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Taking things for granted doesn't have any fun in mathematics.

When we're computing the row space of a matrix, we're using row reduction, which doesn't really change the rowspace of a matrix, but it doesn't change the column space, and again, row reductions don't change the linear independence of the columns of a matrix, but they do change the linear independence of the rows of a matrix. AND! The basis of the row space of the matrix can be found by performing row operations on the transpose of a matrix. Now, these are the basis of the row space but the operations are done on a completely different set of entries and still we find the basis of the row space because, most probably(with mere analogy, as there is probably no theorem for this), this time around the linear dependence of the rows of the matrix is not affected.

Summarizing my large ocean of confusions to four major issues:

  • First of all the problem arises with books having no reasons to whatever they explain, and it starts with not defining why we do row operations at all when we actually change the columspace of a matrix?
  • Then, it continues down with, why do we do row operations? And not column operations? Why isn't there a Reduced Column Echelon Form? As I've explained above, for finding the basis of the row space, it's almost that we're doing Reduced Column Echelon Form, but our doing so doesn't change the results, but there's no theorem for stating that's valid(Atleast, uptill where I've studied), I mean it doesn't change the invertability of a matrix, but again, not all matrices are invertible.

  • And there's more, Why does the column space of a matrix change while we're doing row operations but the linear independence doesn't change but exactly opposite with the row space, i.e. the row space doesn't change but the linear dependence change. Come to think of it, at this last point, if the linear dependence has changed, then the basis that span the row space has changed, but How come the row space doesn't change?

  • What's the connection of row space with column space, and what's the geometrical interpretation of a row space?
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Let me address your issues one by one:

1) I don't know what book you are referencing but I have read Linear Algebra from Gilbert Strang's and he had explained why we do row operations. We do row operations to find a solution to $Ax = b$ among other things because it makes it easy to solve the equations after reducing the matrix to echelon form.

The fact that the row operations change column space has nothing to do with this application because, when you are doing the row operations, you also do the same row operation on right side i.e. on $b$. So After the operations, when you are equating the row equations, you are equating the correct corresponding vector components, so there wont be any problem in that.

2) We do row operations and not column operations is because of the notation that we follow which is vector is represented by a $n$x$1$ matrix i.e the components of a vector are rows in its matrix form. So when we are trying to find the solution for $Ax = b$, it doesn't make sense to do the column operations and make the reduced column echelon form. If you want, you can take transposes of everything and do the operations on columns.

3) Change of basis doesn't mean that the space spanned by that basis has also changed, if the new basis also spans the same space. As for as why column space changes with row operations, consider a simple row exchange operation:

$$ \left[\begin{array}\ 1&1\\ 1&1\\ 0&0 \end{array}\right] \overrightarrow {r_2 \leftrightarrow r_3} \left[\begin{array}\ 1&1\\ 0&0\\ 1&1 \end{array}\right] $$

With this row exchange, the the component of column vectors have changed their position. To simplify, it's similar to saying that you have started denoting y-axis as z and vice-versa, which will have no affect in your answer if you know that this convention has changed. Doing row operations also does the same operations on right hand side, so you don't need to worry about the answer. The linear dependence will remain same because, lets say a vector $v = a_1c_1 + a_2c_2$ where $a_i$ is a scalar. Now after the row operations, the vector components have changed but the vector $v$ is still $a_1c_1+a_2c_2$ thus preserving the linear dependence.

4) Every matrix transforms its row space onto its column space. Or to say it in similar terms, every vector in column space comes from exactly one vector from row space. Geometrically you can visualize row space as the column space of $A^T$ i.e. a matrix $A^T$ projects a vector $x$ onto the row space of $A$ when done $A^Tx$. This is how I have always visualized the row space. May be someone else can give a different visualization.

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  • $\begingroup$ What do you mean by, "every vector in column space comes from exactly one vector from row space"? I mean, the column space of a matrix gets transformed onto the rowspace, thus transforming from the $m^th$ space onto the $n^th$ space. But, your explanation seems counter intuitive. Can you explain a bit more? $\endgroup$ – mathmaniage Aug 22 '18 at 3:02
  • $\begingroup$ I was referencing this diagram. Any vector $x$ can be split into its row space and null space component and when $Ax$ is carried out, null space component is carried to zero vector, and the row space component is carried to the column space (because $Ax_r$ is in column space) $\endgroup$ – artha Aug 22 '18 at 3:19
  • $\begingroup$ did you perhaps read this from a book with better explanation? Gilbert Strang's? I really don't get it, most of it, and also notation of $Ax_r$ $\endgroup$ – mathmaniage Aug 22 '18 at 8:47
  • $\begingroup$ $x_r$ is the row space component from the image that I linked. "Linear Algebra and its Applications by Gilbert Strang" is the book and it is pretty widely known book and I recommend it. And now regarding the explanation, when you are multiplying a vector $x$ by a $m$x$n$ matrix $A$, you agree that the product $Ax$ is in column space right? Now the vector $x$ itself is of m-dimensions just like any row vector or Null vector of $A$. And also row space and null space of $A$ are orthogonal to each with the dimensions $r$ and $m-r$, where $r$ is the rank....coninued $\endgroup$ – artha Aug 22 '18 at 9:47
  • $\begingroup$ .... So the row space and null space combined, span the whole $R^m$ space. So any m-dimensional vector $x$ can be split into its row-space component ($x_r$)and null space component ($x_n$), writing $x = x_r + x_n$. So $Ax = Ax_r + Ax_n = Ax_r + 0 = Ax_r$. And this $Ax_r$ is in column space. So what happened is that this matrix $A$ took a vector in its row space and transformed it into a vector in its column space. If you still have any doubts, I would suggest you to read the above book for stronger fundamentals as these would be essential to move forward in understanding linear algebra. $\endgroup$ – artha Aug 22 '18 at 9:52

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