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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!

Gabriel Parizoto
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  • One approach would be to show that elementary row operations (the steps taken to put a matrix in row echelon form) do not change the dimension of the image of a matrix. – angryavian May 27 '22 at 15:02
  • does this answer your question? https://math.stackexchange.com/questions/3844417/prove-that-ranka-ranktranspose-of-a – Aleksei Kubanov May 27 '22 at 15:11

1 Answers1

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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.

Frank Seidl
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