Why does the rank always hold even when we apply Gaussian elimination?

Question

I have this proof:

For $v_{1},...,v_{k} \in R^{n}$, and $\alpha \in R \setminus \{0\}, \beta \in R$ one can easily verify that \begin{align} \operatorname{rank}(v_{1},...,v_{k}) &= \operatorname{rank}(v_{1},\dots,v_{i-1},\alpha v_{i}, v_{i+1},\dots,v_{k})\\ & = \operatorname{rank}(v_{1},\dots,v_{i-1}, v_{i}, v_{i+1},\dots, v_{j-1},v_{j}+\beta v_{i},v_{j+1},\dots,v_{k}) \end{align}

Could someone explain this proof in laymans terms? It's suppose to conclude that the rank of a matrix is unaffected by row reduction.

Answer 1

It's useful to review the definition of rank.

The $n \times k$ matrix $M$ in your question can be viewed as a function from $\mathbb R^n$ to $\mathbb R^k$: you input a $n$-vector $W = \langle w_1,...,w_n\rangle \in \mathbb R$ thought of as a $1 \times n$ matrix, and you carry out the matrix multiplication $WM$ to obtain a $1 \times k$ matrix thought of as a $k$-vector $WM \in \mathbb R^n$.

The image of this function is the row space of $M$, it is a subspace of the vector space $\mathbb R^k$, and its dimension is, by definition, equal to the rank of $M$.

Row operation is equivalent to multiplying $M$ on the left by an invertible $n \times n$ matrix $E$: replacing $v_j$ with $v_j + \beta v_i$ is equivalent to replacing $M$ by the matrix $EM$ where $E$ has $1$'s down the diagonal, and has a $\beta$ in the $i,j$ entry.

Since the $n \times n$ matrix $E$ is invertible, the matrices $M$ and $EM$ have the exact same row space: the row space itself is unaffected by row operations. In particular, $M$ and $EM$ have the same dimension, and so the ranks of $M$ and $EM$ are equal.

Answer 2

The rank of a family of vectors is the dimension of the subspace they generate.

Here you can easily check that the vector spaces generated by these sets (with $\alpha\ne 0$)

• $V_1=\bigl\langle v_{1},...,v_{k}\bigr\rangle $,
• $V_2= \bigl\langle v_{1},\dots,v_{i-1},\alpha v_{i}, v_{i+1},\dots,v_{k} \bigr\rangle $,
• $V_3=\bigl\langle v_{1},\dots,v_{i-1}, v_{i}, v_{i+1},\dots, v_{j-1},v_{j}+\beta v_{i},v_{j+1},\dots,v_{k}\bigr\rangle $

are identical.

Indeed, it is trivial that $V_2,V_3\subseteq V_1$. Conversely, $V_1$ and $V_3$ have the same generators but the $i$-th. But $v_i=\frac 1\alpha(\alpha v_i)$, whic proves $V_1\subset V_2$.

Similarly, $V_1$ and $V_3$ differ only by their $j$-th generator. Writing $\; v_j=(v_j+\beta v_i) -\beta v_i$ proves that $V_1\subseteq V_3$. So all these subspaces are the same.