53 lines
1.9 KiB
TeX
53 lines
1.9 KiB
TeX
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\chapter{Proofs}\label{ch: proofs}
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\begin{mlemma}\label{proofs: fullcolumn}\label{proof:triangularsystem}
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$\hat{X} \text{ has full column rank } \iff \hat{X}^T\hat{X} \succ 0$
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\end{mlemma}
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\begin{mproof}
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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 $B = XX^T \succeq 0$.
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\begin{align*}
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x^T (B + \lambda^2 I_m) x &= x^{T}Bx + \lambda^2 x^{T}I_{m}x \\
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&= x^{T}Bx + \lambda^2 \norm{x}^2
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\end{align*}
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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 $\hat{X}^T\hat{X} \succ 0$.
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\end{mproof}
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\begin{mlemma}\label{proof:eigenvalues_translation}
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$\alpha \in Sp(A) \iff (\alpha + \lambda) \in Sp(A + \lambda I)$
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\end{mlemma}
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\begin{mproof}
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$Av = \alpha v \iff (A + \lambda I)v = Av + \lambda v = \alpha v + \lambda v = (\alpha + \lambda)v$
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\end{mproof}
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\begin{mlemma}\label{proofs: eigenvalues}
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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 $X$.
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\end{mlemma}
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\begin{mproof}
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Consider the Singular Value Decomposition of the rank $n$ matrix $X$
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\[
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X = U \Sigma V^T
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\]
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$\Sigma = \diag(\sigma_1, \dots, \sigma_n) \in \mathbb{R}^{m \times n}
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$
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Then
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\[
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XX^T = U \Sigma V^T V \Sigma^T U^T = U \Sigma \Sigma^T U
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\]
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with
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\[
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\Sigma \Sigma^T = \diag(\sigma_1^2, \dots, \sigma_n^2, 0, \dots, 0) \in \mathbb{R}^{m \times m}
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\]
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Hence, $XX^T$ has exactly $m$ singular values of which $m-n$ are zeros.
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\end{mproof}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "../main"
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%%% TeX-command-extra-options: "-shell-escape"
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%%% End:
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