathcal{B}'} =
    [\id]_{\mathcal{B}, \mathcal{B}} \equiv I_n \]
\end{remark*}

We changed $\mathcal{B}$ to $\mathcal{B}'$ in $V$.
We can also change basis to $\mathcal{C}$ to
$\mathcal{C}'$ in $W$.

\begin{flashcard}[change-basis-formula]
\begin{proposition*}
    $A = \cloze{[\alpha]_{\mathcal{B},
    \mathcal{C}}}$, $A'
    = \cloze{[\alpha]_{\mathcal{B}',
    \mathcal{C}'}}$, $P = \cloze{
    [\id]_{\mathcal{B}', \mathcal{B}}}$, $Q =
    \cloze{[\id]_{\mathcal{C}', \mathcal{C}}}$. Then
    \[ \cloze{A' = Q^{-1} A P} \]
\end{proposition*}
\end{flashcard}

\begin{proof}
    \[ [\alpha(v)]_{\mathcal{C}} =
    [\alpha]_{\mathcal{B}, \mathcal{C}}
    [v]_{\mathcal{B}} \]
    \[ [\alpha \circ \beta]_{\mathcal{A},
    \mathcal{C}} = [\alpha]_{\mathcal{B},
    \mathcal{C}} [\beta]_{\mathcal{A}, \mathcal{B}} \]
    \[ [v]_{\mathcal{B}} = P[v]_{\mathcal{B}'} \]
    \begin{itemize}
        \item \[ [\alpha(v)]_{\mathcal{C}} =
            Q[\alpha(v)]_{\mathcal{C}} =
            Q[\alpha]_{\mathcal{B}', \mathcal{C}'}
            = QA' [v]_{\mathcal{B}'} \]
        \item $[\alpha(v)]_{\mathcal{C}} =
            [\alpha]_{\mathcal{B}, \mathcal{C}}
            [v]_{\mathcal{B}} =
            AP[v]_{\mathcal{B}'}$.
    \end{itemize}
    So for all $v \in V$,
    \[ QA' [v]_{\mathcal{B}'} =
    AP[v]_{\mathcal{B}'} \]
    hence
    \[ QA' = AP \implies A' = Q^{-1}AP \qedhere \]
\end{proof}

\begin{flashcard}[equivalent-matrices]
\begin{definition*}[Equivalent matrices]
    Two matrices $A, A' \in \mathcal{M}_{m, n}(F)$
    are equivalent if\cloze{:
    \[ A' = Q^{-1} AP \]
    with $Q \in \mathcal{M}_{m, m}, P \in
    \mathcal{M}_{n, n}$, with both invertible.}
\end{definition*}
\end{flashcard}

\begin{remark*}
    This defines an equivalence relation on
    $\mathcal{M}_{m, n}(F)$.
    \begin{itemize}
        \item $A = I_m^{-1} A I_n$
        \item $A' = Q^{-1} AP \implies A =
            (Q^{-1})^{-1} A' P^{-1}$
        \item $A' = Q^{-1}AP$, $A'' = (Q')^{-1}
            A'P'$. Then
            \[ A'' = (QQ')^{-1} A (PP') \]
    \end{itemize}
\end{remark*}

\begin{flashcard}[every-matrix-equivalent-to-I-r-form]
\begin{proposition*}
    Let $V$, $W$ be vector spaces over $F$, with
    $\dim_F V = n$, $\dim_F W = m$. Let $\alpha : V
    \to W$ be a linear map. Then there exists
    $\mathcal{B}$ basis of $V$ and $\mathcal{C}$
    basis of $W$ such that
    \[ [\alpha]_{\mathcal{B}, \mathcal{C}} = \left(
    \begin{tabular}{c|c}
        $I_r$ & $0$ \\
        \hline
        $0$ & $0$
    \end{tabular}
    \right) \]
\end{proposition*}
\prompt{

\begin{proof}
\cloze{
    Let $v_{r + 1}, \ldots, v_n$ be a basis of
    $\ker(\alpha)$. Extend to a basis of $V$:
    \[ \mathcal{B} = (v_1, \ldots, v_r, v_{r + 1},
    \ldots, v_n) \]
    Prove that $(\alpha(v_1), \ldots,
    \alpha(v_r))$ is a basis of $\Image(\alpha)$
    (check that it's free and generating). Extend
    this to a basis of $W$:
    \[ \mathcal{C} = (\alpha(v_1), \ldots,
    \alpha(v_r), w_{r + 1}, \ldots, w_m) \]
    Then $[\alpha]_{\mathcal{B}, \mathcal{C}}$ is
    of the desired form.
}
\end{proof}
}
\end{flashcard}

\begin{proof}
    Choose $\mathcal{B}$ and $\mathcal{C}$ wisely.
    \begin{itemize}
        \item Fix $r \in \NN$ such that $\dim \Ker \alpha =
            n - r$.
        \item $N(\alpha) = \Ker(\alpha) = \{x \in
            V, \alpha(x) = 0\}$
        \item Fix a basis of $N(\alpha)$: $v_{r +
            1}, \dots, v_n$. Extend it to a basis
            of $V$, so
            \[ \mathcal{B} = (v_1, \dots, v_r,
            \ub{v_{r + 1}, \dots, v_n}_{\Ker
            \alpha}) \]
        \item Claim: $(\alpha(v_1), \dots,
            \alpha(v_r))$ is a basis of
            $\Image\alpha$.
            \begin{itemize}
                \item Span:
                    \[ v = \sum_{i = 1}^n
                    \lambda_i v_i \]
                    \[ \implies \alpha(v) =
                    \sum_{i = 1}^n \lambda_i
                    \alpha(v_i) = \sum_{i = 1}^r
                    \lambda_i \alpha(v_i) \]
                    Let $y \in \Image\alpha$ then
                    exists $v \in V$ such that $y
                    = \alpha(v)$ then
                    \[ y = \sum_{i = 1}^r
                    \lambda_i \alpha(v_i) \]
                    \[ \implies y \in \langle
                    \alpha(v_1), \dots,
                    \alpha(v_r) \rangle \]
                \item Free:
                    \[ \sum_{i = 1}^r \lambda_i
                    \alpha(v_i) = 0 \]
                    \[ \implies \alpha \left(
                    \sum_{i = 1}^r \lambda_i v_i
                    \right) = 0 \]
                    \[ \implies \sum_{i = 1}^r
                    \lambda_i v_i \in \Ker \alpha \]
                    \[ \implies \sum_{i = 1}^r
                    \lambda_i v_i = \sum_{i = r +
                    1}^n \mu_i v_i \]
                    \[ \implies \sum_{i = 1}^r
                    \lambda_i v_i - \sum_{i = r +
                    1}^n \mu_I v_i = 0 \]
                    but since $\mathcal{B}$ is
                    free, we must have $\lambda_i
                    = 0, \mu_i = 0$ so it's free.
            \end{itemize}
    \end{itemize}
    Conclusion: $(\alpha(v_1), \dots,
    \alpha(v_r))$ basis of $\Image\alpha$, $(v_{r +
    1}, \dots, v_n)$ basis of $\Ker\alpha$. Let
    \[ \mathcal{B} = (v_1, \dots, v_r, v_{r + 1},
    \dots, v_n) \]
    \[ \mathcal{C} = (\alpha(v_1, \dots,
    \alpha(v_r), w_{r + 1}, \dots, w_m) \]
    Then
    \[ [\alpha]_{\mathcal{B}, \mathcal{C}} =
    (\alpha(v_1), \dots, \alpha(v_r), \alpha(v_{r
    + 1}), \dots, \alpha(v_n)) \]
\end{proof}

\begin{remark*}
    This provides another proof of the rank
    nullity theorem:
    \[ r(\alpha) + N(\alpha) = n \]
\end{remark*}

\begin{corollary*}
    Any $m \times n$ matrix is equivalent to:
    \[ \left(
    \begin{tabular}{c|c}
        $I_r$ & $0$ \\
        \hline
        $0$ & $0$
    \end{tabular} \right) \]
    where $r = r(\alpha)$.
\end{corollary*}
