Homework question:

There are multiple things I do not understand about this question. What is meant by linear combinations in this context? What is $rb1$ and $sb2$? (what do the r and s represent) Why are we getting two 4*1 columns? I understand very little about this question

Please use simple language, as you can probably tell I am not familiar with much mathematical terminology.

  • $\begingroup$ $b_1, b_2$ are going to be the basis vectors for the nullspace of $A$. The numbers $r,s$ are any scalars (in this context you may assume real numbers). I.e. $b_1, b_2$ are vectors with the property that for any $r,s\in\mathbb{R}$ you have $A(rb_1+sb_2) = 0$. By linear combination of $b_1$ and $b_2$ they mean some sum of the form $rb_1 + sb_2$ with $r$ and $s$ scalars. $\endgroup$ – JMoravitz Apr 8 '15 at 20:30

If they are asking this question, I would expect at this point you have learned at least one method to solving the equation $Ax = y$ with $A$ a matrix, and $x,y$ vectors (i.e. single column matrices). Here in this problem, they are taking it a step further to illustrate the problem where $y=0$ (the zero-vector).

Those solutions, $x$, to the equation $Ax = 0$ are said to reside in the nullspace of $A$ (also commonly referred to as the kernel of $A$) which is commonly denoted as $N(A)$ (or as $ker(A)$ respectively). (technically this is the right-nullspace, whereas the left nullspace is for the similar problem $xA = 0$). The nullspace itself satisfies the properties of being a vector space (meaning that for any $x_1, x_2\in N(A)$ you have $x_1+x_2\in N(A)$, you have $0\in N(A)$, and you have $\alpha x_1\in N(A)$ for any scalar $\alpha$).

The nullspace is closely related to other subspaces, as shown in theorems like the fundamental theorem of linear algebra.

As a vector space, the nullspace will have a minimal set of basis vectors, $b_1, b_2, \dots, b_k$ (where $k=\dim N(A)$) such that $N(A) = \text{span}\{b_1,b_2,\dots,b_k\}$. In other words, every vector $x\in N(A)$ can be written as a linear combination of the basis elements.

(note: the dimension of the nullspace is able to be any natural number up to the size of the matrix $A$ and is not necessarily always two)

(by linear combination we mean to say there exists some constants $\alpha_1,\dots,\alpha_k$ such that $x=\alpha_1 b_1 + \alpha_2 b_2 + \dots + \alpha_k b_k$. We call this a linear combination because the terms involved are all of the simple form $\alpha_i b_i$ and terms of a different form such as $b_1 b_2$, $b_1^2$, or $\sin(b_3)$ don't appear).

This exercise is asking you to find a basis for the nullspace of your particular matrix $A$. One such method which should have been learned early on is to set up the matrix:

$$\left[\begin{array}{cccc|c} 1 & 2 & -1 & 0 & 0\\ 1 & 3 & -1 & -2 & 0\\ -1 & 0 & 1 & -4 & 0\\ 2 & 3 & -2 & 2 & 0\end{array}\right]$$

and row reduce to get it into reduced row echelon form.

You will arrive at a matrix looking something like:

$$\left[\begin{array}{cccc|c} 1 & 2 & 0 & 2 & 0\\ 0 & 0 & 1 & -2 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\end{array}\right]$$

(note this is not the actual result of row reduction in this case but an example of what it might look like)

This tells you that a solution to the equation, $x = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix}$, will necessarily have: $x_1 = -2x_2 - 2x_4$ and $x_3 = 2 x_4$, with $x_2$ and $x_4$ as free variables. As such any solution will be of the form, $x_4\cdot \begin{bmatrix} -2 \\ 0 \\ 2 \\ 1\end{bmatrix} + x_2\cdot \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0\end{bmatrix}$.

These vectors, $\begin{bmatrix} -2 \\ 0 \\ 2 \\ 1\end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0\end{bmatrix}$ are precisely our basis vectors, $b_1, b_2$ for the nullspace, and you will have that $Ab_1 = 0, Ab_2 = 0$ and $A(rb_1 + sb_2) = 0$ for any scalars $r$ and $s$.

Again, to solve your exact problem as it is worded, set up your matrix with an extra column of zeroes on the right, row-reduce, and use that information to find in what way each entry of a solution vector depends on the other entries.

(it is useful to note that due to the fact that the righthand column will have all zeroes, after row reduction it will continue to have all zeroes. This is an important fact which is necessary for the nullspace to actually be a subspace and why finding the solutions to the equation $Ax=0$ is as important as it is)


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