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I am reading the book "Introduction to Linear Algebra" by Professor Gilbert Strang and either I have a problem to understand the difference (or similarity) between the complete solution and complements representation, or the information in the book is unclear.

Namely, in the chapter 3.4, the author introduces the complete solution in a very clear and understandable way. We can summarize the chapter in the following way: If we have a system $Ax=b$, the complete solution can be represented as $x=x_p+x_n$, where $x_p$ is one particular solution and $x_n$ is a linear combination of special solutions. The author points out that there is more then one particular solution - if we set the free variables to the values different from the one he uses, we will get a new particular solution for the same system.

Then, in the chapter 4.1, the author explains the orthogonality between the subspaces (column space and left nullspace, row space and nullspace) and he says that any solution $x$ for a system $Ax=b$ can be represented as $x=x_r+x_n$, where $x_r$ is a row space component and $x_n$ is a nullspace component. Then he states that every vector b in the column space comes from one and only one vector in the row space. He did not stress out if he referred to the system with invertible matrix $A$. It would not make sense anyway because he explains the whole problem of subspaces with respect to a rectangular matrix $A$.

The last statement confused me completely. I thought $x_p$ and $x_r$ were basically the same thing expressed in different ways. However, the fact the we can have many particular solutions (just take some other values for the free variables) conflicts with the bolded statement. It is obvious that $x_p$ is a vector from the row space but it is not the only particular solution that solves the system $Ax=b$ (assume the system is consistent). My claim that $x_p=x_r$ seems to be valid just if A is invertible, i.e. when there is just one particular solution.

I am not sure if the author wanted to say that the row component is not multiplied by any arbitrary constant (we have the same with the particular solution). In other words, perhaps he wanted to say:

$x=x_r+x_n=x_r+c_1x_{s_1}+c_2x_{s_2}...c_{n-r}x_{s_{n-r}}$

where $x_{s_1},x_{s_2},...,x_{s_{n-r}}$ are $n-r$ special solutions ($n$ is the number of columns, $r$ is the rank of matrix $A$). However, even if he wanted to say that, there is now way we can say that $b$ comes just from one vector in the row space since all particular solutions (row components) are in the row space of matrix $A$. Correct me if I am wrong.

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He says that any solution $x$ for a system $Ax=b$ can be represented as $x=x_r+x_n$, where $x_r$ is a row space component and $x_n$ is a nullspace component.

I checked and I think he meant that every $\mathbf{x} \in \mathbf{R}^n$ can be represented as $\mathbf{x}=\mathbf{x}_r+\mathbf{x}_n$. How?

The author stated this after he deduced that the row space $\mathbf{C}(A^T)$ and nullspace $\mathbf{N}(A)$ are orthogonal, which means they can't share any vector other than $0$, which means $\text{dim } \mathbf{C}(A^T) \cap \mathbf{N}(A)=0$. We also know that $\text{dim } \mathbf{C}(A^T)+ \text{dim }\mathbf{N}(A)=r+n-r=n$. Thus, $\mathbf{C}(A^T)+\mathbf{N}(A)$ is essentially $\mathbf{R}^n$ according to the identity $$\text{dim }(U+V)=\text{dim }U+\text{dim }V- \text{dim }(U \cap V)$$ for any subspaces $U,V$ of a vector space.

The last statement confused me completely. I thought $x_p$ and $x_r$ were basically the same thing expressed in different ways.

Every $\mathbf{x} \in \mathbf{R}^n$ can be represented as $\mathbf{x}=\mathbf{x}_r+\mathbf{x}_n$, but only $\mathbf{x}$ so $\mathbf{Ax=b}$ then $\mathbf{x}=\mathbf{x}_p+\mathbf{x}_n$. So $\mathbf{x}_r$ and $\mathbf{x}_p$ are not the same thing, i.e. not every vector in the row space is a solution to $\mathbf{Ax=b}$, and not every solution is in the row space.

The author stated that for every $\mathbf{x}$, if we multiply $\mathbf{Ax}$ we obtain a vector in a column space. Indeed, if let $\mathbf{x}=(x_i)_{n \times 1}$ and $\mathbf{a}_i$ as the $i$th column vector of $\mathbf{A}$ then we can represent $$\mathbf{Ax}=x_1\mathbf{a_1}+x_2\mathbf{a_2}+ \ldots+ x_n\mathbf{a_n},$$ which is a linear combination of vectors in the column space, which follows that $\mathbf{Ax}$ belongs to the column space. Thus, we can state that every $\mathbf{x} \in \mathbf{R}^n$ maps to a vector $\mathbf{b}$ in the column space ($\mathbf{b}$ is obtained by doing $\mathbf{Ax}$). We also know that $\mathbf{x}=\mathbf{x}_r+\mathbf{x}_n$ and $\mathbf{Ax}_n=0$ so that means $\mathbf{Ax}_r=\mathbf{b}=\mathbf{A(x_r+x_n)}=\mathbf{Ax}$.

Next, the author stated that

Every vector b in the column space comes from one and only one vector in the row space.

What it wants to say is that there can't be no two vectors $\mathbf{x}_r$ and $\mathbf{x}_r'$ from the row space so they map to the same vector $\mathbf{b}$ in the column space, i.e. $\mathbf{Ax}_r=\mathbf{Ax}_r'=\mathbf{b}$. Indeed, assume the contrary there is such two different vectors. This deduce $\mathbf{A(x_r-x_r')}=0$ so $\mathbf{x}_r-\mathbf{x}_r'$ is in the nullspace. But $\mathbf{x}_r-\mathbf{x}_r'$ is also in the row space (sum of two vectors in a same vector space belongs to that vector space). This follows $\mathbf{x}_r-\mathbf{x}_r'=0$, a contradiction since they are different.

One more thing, I believe the author meant that all of the above is true for any matrix $\mathbf{A}$, not just invertible $\mathbf{A}$ or rectangular $\mathbf{A}$. However, there is a invertible matrix $r \times r$ hiding inside $A$, i.e. after throwing out two nullspaces (in the figure of the book), we only see row space and column space. That's why the author wrote

From the row space to the column space, $\mathbf{A}$ is invertible.

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  • $\begingroup$ nice............... +1 $\endgroup$ – jeanne clement Oct 23 '17 at 22:08
  • $\begingroup$ Where did Prof. Strang prove $\text{dim }(U+V)=\text{dim }U+\text{dim }V- \text{dim }(U \cap V)$ in his book? $\endgroup$ – tchappy ha May 11 '19 at 9:42
  • $\begingroup$ I found the above formula in the challenge problem 43 on p.183. He didn't prove the above formula in his book. $\endgroup$ – tchappy ha May 11 '19 at 9:45

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