- Show that adding a scalar multiple of one column to another leaves both the column rank and the row rank of a matrix unaltered.
- Deduce from 1 that the row rank equals the column rank.
HINT: 1. Use the fact that $\varphi((x_1,x_2,...,x_n))=(x_1+\beta x_2,x_2,...x_n)$ (where $\beta$ is a fixed scalar) is an isomorphism from $F^n$ to itself.
2. Reduce the columns not belonging to a column basis to null vectors.
My attempt:
$1\le k,p\le n$ are fixed numbers. Let $A_{m\times n}$ and $B_{m\times n}$ be matrices such that
$B_{*j}=
\begin{cases}
A_{*j} &\quad\text{if, } j\ne k\\
A_{*k}+A_{*p} &\quad\text{if, } j=k\\
\end{cases}$ $\quad\forall j=1,2,...n$
Let $S_1=\mathcal{R}(A)$ and $S_2=\mathcal{R}(B)$ where $\mathcal{R}$ represents the rowspace of a matrix. $V_1,V_2$ are vector spaces over $F^n$.
We define $\varphi: F^n\longrightarrow F^n$ such that
$$\varphi((a_{i1},a_{i2},...,a_{ik},...,a_{in}))=(a_{i1},a_{i2},...,a_{ik}+a_{ip},...,a_{in})$$
$\Rightarrow\varphi(A_{i*})=B_{i*}$
Using the hint, we know that $\varphi$ is an isomorphism from $F^n$ to itself
$\Rightarrow S_1$ is a subspace of $F^n$ so $\varphi(S_1)=S_2$ is a subspace of $F^n$ in the same dimension
$\Rightarrow \dim(\mathcal{R}(A))=\dim(S_1)=\dim(S_2)=\dim(\mathcal{R}(B))$
Similarly, $\dim(\mathcal{C}(A))=\dim(\mathcal{C}(B))$ where, $\mathcal{C}$ represents the column space of a matrix.
a) Is my attempt correct? Is there a better approach?
b) I know a general proof to the 2 but can't understand how to deduce it from 1.
a) Don't seem to have problem.
b) Since row operation is dual to column operation, reduce $A$ to its RREF, which has equal row rank and column rank, or further column-reudce its RREF to most column-reduced form $C$ such that it is eminant that its column rank equals its row rank (by counting remaining non-zero entries in the matrix), hence column rank of A= column rank of C= row rank of C= row rank of A.
Note that though only one type of operations (adding a scalar multiple to others) is available, it still convert $A$ to its RREF except all leading entries may not be 1.