# What is the difference between the rowspace and the columnspace in linear algebra?

The question is essentially all in the title, What is the difference between the rowspace and the columnspace?

Additionally, do they have the same solution space?

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I think the definitions of row space and column space are simple and pretty clear. Given a matrix $A$, denote $C(A)$ and $R(A)$ as its column space and row space, respectively. Because $R(A)=C(A^T)$, a row space can always be converted to a column space. As such, we usually consider column spaces. – Shiyu Sep 14 '11 at 3:21

I had hoped that the other answer would answer all of your questions, but as I'm pretty sure my last answer to your last question was the source of this question (which is the proper way for things to go - much better than trying to fit everything into one question).

In general, row spaces and column spaces can be very different. For non-square matrices, they will be in different dimension ambient spaces (although the dimension of the subspace spanned by each will be the same - the dimension of this subspace is usually called the rank of the transformation). And so this is very, very different.

Now, when you refer to 'the values that you can set the syste equal to and have a valid solution for,' you imply more than just a matrix. In particular, you assume a whole equation, and this is important. I suspect that you refer to cases like $\left( \begin{array}{cc} a & b \\ c&d \\ e&f \end{array} \right) \left( \begin{array}{cc} x\\ y \\ \end{array} \right) = \left( \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right)$. In these cases, the 'sets of $\left( \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right)$ ' that have solutions, known as the Image of the transformation (this is an important word - remember Image) is the column space. I used a non-square matrix to show it very easily: the dimension is even different.

But here is something confusing: the Image is not affected by elementary row operations. Whoa - that's cool. Perhaps nonintuitive. In other words, the set of vectors that can be 'reached' by this matrix do not change. And here's something else important: these row operations do not even change the solution for each particular element in the Image. That's why they're so great - but this is weird. Changing the rows doesn't affect the column space, and though the coefficients of the x and y might change, these direct solution for any $x_1, x_2, x_3$ vector do not change. This is, I think, the confusing part, because column operations do change these solutions, even though the Image itself isn't effected.

So how does one get a case where column operations don't change the solutions themselves? By left multiplication. $\left( \begin{array}{ccc} x & y & z \end{array} \right) \left( \begin{array}{cc} a & b \\ c&d \\ e&f \end{array} \right) = \left( \begin{array}{cc} x_1 & x_2\end{array} \right)$. But this is just something that we usually don't touch - it looks funny to me, even. Why? Usually we deal with the familiar right multiplication and transpose things if we need to.

Alright, so I gave a quick guide through some of the ideas here. Now I want to recommend something to you, so that it feels more real. Just like in the last question, I advised you to directly solve the system before and after a couple changes, I advise you to try new and different changes, and see if vectors can still be hit (are still in the image) and if they can, see if the solutions are the same. Then you'll have quick reference for your thoughts on different actions you might try.

I didn't quite answer your question proper, but I tried to hit at what I think is the heart of the issue. Is that right?

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That was by far one of the best explanations of linear algebra I have found. Thanks very much. I hope that for the next century, people trying to learn linear algebra come to this site, and dig up this answer. – weezybizzle Sep 14 '11 at 4:45
GREAT answer,mixed! – Mathemagician1234 Sep 14 '11 at 4:49

The row space is the set of all possible linear combinations of the row vectors, e.g. vectors of the form $$c_{1}row_{1} + c_{2}row_{2} +...+c_{m}row_{m}$$

Where c is a constant & for some A such that $$A = \left( \begin{array}{ccc} a_{1,1} & \cdots & a_{1,n} \\ \vdots & \ddots & \vdots \\ a_{m,1} & \cdots & a_{m,n} \end{array} \right)$$

and $$r_{i} = \langle a_{i,1}, a_{i,2} ,\ldots a_{i,n}\rangle$$

The column space is linear combinations of the column vectors for the same matrix A e.g. $$c_{1}col_{1} + c_{2}col_{2} +...+c_{m}col_{m}$$

for $$col_{i} = \langle a_{1,i} , a_{2,i} ,\ldots , a_{m,i}\rangle$$

In more plain english, for any given matrix, make vectors from the columns of any matrix, and add them up with any scalar multiplier on each vector (the scalars may be different for each vector). The set of vectors you can create via that process is the column space. The row space is the same thing, but for row vectors.

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So in terms of their solution space, are they the same or different? – weezybizzle Sep 14 '11 at 2:59
The solution space for the column vectors are inside the row space, am I right to say this? – xenon Sep 14 '11 at 3:01
@weezybizzle Define what what you mean by solution space. Do you mean the space the row/column space lives in? In that case, column space is a subspace of $$\mathbb{R}^{m}$$ for some m x n matrix, and the row space is a subspace of $$\mathbb{R}^{n}$$ – Drew Christianson Sep 14 '11 at 3:10
thinking in terms of systems of linear equations, I am thinking of the values that you can set the system equal to and have a valid solutions for. Would that be the columnspace or the rowspace? – weezybizzle Sep 14 '11 at 3:17
The solutions of $Ax=0$ form a vector space called the nullspace of $A$; it is not the same as the row space of $A$, and it is not the same as the column space of $A$. Why don't you take a simple example, like $A=\pmatrix{0&1&0\cr0&0&0\cr}$, and work out for yourself what the row space is, what the column space is, and what the null space is, and see how they differ. – Gerry Myerson Sep 14 '11 at 4:44