Solving Ax = b: Row Reduced Form R

May 26, 2019

Remember we have talked about other values of b on lecture 6. Today we will explore further how a non-zero $\mathbf b$ lead to not-a-space solution, and discuss what values of $\mathbf b$ can have a solution, many solutions, or no solution, extending from lecture 1 and lecture 7.

Solvability conditions on b

Let’s again use the example:

$A=\begin{bmatrix} 1 & 2 & 2 & 2\\ 2 & 4 & 6 & 8\\ 3 & 6 & 8 & 10 \end{bmatrix}\tag{1}$

Actually not many b’s can have a solution. You can try throw some random numbers in. We have known that the r3 of $A$ is r1 + r2. When we have a much more larger matrix, that will not come in a very clear way, say maybe r8 = 2*r4+1/2*r. So we need elimination to discover what exactly the relations on rows of $A$ and then find out what conditions we need for $\mathbf b$ . Augment the matrix $A$ with $\mathbf b$ , and do elimination real quick:

$\left[\left.\begin{matrix} 1 & 2 & 2 & 2\\ 2 & 4 & 6 & 8\\ 3 & 6 & 8 & 10\ \end{matrix}\right\rvert \begin{matrix} b_1\\ b_2\\ b_3 \end{matrix} \right]\rightarrow...\rightarrow\left[\left.\begin{matrix} 1 & 2 & 2 & 2\\ 0 & 0 & 2 & 4\\ 0 & 0 & 0 & 0\ \end{matrix}\right\rvert \begin{matrix} b_1\\ b_2-2b_1\\ b_3-b_2-b_1 \end{matrix} \right]$

By elimination we see that the constant vector $\mathbf b$ must follow the rules on the augemented side. Picking the zero row is most convenient to phrase the condition. If and only if $b_3-b_2-b_1=0$ , we have a solution to $A\mathbf x=\mathbf b$ . A particular one will be $\mathbf b=\begin{bmatrix}1\\5\\6\end{bmatrix}$ . The zeros rows are conditions for solving the equations.

Note that this is equivalent to say $\mathbf b$ is in $C(A)$ which we have discussed before(idk formally why yet, probably this is a way of checking if b is in the column space of A).

Solutions to Ax=b if there’s one

To find the solution to $A\mathbf x=\mathbf b$ , we first set the free variables to zero to get $\mathbf x_{particular}$ and then get $\mathbf x_{null}$ by setting one free variable as 1 each time. For our example $A$ , we let $x_2=x_4=0$ .

$\begin{align} 2x_3&=3\Rightarrow x_3=3/2\\ x_1+2x_3&=1\Rightarrow x_1=-2\\ \end{align}$

then

$\mathbf x_{particular}= \begin{bmatrix} -2\\ 0\\ 3/2\\ 0 \end{bmatrix}$

But why we need to add the solutions in nullspace? It’s because

$\begin{align} A\mathbf x_{particular}&=\mathbf b\\ A\mathbf x_{null}&=0\\ A(\mathbf x_{particular}+\mathbf x_{null})&=\mathbf b \end{align}$

Copy the null solutions from the last lecture notes,

$\mathbf x_{complete}=\begin{bmatrix} -2\\ 0\\ 3/2\\ 0 \end{bmatrix}+c_1\begin{bmatrix} -2\\ 1\\ 0\\ 0 \end{bmatrix}+c_2\begin{bmatrix} 2\\ 0\\ -2\\ 0 \end{bmatrix}$

Note that we don’t have a constant in front of $\mathbf x_{particular}$ because it solves $A\mathbf x=\mathbf b$ for the particular $\mathbf b$ . Remember the null solution is a nullspace. But this solution to the particular $\mathbf b$ does not have a space. It’s the plane formed by the $\mathbf x_{null}$ shifted by the vector $\mathbf x_{particular}$ . It does not go through the origin.

Rank

$r$ is ranks and equal to the number of pivots in the matrix.

Full Column rank: r=n

Full column rank means there’s a pivot in every column. For example

$A=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ 0 & 0 & 0 \end{bmatrix}$

In this case the $N(A)$ is just the $\mathbf 0$ because we have no free variables can be assigned to the nullspace. The solution to $A\mathbf x=\mathbf b$ is just $\mathbf x_{particular}$ , if there’s one, determined by the checking solvability technique mentioned above. We have either 0 or 1 solution.

In addition, the rref(A) or $R$ for full column rank matrix will always be $\begin{bmatrix} I\\ 0\\ \end{bmatrix}$ .

Full Row rank: r=m

Every row will have a pivot. For example

$A=\begin{bmatrix} 1 & 2 & 3 & 0\\ 4 & 5 & 6 & 0\\ 7 & 8 & 9 & 0\\ \end{bmatrix}$

You see first there will not be any zero rows after elimination. This promises us there will always be a solution. Also since we have at least one free variable in this case, there will be infinitely many solutions(check out how to find the null solutions). And the rref(A) will be $\begin{bmatrix} I & F\\ \end{bmatrix}$ (Or there can be a mix between columns).

Full rank r=m=n

$A=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\\ \end{bmatrix}$

Since $r=n$ , no free variables, the nullspace for this matrix is just zero only. Since $r=m$ , no zero rows, there won’t be any constraint on $\mathbf b$ . There will always be one and only one solution to full rank matrix. rref(A) = $I$ .

r<m, r<n

The $R$ will be $\begin{bmatrix} I & F\\ 0 & 0 \end{bmatrix}$ . The $A$ in (1) is this case. Remember from last lectures we may not have a solution or there’re infinitely many solutions to this one.