\chapter{Proofs}\label{ch: proofs}

\begin{mlemma}\label{proofs: fullcolumn}\label{proof:triangularsystem}
    $\hat{X} \text{ has full column rank } \iff \hat{X}^T\hat{X} \succ 0$
\end{mlemma}

\begin{mproof}
To show that $\hat{X}^T\hat{X} = XX^T + \lambda^2 I_m \succ 0$, $\lambda > 0$, we can consider the quadratic form $x^T (X^{T}X + \lambda^2 I_m) x$. Let $B = XX^T \succeq 0$. 
\begin{align*}
    x^T (B + \lambda^2 I_m) x &= x^{T}Bx + \lambda^2 x^{T}I_{m}x \\
    &= x^{T}Bx + \lambda^2 \norm{x}^2
\end{align*}

Since $B$ is positive semidefinite, we have $x^{T}Bx \geq 0\ \ \forall x \in \mathbb{R}^m$. Additionally, $\lambda^2 \|x\|^2 > 0\ \ \forall x \neq 0$. Therefore, $x^T (B + \lambda^2 I_m) x > 0$ for all non-zero vectors $x$, meaning that $\hat{X}^T\hat{X} \succ 0$.

\end{mproof}

\begin{mlemma}\label{proof:eigenvalues_translation}
    $\alpha \in Sp(A) \iff (\alpha + \lambda) \in Sp(A + \lambda I)$
\end{mlemma}

\begin{mproof}
    $Av = \alpha v \iff (A + \lambda I)v = Av + \lambda v = \alpha v + \lambda v = (\alpha + \lambda)v$
\end{mproof}

\begin{mlemma}\label{proofs: eigenvalues}
    The singular values of the matrix $XX^T, X \in \mathbb{R}^{m \times n}$\newline are $\{\sigma_1^2 \dots \sigma_n^2, 0 \dots 0\}$, with $\sigma_1 \dots \sigma_n$ being the singular values of $X$.
\end{mlemma}

\begin{mproof}
    Consider the Singular Value Decomposition of the rank $n$ matrix $X$
    \[
    X = U \Sigma V^T
    \]
    $\Sigma = \diag(\sigma_1, \dots, \sigma_n) \in \mathbb{R}^{m \times n}
    $
    Then
    \[
    XX^T = U \Sigma V^T V \Sigma^T U^T = U \Sigma \Sigma^T U
    \]
    with
    \[
    \Sigma \Sigma^T = \diag(\sigma_1^2, \dots, \sigma_n^2, 0, \dots, 0) \in \mathbb{R}^{m \times m}
    \]
    Hence, $XX^T$ has exactly $m$ singular values of which $m-n$ are zeros.
\end{mproof}

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