# Can you determine the range of a linear transformation Ax, without finding the reduced row echelon form of A?

Say we want to determine the range of the transformation A x. And we call the solution vectors b. So if we want to determine all possible b's, are we going to have to put A into reduce row echelon form? or is there another way?

Or alternatively, given a transformation A and a b, is there a way to tell whether b is in the range without reducing Ax=b into reduced row echelon form?

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What does "we call the solution $\mathbf{b}$" mean? The range of a linear transformation $A$ is a subspace of a certain vector space. And $A\mathbf{x}$ is not generally a transformation, but the value of the transformation $A$ at the vector $\mathbf{x}$. What exactly are you trying to ask? Are you asking, given a transformation $A$ and a $\mathbf{b}$, is there a way to tell whether $\mathbf{b}$ is in the range without reducing $A\mathbf{x}=\mathbf{b}$ into reduced row echelon form? –  Arturo Magidin Sep 11 '11 at 21:33
essentially, yes that is what I am asking. –  weezybizzle Sep 11 '11 at 21:45
Depends on $A$. Sometimes it is clear what the range is by other considerations; often it is not. The standard way to determine the range of a linear transformation is to try to compute the span of $A\mathbf{e}_1,\ldots,A\mathbf{e}_n$, where $\mathbf{e}_1,\ldots,\mathbf{e}_n$ are the standard basis vectors; but that amounts to precisely trying to figure out the columnspace of $A$, and the simplest way of doing that is to perform at least some row-reduction on $A$ so that the columnspace is "obvious". You often don't have to go all the way to reduced row echelon form. –  Arturo Magidin Sep 11 '11 at 21:47
I see, so no shortcuts really. Thank you very much for your help. –  weezybizzle Sep 11 '11 at 21:52
Although elementary row operations do not respect the column space, they do respect linear dependence relations among the columns; columns 2, 4, and 7 (say) are linearly dependent after you do an elementary row operation if and only if those columns were linearly dependent before the operation. So you can go to row echelon form (you don't have to go all the way to reduced rwo echelon form), pick out the columns that form a basis for the column space in the echelon matrix (namely, those with a leading non-zero entry), and the corresponding columns in the original matrix form a basis for its... –  Gerry Myerson Sep 12 '11 at 3:54

You can check the rank of $$\left[ \begin{array}{c|c}A &b \end{array} \right]$$ if that is an alternative way that you are looking for (note that row reduced echelon form is also a way of to compute the rank).
In particular, if $A$ is fat i.e. number of rows $m$ is less than the number of columns $n$, then infinitely many solution exists. If $m=n$ then either the matrix is invertible and every $b$ has a solution in the form of $x=A^{-1}b$ or $A$ is singular and you can reduce it to either a fat or tall matrix with full row/column rank.
The last option is the tall matrix case. Then you can simply check if $$\operatorname{rank}\left[ \begin{array}{c|c}A &b \end{array} \right]>\operatorname{rank}A$$ In case of a positive answer : No $x$ can lead to that particular $b$ (which is usually the case). Hence the popularity of least squares solution.