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Why is the "column space" on the vertical in a matrix? In my mind the column space is that space that the vectors in the matrix have created. I mean, for example take the equations:

3x + 4y = 5

2x + 8y = 6

Then the matrix will be:

\begin{pmatrix} 3 & 4 \\ 2 & 8 \end{pmatrix}

But why is the space defined by this matrix on the vertical?

Aren't the two vectors:

3i + 4j

and

2i + 8j 

defining the space we're working in?

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There is also a space defined by the rows of the matrix and it is called (unsurprisingly) the "row space". When you construct a matrix from a linear system of equations, you are indeed constructing a matrix "row by row" and not "column by column" in the sense that each equation defines a row and not a column and so it might seem that you shouldn't care about the columns. However, when you begin solving the equation, you see that the columns also play an important role. For example, $Ax = b$ will have a solution if and only if $b$ belongs to the span of the columns of $A$ (the column space of $A$). Both spaces are important and are related by duality and/or the fact that you can always convert rows to columns by performing the transpose operation.

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  • $\begingroup$ "important row" ​ -> ​ "important role" ​ ​ ​ ​ $\endgroup$ – user57159 Jun 11 '16 at 19:44
  • $\begingroup$ @RickyDemer Freudian slip, corrected, thanks! $\endgroup$ – levap Jun 11 '16 at 20:08
  • $\begingroup$ So what is the transpose doing geometrically speaking? $\endgroup$ – user963319 Jun 12 '16 at 22:02
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    $\begingroup$ @user963319 If you want to understand transpose of matrices geometrically, it might be useful to read math.stackexchange.com/questions/37398/…. If you want to understand the relation of the transpose to equation solving and the various different spaces associated to a matrix (row space, column space, left kernel, right kernel), take a look at Strang's linear algebra book. $\endgroup$ – levap Jun 13 '16 at 10:04
  • $\begingroup$ I'm going through that book right now :) But I think it's best for me to view it as a geometry problem. I can imagine and internalize that perspective better I'd like to think. $\endgroup$ – user963319 Jun 13 '16 at 16:19
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I think what you are meaning to ask is:

Why is the collumn space the range of the matrix, in the appropriate bases?

Otherwise, the collumn space is just a definition per se.

Therefore, we must see who is the image of the basis. That is, compute

$\begin{pmatrix}a_{1,1} & a_{1,2} & a_{1,3} &\cdots & a_{1,m} \\ a_{2,1} & a_{2,2} & a_{2,3} &\cdots & a_{2,m} \\ \cdots & \cdots & \cdots &\cdots & \cdots \\ \cdots & \cdots & \cdots &\cdots & \cdots \\ a_{n,1} & a_{n,2} & a_{n,3} &\cdots & a_{n,m} \end{pmatrix} \cdot \begin{pmatrix}0 \\ \cdots \\ 1 \\ \cdots \\ 0 \end{pmatrix},$

where the $1$ appears on the $i-$th coordinate. But note that this is precisely

$\begin{pmatrix}a_{1,i} \\ a_{2,i} \\ \cdots \\ a_{(n-1),i} \\ a_{n,i} \end{pmatrix}.$

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  • $\begingroup$ I agree. When you do Ab=x, then i will multiply column 1, and j column 2... The question then can be formulated as: why do we choose to transpose the column vector x in Ax=b? $\endgroup$ – user963319 Jun 12 '16 at 21:06

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