0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.