I'm looking for someone who knows a theorem who proves that this two definitions of rank of Linear Transformations/Matrices are equivalent.
$rk(T) = \dim \operatorname{Span}( \{(a_{i1}, \dots, a_{in}), i = 1, \dots, m \})$ (is the number of nonzero lines on the matrix representation for $T$ in its row echelon form).
$rk(T) = \dim \operatorname{Im}(T)$
I was using this as equivalent, but i found in a book that this equivalence should be proved if not given.
My attempt is:
Let $A: V \longrightarrow W$ a linear transformation that maps $v \mapsto Av = w$. Let's assume that $\dim \operatorname{V} = n$ and $\dim \operatorname{W} = m$. Since we can rethink this as a linear system, we can construct this system and then put it in his row echelon form. Now, we have an equivalent system, and let's say that $r$ rows are nonzeros on this form.That could again be rethinked as a new linear transformation $\bar{A}: V \longrightarrow \bar{W}$; $v \mapsto \bar{A}v = \bar{w}$. Since $\bar{A}_{r \times n}$, and all lines of $\bar{A}$ are linear independent, $\dim \operatorname{Im}(\bar{A})$ = r.
I'm not sure that if this "proof" sounds rigorous, so any hint is helpful.
Thanks!
Each row operation can be achieved by multiplying by an invertible matrix. For instance, the operation of adding the top row of a $2 \times 2$ matrix to the bottom row is the same as left-multiplying by \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} Thus, if $R$ is the reduced form of a matrix $A$, then we may write $R = E_1 E_2 \ldots E_k A$ where each $E_i$ is invertible. Therefore the image of $R$ is equal to the image of $A$ after a linear transformation by $E_1E_2\ldots E_k$. This transformation is invertible, which means that Im$A$ has the same dimension as Im$R$. But $R$ is in reduced row echelon form, so the dimension of its image is equal to the number of nonzero rows of $R$. This is the rank of $A$ by definition.