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?

  • $\begingroup$ Do you know what it means to say that a set of vectors spans $\mathbf R^n?$ $\endgroup$ – saulspatz Mar 24 '18 at 1:43
  • $\begingroup$ @saulspatz Hi thanks for the quick reply. When we say a set of vectors are spanning Rn, I think we are equivalently saying that these are vectors are linearly independent and we exactly have n such vectors. $\endgroup$ – Nicholas Humphrey Mar 24 '18 at 1:46
  • $\begingroup$ I think I'm close to get the idea of splitting X. Let's say Ax = b, and A is invertible. Axn=0 of course only has a zero solution. And Axr=b=Ax. What I don't understand is why he says "The row space component goes to the column space". What does "go to" mean? Very confusing... $\endgroup$ – Nicholas Humphrey Mar 24 '18 at 1:50
  • $\begingroup$ No, that's not really it. When we say that a set of vectors spans a space, it means that every vector in the space can be written as a linear combination of the vectors in the spanning set. A basis is a spanning set that is also linearly independent. That's probably what's confused you. It is true that a linearly independent set of $n$ vectors will be a spanning set, but that's not the definition, which is what the author is refering to. $\endgroup$ – saulspatz Mar 24 '18 at 1:51
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    $\begingroup$ What is a "row column vector"? Is it a row vector or a column vector?? $\endgroup$ – bof Mar 24 '18 at 2:14

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.

  • $\begingroup$ Thanks. I think I know what's blocking me, it's the n=n-r+r part. I know this fact but do not really understand it. I'll go back to Chapter 3 and review the material. Hopefully I can make the connection. $\endgroup$ – Nicholas Humphrey Mar 24 '18 at 2:15
  • $\begingroup$ Ah I finally get it, pretty easy actually. Since A is m*n, X's columns must be of Rn, and because the row space and null space of A span Rn, X's columns are linear combinations of row space and null space of A. This means that X can be split into two vectors, one of row space of A and another of the null space of A. $\endgroup$ – Nicholas Humphrey Mar 24 '18 at 12:54

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