Determine $\mathrm{Col(A)}$ with only matrix $B$ given which is row equivalent to matrix $A$

Question

I have a question about determining a basis of $\mathrm{Col(A)}$.

If a matrix $A$ is given, I understand that I have to row reduce matrix $A$ into a reduced echelon form (call this $U$). By looking at the pivots, I can determine which columns of matrix $A$, must be in the span of $\mathrm{Col(A)}$.

But what if I don't get the matrix $A$ given, but only a matrix $B$ which is row equivalent to $A$?

Then there are two options. First row reduce the matrix $B$ and look at the pivot columns of the reduced echelon form of $B$.

• If $\mathrm{Col(A)}$ has dimension $2$ for example in $\Bbb R^3$ (i.e. $\Bbb R^m$ of matrix of $m \times n$), then we can't determine $\mathrm{Col(A)}$? Is this true? And why is this true? And what if matrix $A$ is given in this situation? May we then still take 2 columns of $A$ in $\Bbb R^3$ as $\mathrm{Col(A)}$?

• If $\mathrm{Col(A)}$ has dimension, that is the number of pivots, $m$ in $\Bbb R^m$ (with $m$ is the number of rows of matrix $A$), then we may take the columns of the reduced echelon form of $B$? Is this true? Or can you only do this if these are vectors of the identity matrix for example $[1,0,0]$? So is this not the case for another vector/column?

So, summarizing, when may you take the columns of the reduced echelon form of $A$ for $\mathrm{Col(A)}$? Is it only if the reduced echelon form consist of vectors of the identity matrix or if $\mathrm{Col(A)}$ has dimension $m$ in $\Bbb R^m$ or both?

I am a beginner in linear algebra, so I would be really thankful if someone could explain this to me.

Answer 1

I think I understand it now, thank you guys very much! And I indeed meant the basis for the Col(A), my apologies. Luckily you understood my question.

Thus,

If matrix A is given, I can decide which columns of A form a basis of Col(A) by looking at the pivot columns of the reduced echelon form of A (which is the same as the reduced echelon form of B because A and B are row equivalent).

If matrix A is not given and Col(A) is not equal to ℝm, then the basis of Col(A) can not be determined.

If matrix A is not given and Col(A)=ℝm (i.e. the amount of pivots equals the amount of rows of A / dim Col(A) = m), then the pivot columns of the reduced echelon form form a basis of Col(A) which is {e1, e2, ... , en} (with ei as unit vectors/columns of the identity matrix).

Correct me if I am wrong.

Answer 2

You cannot get a basis for the column space of $A$ from a matrix $B$ which is row equivalent to $A$ because row operations change the column space.

For example,

$$ A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$

are row equivalent but they have different column spaces so if you are only given $B$, you cannot reconstruct the column space of $A$ (but you can reconstruct the row space because row operations don't change the row space).

Answer 3

This phrase in your question worries me:

which columns of matrix A, must be in the span of $\mathrm{Col}(A)$

First of all, $\mathrm{Col}(A)$ is already a subspace (of $\mathbb{R}^m$), defined as the span of the columns of the matrix $A$. Although we certainly can form the span of $\mathrm{Col}(A)$, it'll be $\mathrm{Col}(A)$ itself, because it's already a subspace. So while not technically wrong, it's very uncommon that anybody would want to set up the "span of $\mathrm{Col}(A)$".

Secondly, by definition all columns of the matrix $A$ are in $\mathrm{Col}(A)$ — because, as I said above, it's precisely the span of all those columns.

What you're probably talking about (but misinterpreting) is the question to find a basis for the column space $\mathrm{Col}(A)$. A standard method taught in linear algebra courses is to row-reduce $A$, and the pivot columns of the (reduced) row echelon matrix $U$ will tell you which columns of the original matrix $A$ form a basis for $\mathrm{Col}(A)$.

If $B$ is row-equivalent to $A$, they have the same reduced row echelon matrix. So you will be able to tell which columns of $A$ form a basis of $\mathrm{Col}(A)$. But without having $A$ itself, you can't actually present that basis nor determine $\mathrm{Col}(A)$. The only exception would be when you find that you have a basis consisting of $m$ vectors in $\mathbb{R}^m$ (i.e. your matrix is of size $m\times n$, so that all columns are elements of $\mathbb{R}^m$, and there also happen to be $m$ pivots) — then you can claim that $\mathrm{Col}(A)=\mathbb{R}^m$, the entire space.