# Show that $(e_1,\dots, e_r)$ is a base of $\operatorname{im}(A) \subseteq \mathbb{K}^m$

Let $$A\in M(m\times n,\mathbb{K})$$ be a matrix given in the row echelon form with rank $$r$$. Let $$e_i$$ be the $$i$$-th unit vector of $$\mathbb{K}^m$$. Show that $$(e_1,\dots, e_r)$$ is a base of $$\operatorname{im}(A) \subseteq \mathbb{K}^m$$

In my opinion, $$A$$ should look like

$$\begin{pmatrix} a_{11} & \\ 0 & a_{22} & &*\\ \vdots & \ddots & \ddots \\ 0 & \dots & 0 & a_{rr} \end{pmatrix}$$ with $$*$$ I indicate arbitrarily coefficients of $$A$$. Because we don't change the matrix by multiplying the inverse elements to every $$a_{ij}$$, we obtain

$$\begin{pmatrix} 1 & \\ 0 & 1 & &*\\ \vdots & \ddots & \ddots \\ 0 & \dots & 0 & 1 \end{pmatrix}$$ which actually are all $$r$$ unit vectors $$(e_1,\dots,e_r)$$

Since $$\operatorname{im(A)}:=\operatorname{span}\{(a_{11},\dots,a_{m1}),\dots,(a_{n1},\dots,a_{mn})\}$$ is the span of all column vectors of the initial matrix, we found a basis for $$\operatorname{im}(A)\quad _\blacksquare$$

Is this proof ok?

The matrix is an $$m\times n$$ matrix mapping from $$\Bbb K^n\to \Bbb K^m$$, it's not an $$r\times r$$ matrix, it just has rank $$r$$, so your $$A$$ isn't quite right (unless $$r=n=m$$). I'm not entirely sure what you mean by multiplying by the inverse entries of the $$a_{ij}$$, do you mean to multiply by the diagonal matrix $$\text{diag}(a_{11},a_{22},\dots,a_{rr})$$?
If you put the rank $$r$$ matrix $$A$$ in row echelon form, then with the ordered basis $$(e_1,\dots,e_m)$$ on $$\Bbb K^m$$, the image of $$A$$ will lie in the span of the first $$r$$ standard basis vectors (small example so it's easy to typeset): $$\begin{bmatrix}a&b&c&d&e\\0&f&g&h&i\\0&0&j&k&l\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix}.$$ (Assume the first three rows are linearly independent) This is a linear map from $$\Bbb K^5\to\Bbb K^6$$ and it sends \begin{align}e_1&\mapsto ae_1,\\e_2&\mapsto be_1+fe_2,\\e_3&\mapsto ce_1+ge_2+je_3,\\e_4&\mapsto0\end{align} and so on. You can see each element of the image is written in terms of the first $$r$$ standard basis vectors of $$\Bbb K^m$$. You can lift this to the general case easily.
• I mean that I want to multiply ever row containing some $a_{ij}$ with $\frac{1}{a_{ij}}$ – Doesbaddel Jun 14 at 10:11
• I didn't want to express (but unfortunately I did), that we have an $r\times r$ matrix, I wanted to express, that our matrix in row echelon form has $r$ pivots that will indicate the first $r$ standard vectors. – Doesbaddel Jun 14 at 10:17