An Introductory Random Matrix Theory Part I: Matrix Elementary Revisited
Table of Contents 目录
Vector Space
(Linear Dependence Lemma) Suppose \(v_{1}, v_{2}, \dots, v_{m}\) is a linearly dependent list in \(V\). Then there exists \(k \in \{1, 2, \dots, m\}\) such that
\[v_{k} \in \mathtt{span}\{v_{1}, v_{2}, \dots, v_{k-1}\}.\]Furthermore, if \(k\) satisfies the condition above and the \(k^{\mathtt{th}}\) term is removed from \(v_{1}, v_{2}, \dots, v_{m}\), then the span of the remaining list equals to \(\mathtt{span}\{v_{1}, v_{2}, \dots, v_{m}\}\).
Proof
(1) Since \(v_{1}, v_{2}, \dots, v_{m}\) are linearly dependent, there exists a list of \(a_{1}, a_{2}, \dots, a_{m} \in \mathbb{F}\) and they are not all zero, such that \(a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{m}v_{m} = 0\)
Let \(k\) be the largest index of a non-zero \(a_{i}\), then we have \(a_{i}=0\) for all \(k>i\) and
\[v_{k} = -\frac{a_{1}}{a_{k}}v_{1} - \frac{a_{2}}{a_{k}}v_{2} -\cdots \frac{a_{k-1}}{a_{k}}v_{k-1},\]i.e., \(v_{k} \in \mathtt{span}\{v_{1}, v_{2}, \dots, v_{k-1}\}.\)
(2) If \(v_{k}\) is removed from the list, we have \(\mathtt{span}\{v_{1}, \dots, v_{k-1}, v_{k+1}, v_{m}\} \subset \mathtt{span}\{v_{1}, \dots, v_{m}\}\). Let \(v \in \mathtt{span}\{v_{1}, \dots, v_{m}\}\), there exists a list of \(\beta_{1}, \dots, \beta_{m}\) such that
\[\begin{aligned} v & = \beta_{1}v_{1} + \cdots + \beta_{m}v_{m} \\ & = \beta_{1}v_{1} + \cdots + \beta_{k-1}v_{k-1}+ \beta_{k}\left(-\frac{a_{1}}{a_{k}}v_{1} - \frac{a_{2}}{a_{k}}v_{2} -\cdots \frac{a_{k-1}}{a_{k}}v_{k-1}\right)+ \beta_{k+1}v_{k+1} +\cdots + \beta_{m}v_{m} \\ & = \left(\beta_{1} -\frac{a_{1}}{a_{k}}\right)v_{1} + \cdots + \left(\beta_{k-1} -\frac{a_{k-1}}{a_{k}}\right)v_{k-1} + \beta_{k+1}v_{k+1} +\cdots + \beta_{m}v_{m}, \end{aligned}\]which implies \(v \in \mathtt{span}\{v_{1}, \dots, v_{k-1}, v_{k+1}, v_{m}\} \subset \mathtt{span}\{v_{1}, \dots, v_{m}\}\).
(Every Spanning List Contains a Biasis) Every spanning list in a vector space can be reduced to a basis of the vector space.
Proof
Let \(V = \mathtt{span}\{v_{1}, v_{2}, \dots, v_{m}\}\) and \(B:=\{v_{1}, v_{2}, \dots, v_{m}\}\).
If \(v_{1} = 0\), remove \(v_{1}\) from \(B\), otherwise keep \(v_{1}\) in \(B\).
for \(k = 2,3, \dots, m\), if \(v_{k} \in \mathtt{span}\{v_{1}, v_{2}, \cdots, v_{k-1}\}\), remove \(v_{k}\) from \(B\), otherwise keep \(B\) unchanged.
After the removing procedure, we obtain a list of vectors \(B\) satisifed that
\[v_{k_{i}} \notin \mathtt{span}\{v_{k_{1}}, v_{k_{2}}, \dots, v_{k_{i-1}}\}, \forall v_{k_{i}} \in B.\]By the Linear Dependence Lemma, the vectors in \(B\) are linearly independent and these vectors span \(V\). Hence we get a basis of \(V\).