N-dimensional Vector Space

1. Exercises

1. Suppose the \(s \times n\) matrix on field \(K\)
   
   \begin{equation*} A = {\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & & \vdots\\ a_{s1} & a_{s2} & \cdots & a_{sn}\\ \end{pmatrix}} \end{equation*}

   satisfies \(s \le n\), and \(2|a_{ii}|>\sum_{j=1}^{n}{|a_{ij}|},\ \ (i = 1, 2, \cdots, s)\), prove: rank of \(A\)'s row vector group, \(\gamma_{1}, \gamma_{2}, \cdots, \gamma_{s}\), equals to \(s\).

   \(Proof\) Let

   \begin{equation*} A' = {\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1s}\\ a_{21} & a_{22} & \cdots & a_{2s}\\ \vdots & \vdots & & \vdots\\ a_{s1} & a_{s2} & \cdots & a_{ss}\\ \end{pmatrix}} \end{equation*}

   if we can prove that rank of \(A'\)'s row vector group equals to \(s\), then by appending the row vectors, we can show that rank of \(A\)'s row vector group equals to \(s\).

   Now, let's prove that \(\det{A'} \ne 0\).

   If \(\det{A'} = 0\), then there exists a non-zero \(X = (x_1, \cdots, x_s)^T\), such that \(A'X = 0\). Then suppose \(|x_k| = \max{|x_{i}|, i = 1, 2, \cdots, s}\),

   \begin{align*} \sum_{j=1}^sa_{kj}x_{j} &= 0 \\ a_{kk}x_k &= -\sum_{j \ne k}{a_{kj}x_j} \\ |a_{kk}x_k| &= |\sum_{j \ne k}{a_{kj}x_j}| \\ |a_{kk}||x_k| &\le \sum_{j \ne k}(|a_{kj}||x_j|) \\ |a_{kk}| &\le \sum_{j \ne k}({|a_{kj}|\frac{|x_j|}{|x_k|}}) \\ |a_{kk}| &\le \sum_{j \ne k}{|a_{kj}|} \end{align*}

   a contradiction. Thus, \(\det{A'} \ne 0\).

   So rank of \(A'\)'s row vector group equals to \(s\). \(\blacksquare\)