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I know what is the column space of a matrix: it is basically the subspace formed by the linear combinations of the columns (vectors) of a matrix. From wikipedia, we have the following nice picture:

enter image description here

Which shows up the vectors that are used to make any linear combination to form the column space.

So, my problem is not strictly related to the concept of column space of a matrix, but why is it useful?

Before knowing the concept of column space, I already was used to find, for example, the RREF of a matrix, to see if the columns for a basis or are linearly independent or whatever... What's in the connection between column space and all those things?

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    $\begingroup$ Well, the column space of your matrix is exactly the image of your matrix. Seems to me this is pretty important. $\endgroup$ – Timbuc Mar 27 '15 at 13:46
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    $\begingroup$ If $x$ is the column vector with components $x_1,x_2,...,x_n$, then $Ax=x_1a_1+a_2x_2+...+x_na_n$, where $a_i$ are the columns of $A$. $\endgroup$ – Nathanson Mar 27 '15 at 13:47
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An important property:

The linear system $Ax=b$ has a solution if and only if $b$ belongs to the column space of $A$.

Since linear systems of equations arise often in practice (particularly when working with computers), knowing when a linear system of equations has a solution may be very useful.

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The columns of a given matrix span the range of that matrix. In other words, if $(a_1,...,a_n)$ are the column vectors of $A$, then $Range(A)=span(a_1,...,a_n)$. The number of linearly independent column vectors gives thus the rank of the matrix $A$. And knowing whether $Ax=b$ has a solution, i.e. if $b\in Range(A)$, we only have to check whether $b\in span(a_1,...,a_n)=Col(A)$.

Through the rank-nullify theorem this also gives information of the kernel and injectivity of the matrix, and thus uniqueness of solutions whenever they exist. The Rank-Nullity theorem states that $Rank(A)+Null(A)=dim(domain(A))$, where $Null(A)$ denotes here for short the dimension of the kernel of $A$ (or the null space). Thus if $A$ has the same amount of linearly independent column vectors as the domain's dimension, then the matrix is injective as a linear operator, so whenever $Ax=b$ is solvable then the solution is unique. If there are less linearly independent column vectors as the domain's dimension, then whenver that equation is solvable, there are infinitely many solutions.

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