Solving $A\mathbf{x}=\mathbf{b}$ : row reduced form $R$

讲座摘要（lecture summary）​

When does $A\mathbf{x}=\mathbf{b}$ have solutions $\mathbf{x},$ and how can we describe those solutions?

Solvability conditions on $\mathbf{b}$​

We again use the example:

The third row of $A$ is the sum of its first and second rows, so we know that if $A\mathbf{x}\;=\;\mathbf{b}$ the third component of $\mathbf{b}$ equals the sum of its first and second components. If b does not satisfy $b_{3}=b_{1}+b_{2}$ the system has no solution. If a combination of the rows of $A$ gives the zero row, then the same combination of the entries of $\mathbf{b}$ must equal zero.

One way to find out whether $A\mathbf{x}=\mathbf{b}$ is solvable is to use elimination on the augmented matrix. If a row of $A$ is completely eliminated, so is the corresponding entry in b. In our example, row $3$ of $A$ is completely eliminated:

If $A\mathbf{x}=\mathbf{b}$ has a solution, then $b_{3}-b_{2}-b_{1}=0$. For example, we could choose $\mathbf{b}={\left[\begin{array}{l}{1}\\ {5}\\ {6}\end{array}\right]}.$

From an earlier lecture, we know that $A\mathbf{x}=\mathbf{b}$ is solvable exactly when $\mathbf{b}$ is in the column space $C(A)$. We have these two conditions on $\mathbf{b}$; in fact they are equivalent.

Complete solution​

In order to find all solutions to $A\mathbf{x}\;=\;\mathbf{b}$ we first check that the equation is solvable, then find a particular solution. We get the complete solution of the equation by adding the particular solution to all the vectors in the nullspace.

A particular solution​

One way to find a particular solution to the equation $A\mathbf{x}=\mathbf{b}$ is to set all free variables to zero, then solve for the pivot variables.

For our example matrix $A$ , we let $x_{2}=x_{4}=0$ to get the system of equa tions:

which has the solution $x_{3}=3/2,x_{1}=-2$ . Our particular solution is:

Combined with the nullspace​

The general solution to $A\mathbf{x}=\mathbf{b}$ is given by $\mathbf{\boldsymbol{x}}_{\mathrm{complete}}=\mathbf{\boldsymbol{x}}_{p}+\mathbf{\boldsymbol{x}}_{n},$ where $\mathbf{x}_{n}$ is a generic vector i n the nullspace. To see this, we add $A\mathbf{x}_{p}=\mathbf{b}$ to $A\mathbf{x}_{n}=\mathbf{0}$ and get $A\left(\mathbf{x}_{p}+\mathbf{x}_{n}\right)=\mathbf{b}$ for every vector ${\bf x}_{n}$ in the nullspace.

Last lecture we learned that the nullspace of $A$ is the collection of all combinations of the special solutions $\left[\begin{array}{r}{-2}\\ {1}\\ {0}\\ {0} \end{array}\right]$ and $\left[\begin{array}{r}{2}\\ {0}\\ {-2}\\ {1}\end{array}\right]$. So the complete solution to the equation $A{\mathbf{x}}={\left[\begin{array}{l}{1}\\ {5}\\ {6}\end{array}\right]}$ is:

where $c_{1}$ and $c_{2}$ are real numbers.

The nullspace of $A$ is a two dimensional subspace of $\mathbb{R}^{4}$ , and the solutions to the equation $A\mathbf{x}=\mathbf{b}$ form a plane parallel to that through $x_{p}={\left[\begin{array}{r}{\,\,-2}\\ {\,\,\,\,0}\\ {\,\,\,3/2}\\ {\,\,\,\,0}\end{array}\right]}.$

Rank​

The rank of a matrix equals the number of pivots of that matrix. If $A$ is an $m$ by $n$ matrix of rank $r_{.}$ , we know $r\leq m$ and $r\leq n$ .

Full column rank​

If $r=n$ , then from the previous lecture we know that the nullspace has dimension $n-r=0$ and contains only the zero vector. There are no free variables or special solutions.

If $A\mathbf{x}=\mathbf{b}$ has a solution, it is unique; there is either 0 or 1 solution. Examples like this, in which the columns are independent, are common in applications.

We know $r\,\leq\,m,$ so if $r\,=\,n$ the number of columns of the matrix is less than or equal to the number of rows. The row reduced echelon form of the matrix will look like $\begin{array}{r}{R\,=\,\left[\begin{array}{c}{I}\\ {0}\end{array}\right]}\end{array}$ For any vector $\mathbf{b}$ in $\mathbb{R}^{m}$ that’s not a linear combination of the columns of $\bar{A}$ , there is no solution to $A\mathbf{x}=\mathbf{b}$ .

Full row rank​

If $r\,=\,m_{.}$, then the reduced matrix $R\,=\,{\left[\begin{array}{l l}{I}&{F}\end{array}\right]}$ has no rows of zeros and so there are no requirements for the entries of $\mathbf{b}$ to satisfy. The equation $A\mathbf{x}=\mathbf{b}$ is solvable for every $\mathbf{b}$ . There are $n-r=n-m$ free variables, so there are $n-m$ special solutions to $A\mathbf{x}=\mathbf{0}$ .

Full row and column rank​

If $r=m=n$ is the number of pivots of $A$ , then $A$ is an invertible square matrix and $R$ is the identity matrix. The nullspace has dimension zero, and $A\mathbf{x}=\mathbf{b}$ has a unique solution for every $\mathbf{b}$ in $\mathbb{R}^{m}$ .

Summary​

If $R$ is in row reduced form with pivot columns first (rref), the table below summarizes our results.

Problems and Solutions（习题及答案）​

Problem 8.1: (3.4 #13.(a,b,d) Introduction to Linear Algebra: Strang) Explain why these are all false:

a) The complete solution is any linear combination of $\mathbf{x}_{p}$ and $\mathbf{x}_{n}$ .
b) The system $A\mathbf{x}=\mathbf{b}$ has at most one particular solution.
c) If $A$ is invertible there is no solution $\pmb{x}_{n}$ in the nullspace.

Problem 8.2: (3.4 #28.) Let

Use Gauss-Jordan elimination to reduce the matrices $\left[U\ 0\right]$ and $\textbf{[}U\textbf{c}\bigr]$ to $\left[R\,\mathrm{~}0\right]$ and $\left[R\,\textbf{d}\right]$ . Solve $R\mathbf{x}=\mathbf{0}$ and $R\mathbf{x}=\mathbf{d}$ .

Check your work by plugging your values into the equations $U\mathbf{x}=\mathbf{0}$ and $U\mathbf{x}=\mathbf{c}$ .

Problem 8.3: (3.4 #36.) Suppose $A\mathbf{x}=\mathbf{b}$ and $C\mathbf{x}\,=\,\mathbf{b}$ have the same (complete) solutions for every b. Is it true that $A=C$?