Today we start Section 2.3. Continue reading Section 2.3 for next class. Lots of new concepts in 2.3, so read the text. Work through suggested exercises.
Homework 3 is due today at 11:55pm on WeBWorK.
Homework 4 will be on Gradescope. I'll give a quick demo on Wednesday.
No classes on Monday.
Math Help Centre: M-F 12:30-5:30 in PAB48/49 and online 6pm-8pm.
My next office hour is today 2:30-3:20 in MC130. Also on Monday, 3:30-4:00 in MC130.
Performing the following elementary row operations on the augmented matrix doesn't change the solution set:
Definition: A matrix is in row echelon form (REF) if it satisfies:
Definition: A matrix is in reduced row echelon form (RREF) if:
We can always use the elementary row operations to get a matrix into REF and RREF:
Note: Row echelon form is not unique, but reduced row echelon form is.
Gaussian elimination: This means to do row reduction on the augmented matrix until you get to row echelon form, and then use back substitution to find the solutions.
Gauss-Jordan elimination: This means to do row reduction on the augmented matrix until you get to reduced row echelon form, and then use back substitution to find the solutions.
Back substitution: We call the variables corresponding to a column with a leading entry the leading variables, and the remaining variables the free variables. We solve for the leading variables in terms of the free variables, and assign parameters $s$, $t$, etc. to the free variables.
Definition: For any matrix $A$, the rank of $A$ is the number of nonzero rows in its row echelon form. It is written $\rank(A)$. (We'll see later that this is the same for all row echelon forms of $A$.)
Note: The number of leading variables equals the rank of the coefficient matrix.
Theorem 2.2: Let $A$ be the coefficient matrix of a linear system with $n$ variables.
If the system is consistent, then
$$
\text{number of free variables} = n - \rank(A) .
$$
When there are 0 free variables, we have a unique solution.
When there are 1 or more free variables, we have infinitely many solutions.
Consistency: You can tell whether the system is consistent or inconsistent from the row echelon form of the augmented matrix:
Definition: A system of linear equations is homogeneous if the constant term in each equation is zero.
Theorem 2.3: A homogeneous system $[ A \mid \vec 0 \, ]$ is always consistent. Moreover, if there are $m$ equations and $n$ variables and $m < n$, then the system has infinitely many solutions.
Note: If $m \geq n$ the system may have infinitely many solutions or it may have only the zero solution.
Here is an applet for practicing row reduction.
We aren't covering solving systems over $\Z_p$.
Recall: A vector $\vv$ is a linear combination of vectors $\vv_1, \vv_2, \ldots, \vv_k$ if there exist scalars $c_1, c_2, \ldots, c_k$ (called coefficients) such that $$ c_1 \vv_1 + \cdots + c_k \vv_k = \vv . $$
Example: Is $\colll 4 8 6$ a linear combination of $\colll 4 5 6$ and $\colll 2 1 3$?
That is, can we find scalars $x$ and $y$ such that $$x \colll 4 5 6 + y \colll 2 1 3 = \colll 4 8 6 ?$$
Expanding this into components, this becomes a linear system $$ \begin{aligned} 4 x + 2 y\ &= 4 \\ 5 x + \ph y\ &= 8 \\ 6 x + 3 y\ &= 6 \end{aligned} $$ and we already know how to determine whether this system is consistent: use row reduction!
The augmented matrix is $$ \kern-8ex \bmat{rr|r} 4 & 2 & 4 \\ 5 & 1 & 8 \\ 6 & 3 & 6 \emat \quad \Leftarrow \small\text{Note that the vectors appear as the columns here.} $$ This has row echelon form (work omitted): $$ \bmat{rr|r} 1 & 1/2 & 1 \\ 0 & -3 & 6 \\ 0 & 0 & 0 \emat . $$ From this, we can already see that the system is consistent, so the answer is YES.
If we want to find $x$ and $y$, we can use back substitution (maybe first going to RREF), and we find that $x = 2$ and $y = -2$ is the unique solution. (Do this at home.)
Example: Is $\colll 4 8 8$ a linear combination of $\colll 4 5 6$ and $\colll 2 1 3$?
Solution: The augmented matrix $$ \bmat{rr|r} 4 & 2 & 4 \\ 5 & 1 & 8 \\ 6 & 3 & \red{8} \emat $$ has row echelon form $$ \bmat{rr|r} 1 & 1/2 & 1 \\ 0 & -3 & 6 \\ 0 & 0 & \red{2} \emat $$ and so the system is inconsistent and the answer is NO.
Theorem 2.4: A system with augmented matrix $[A \mid \vb \,]$ is consistent if and only if $\vb$ is a linear combination of the columns of $A$.
This gives a different geometrical way to understand the solutions to a system. For example, consider the following system from Lecture 7: $$ \href{javascript:divShow("system1");divHide("system2");divHide("system3")}{ \begin{aligned} \phm x + y\ &= 2\\ - x + y\ &= 4 \end{aligned} } $$ We already know that we can interpret this as finding the point of intersection of two lines in $\R^2$, and so in this case we get a unique solution ($x = -1$, $y = 3$).
But we can also interpret this as writing $\coll 2 4$ as a linear combination of $\coll 1 {-1}$ and $\coll 1 1$, which has a different geometric interpretation.
$ \qquad\begin{aligned} \phm x + y\ &= 2\\ - x + y\ &= 4 \end{aligned} $
$\quad x \coll 1 {-1} + y \coll 1 1 = \coll 2 4$
$\quad \query{x = -1, \quad y = 3}$
If $\span(S) = \R^n$, then $S$ is called a spanning set for $\R^n$.
Example: The vectors $\ve_1 = \coll 1 0$ and $\ve_2 = \coll 0 1$ are a spanning set for $\R^2$, since for any vector $\vx = \coll a b$ we have $$ a \coll 1 0 + b \coll 0 1 = \coll a b . $$ Another way to see this is that the augmented matrix associated to $\ve_1$, $\ve_2$ and $\vx$ is $$ \bmat{rr|r} 1 & 0 & a \\ 0 & 1 & b \emat $$ which is already in RREF and is consistent.
Similarly, the standard unit vectors in $\R^n$ are a spanning set for $\R^n$.
Example: Find the span of $\vu = \colll 1 2 3$.
Solution: The span consists of every vector $\vx$ that can be written as $\vx = s \vu$ for some scalar $s$. So it is the line through the origin with direction vector $\vu$.
Example: Find the span of $\vu = \colll 1 2 3$ and $\vv = \colll 4 5 6$.
Solution: The span consists of every vector $\vx$ that can be written as $$ \vx = s \vu + t \vv $$ for some scalars $s$ and $t$. Since $\vu$ and $\vv$ are not parallel, this is the plane through the origin in $\R^3$ with direction vectors $\vu$ and $\vv$.
Example 10-1: What is the span of $\coll 1 3$ and $\coll 2 4$? They are not parallel, so intuitively their linear combinations should fill out all of $\R^2$. On board: We can see this algebraically, by row reducing the augmented matrix $$ \bmat{rr|r} 1 & 2 & a \\ 3 & 4 & b \emat $$
Note: The word "span" is really just a fancy way of saying "all linear combinations of these vectors".
Question: What is $\span(\coll 1 2)$? What is $\span(\coll 1 2, \coll 2 4)$? $$\querytext{Both are a line through the origin in the direction $[1, 2]$.}$$
Question: We saw that $\span(\coll 1 0, \coll 0 1) = \R^2$. What is $\span(\coll 1 0, \coll 0 1, \coll 2 4)$? $$\querytext{Also $\R^2$.}$$
Question: What vector is always in $\span(\vv_1, \vv_2, \ldots, \vv_k)$? $$\querytext{The zero vector.}$$
Question: Find some vectors with span equal to $\{ \collll x {2x} {3x} {4x} \mid x \in \R \}$. $$\query{\span(\collll 1 2 3 4) = \{ \collll x {2x} {3x} {4x} \mid x \in \R \}}$$ $$\querytext{Also, $\span(\collll 2 4 6 8) = \{ \collll x {2x} {3x} {4x} \mid x \in \R \}$}$$
Question: Find some vectors with span equal to $\{ \coll x y \mid y = x + 1 \}$.