The homework question I am working on is:

Suppose $A$ is an $m \times n$ matrix in which $m \le n$. Suppose also that the rank of $A$ equals $m$. Show that the transformation $T$ determined by $A$ maps $\Bbb R^n$ onto $\Bbb R^m$. Hint: The vectors $\overrightarrow{e_1},...,\overrightarrow{e_m}$ occur as columns in the reduced row-echelon form for $A$.

Things I have tried:

  1. I wrote down things that the problem statement implies:

The number of pivot columns equals the numbers of rows since the rank of $A=m$.

  1. I looked at this question and answers here for a while: Is a linear tranformation onto or one-to-one? but didn't have any additional conclusions after that.

When you reduce a matrix to row echelon form, each operation amounts to multiplying the matrix on the left by some matrix $B_i$. If $B$ is the product of these matrices, we have $C = BA$, where $C$ is the row echelon form of $A$.

You must prove that the vector equation $Ax = y$ has a solution in $\mathbb{R}^n$ for any $y\in\mathbb{R}^m$. This is equivalent to:

$$(BA)y = Cx = z = By$$

Because $C$ has a pivot on each row, the system of equations $Cx = z$ has always a solution, and this is what you needed to prove.


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