I'm reading Strang's Introduction to Linear Algebra, and cannot get why I can split every vector into a row vector and a null space vector.
In fact, I don't why the book suddenly comes to this conclusion. Half a page before (Page 200), the books explains that if $A$ is invertible then $Ax=b$ has a unique solution, which is easy to understand. Then he mentions "With bases for the row space and nullspace, we have $r+(n-r)=n$ vectors. This is the right number. Those $n$ vectors are independent. Therefore they span $\mathbb{R}^n$". OK that's easy, too.
Then he suddenly comes to the conclusion that each $x$ is the sum of $x_n$ and $x_r$, in which $x_r$ is a row space vector, $x_n$ is a nullspace vector.
My thought is, from previous chapters, if $Ax=b$ and $A$ is invertible, then $X=X_n+X_p$, with $A$ invertible then $X_n$ is zero vector, and $X_p$ is the sole solution. This, however, seems to have nothing to do with what he is talking here.
Any thought how should I approach this problem?
To say that a set $S$ of vectors spans $\mathbf R^n$ is to say that every vector in $\mathbf R^n$ can be written as a linear combination of vectors in $S$. Now,$\mathbf R^n$ is $n-$dimensional which means it has a basis of $n$ elements. Recall that a basis is a set of linearly independent vectors that spans the space. Furthermore, if there is a basis with $n$ elements, then there are two important facts which have surely been covered in your textbook. First, every linearly independent set of $n$ vector is a basis (that is it spans the space.) Every spanning set of $n$ vectors is a basis (that is, it is linearly independent.)
You should find these facts in your text, and review the relevant material, I would say. They are really fundamental.
Once you realize that the $r$ basis vectors for the row space, together with the $n-r$ basis vectors for the null space span $\mathbf R^n,$ the statement should be evident.