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My question is related to what I learn watching the MIT course of Prof. Strang.

Suppose I have a matrix $m \times n$ with non-independent columns. The column space will be in $\Bbb R^m$ while the null space will be in $\Bbb R^n$.

For the column space, I have a $m$-dimensional vector, so it is natural that it will be in $\Bbb R^m$.

For the null space it is explained that the space is in $\Bbb R^n$ because the solution vector of $x$ $(Ax = 0)$ is of dimension $n$.

So, if I understand it correctly, the column space is defined based on vectors from the original matrix, while the null-space is defined not based on our original vectors, but on vector $x$. What confuses me further, is that when we construct a complete solution, it is in the form of particular solution plus the null space solutions. But in the complete solution, the vectors are m-dimensional (both particular and null spaces solutions). How should this be reconciled with the fact that the null space solution is in $\Bbb R^n$? Is it somehow related to the nulls in vectors we have in null-space part of the complete solution?

Many thanks!

UPDATE: It was an erroneous statement that "in the complete solution, the vectors are m-dimensional (both particular and null spaces solutions)". The complete solution is a combination of a particular solution (in $\Bbb R^n$) and null space solutions (also in $\Bbb R^n$). So, the vectors in the original matrix are in $\Bbb R^m$ and the solution vectors (including those in the null space) are in $\Bbb R^n$.

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  • $\begingroup$ what is a "whole solution"? A matrix $m\times n$ with real coefficients represent a linear map $T$ with domain $\Bbb R^n$ and codomain $\Bbb R^m$. The null space of the matrix is the kernel of $T$, that is, the subspace $A\subseteq\Bbb R^n$ such that if $v\in A$ then $Tv=0$. $\endgroup$ Commented Jul 28, 2017 at 12:42
  • $\begingroup$ by "whole solution" I meant "complete solution" (I fixed this) $\endgroup$ Commented Jul 28, 2017 at 12:47
  • $\begingroup$ This playlist by the youtuber 3blue1brown might be of help to you. In my opinion, it's a good way to spend a short afternoon regardless of your relationship with linear algebra, but if you're just beginning, I believe the benefit might be substantial. $\endgroup$ Commented Jul 28, 2017 at 12:53
  • $\begingroup$ Thanks. Yes, I know about this nice resource. But it does not address my specific question... $\endgroup$ Commented Jul 28, 2017 at 13:08
  • $\begingroup$ I actually have trouble to understand your question. " for the vector space is defined based on vector as is, while for the null-space we need yet to find a solution?" What does this first part mean? Which vector space are you talking about and how is it defined "based on vector as is"? Also: But the dimension of the complete solution is m". What should that mean? Any solution vector has $n$ entries. $\endgroup$ Commented Jul 28, 2017 at 13:15

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A solution to the equation $$Ax=b$$ is a vector in $\mathbb{R}^n $ because only then the matrix-vector multiplication $Ax$ is defined. Don't confuse the solution of the equation $Ax=b $ for a given vector $ b \in \mathbb{R}^m$ with the solution of the multiplication $Ax$ with a given vector $x \in \mathbb{R}^n$! The first on will be $n $- dimensional, the second one $m$- dimensional.

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  • $\begingroup$ OK, I see. Are the column space and null space are two incompatible concepts because they are in different R. If so, how then we use them together for the complete solution? $\endgroup$ Commented Jul 28, 2017 at 14:07
  • $\begingroup$ We don't! The 'complete solution ' as you call it is the sum of the nullspace wih one particular solution of the equation $ Ax = b $. Both have dimension $n $, as stated in the answer. The vector $b $ has dimension $m $ and is in the column space. $\endgroup$ Commented Jul 28, 2017 at 14:10
  • $\begingroup$ Ah...........Got it. For some reason I thought that the particular solution is related to column space. But it is not at all. So, how you would define the relationship between column space and null space? $\endgroup$ Commented Jul 28, 2017 at 14:21
  • $\begingroup$ There is a dimensional relation. If you know the definition of the dimension of a subvectorspace: the dimension of the nullspace plus the dimension of the column space equals $n$ $\endgroup$ Commented Jul 28, 2017 at 14:47
  • $\begingroup$ @student: There is a relationship -- each solution vector can be interpreted as the coefficients of a linear combination expressing $b$ as a linear combination of the columns of $A$. (and thus witnessing $b$ as an element of the column space) $\endgroup$ Commented Jul 28, 2017 at 14:54

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