how do I explicitly prove that a matrix of this kind:
$$A=\begin{pmatrix}1&1&0&0&0\\ \:1&1&0&0&0\\ \:0&0&0&0&0\\ \:0&0&0&0&0\\ \:0&0&0&0&0\end{pmatrix}$$
for any base : $$ B={b_1, b_2, b_3, b_4, b_5} $$
$A_{\left\{B\right\}}$ will have a row $r_1$ that every other row $r_2$ can be expressed as:
$$ r_2 = \alpha \cdot r_1$$
One proof is as follows. Let $n$ be the size of $A$. Note that $A$ can be written in the form $A = A_1 A_2$, where $$ A_1 = \pmatrix{1\\1\\0\\ \vdots \\ 0}, \quad A_2 = \pmatrix{1&1&0&\cdots & 0}. $$ Let $S$ be the change of basis matrix such that $S[x]_{\mathcal B} = x$ for all $x \in \Bbb R^n$, where $[x]_{\mathcal B}$ denotes the coordinate vector of $x$ relative to the basis $\mathcal B = \{b_1,\dots,b_n\}$. The matrix of $A$ relative to a new base will be given by $$ S^{-1}AS = (S^{-1}A_1)(A_2 S^{-1}). $$ Using the fact that $S^{-1}A_1$ is a $1 \times n$ matrix and $A_2 S^{-1}$ is an $n \times 1$ matrix, show that every row of the product has the form $\alpha(A_2 S^{-1})$.