Read Section 2.4 for next class. Work through recommended homework questions.
Midterm 1 is tomorrow 7-8:30pm. It covers until the end of Section 2.2, except for linear systems over $\Z_m$. A practice exam is available from the course home page. Last name A-Q must write in NS1, R-Z in NS7. See the missed exam section of the course web page for policies, including for illness.
Tutorials: No quiz, focused on review. Take advantage of them! No quizzes next week either.
Office hour: today, 12:30-1:30, MC103B.
Help Centers: Monday-Friday 2:30-6:30 in MC 106.
These lecture notes now available in pdf format as well, a day or two after each lecture. Be sure to let me know of technical problems.
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 $$ \begin{aligned} 4 x + 2 y &= 4 \\ 5 x + \ph y &= 8 \\ 6 x + 3 y &= 6 \end{aligned} \qquad\text{with augmented matrix}\qquad \bmat{rr|r} 4 & 2 & 4 \\ 5 & 1 & 8 \\ 6 & 3 & 6 \emat $$ and we already know how to determine whether this system is consistent: use row reduction!
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.
Example: $\span(\ve_1, \ve_2, \ldots, \ve_n) = \R^n$.
Example: The span of $\vu = \colll 1 2 3$ and $\vv = \colll 4 5 6$ 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 $\vv$.
Question: We saw that $\span(\coll 1 0, \coll 0 1) = \R^2$. What is $\span(\coll 1 0, \coll 0 1, \coll 2 4)$?
Question: What vector is always in $\span(\vv_1, \vv_2, \ldots, \vv_k)$?
Definition: A set of vectors $\vv_1, \ldots, \vv_k$ is linearly dependent if there are scalars $c_1, \ldots, c_k$, at least one of which is nonzero, such that $$ c_1 \vv_1 + \cdots + c_k \vv_k = \vec 0 . $$ Since at least one of the scalars is non-zero, the corresponding vector can be expressed as a linear combination of the others.
Example: $ \coll {-2} 4 - 2 \coll {-1} 2 + 0 \coll 5 6 = \coll 0 0 $, so the vectors $\coll {-2} 4$, $\coll {-1} 2$ and $\coll 5 6$ are linearly dependent.
Note that either of the first two can be expressed as a linear combination of the other, but the third one is not a linear combination of the first two.
Example: Are the vectors $\ve_1 = \coll 0 1$ and $\ve_2 = \coll 1 0$ linearly dependent?
Solution: If $c \, \ve_1 + d \, \ve_2 = \vec 0$ and $c \neq 0$, then $\ve_1 = -\frac d c \ve_2$, which is not possible. Similarly, if $d \neq 0$, then $\ve_2$ is a multiple of $\ve_1$. So the only way to have $c \, \ve_1 + d \, \ve_2 = \vec 0$ is with $c = d = 0$.
Theorem 2.5: The vectors $\vv_1, \ldots, \vv_k$ are linearly dependent if and only if at least one of them can be expressed as a linear combination of the others.
Proof: We've seen one direction. For the other, if $\vv_k = c_1 \vv_1 + \cdots c_{k-1} \vv_{k-1}$, then $c_1 \vv_1 + \cdots c_{k-1} \vv_{k-1} - \vv_k = \vec 0$, so the vectors are linearly dependent. The same argument works if it is a different vector that can be expressed in terms of the others.
Example: What about the vectors $\ve_1$, $\ve_2$ and $\coll 0 0$?
Solution: They are linearly dependent, since $$ 0 \coll 1 0 + 0 \coll 0 1 + 1 \coll 0 0 = \coll 0 0 . $$ Fact: Any set of vectors containing the zero vector is linearly dependent.
Definition: A set of vectors $\vv_1, \ldots, \vv_k$ is linearly independent if it is not linearly dependent.
Another way to say this is that the system $$ c_1 \vv_1 + \cdots + c_k \vv_k = \vec 0 . $$ has only the trivial solution $c_1 = \cdots = c_k = 0$.
This is something we know how to figure out! Use row reduction!
Example: Are the vectors $\vu = \colll {-1} 3 2$, $\vv = \colll 2 1 1$ and $\vw = \colll 6 {-4} {-2}$ linearly independent?
That is, does the system $$ c_1 \colll {-1} 3 2 + c_2 \colll 2 1 1 + c_3 \colll 6 {-4} {-2} = \vec 0 $$ have a non-trivial solution?
The augmented matrix is $$ \bmat{rrr|r} -1 & 2 & 6 & 0 \\ 3 & 1 & -4 & 0 \\ 2 & 1 & -2 & 0 \emat \qquad \text{which row reduces to} \qquad \bmat{rrr|r} -1 & 2 & 6 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \emat $$ So what's the answer? There are 3 variables and 2 leading variables (the rank is 2), so there is one free variable, which means there are non-trivial solutions. Therefore, the vectors are linearly dependent.
Example: Are the vectors $\vu = \colll {-1} 3 2$, $\vv = \colll 2 1 1$ and $\vw = \colll 6 {-4} {\red{3}}$ linearly independent?
That is, does the system $$ c_1 \colll {-1} 3 2 + c_2 \colll 2 1 1 + c_3 \colll 6 {-4} {\red{3}} = \vec 0 $$ have a non-trivial solution?
The augmented matrix is $$ \bmat{rrr|r} -1 & 2 & 6 & 0 \\ 3 & 1 & -4 & 0 \\ 2 & 1 & \red{3} & 0 \emat \qquad \text{which row reduces to} \qquad \bmat{rrr|r} -1 & 2 & 6 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & \red{1} & 0 \emat $$ So what's the answer? There are 3 variables and 3 leading variables (the rank is 3), so there are no free variables, which means there is only the trivial solution. Therefore, the vectors are linearly independent.
Example 2.24: Are the standard unit vectors $\ve_1, \ldots, \ve_n$ in $\R^n$ linearly independent?
Solution: The augmented matrix is $$ \bmat{cccc|c} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & & & & 0 \\ 0 & 0 & \cdots & 1 & 0 \emat $$ with $n$ rows and $n$ variables. The rank is $n$, so there is only the trivial solution. So the standard unit vectors are linearly independent.
Note: You can sometimes see by inspection that some vectors are linearly dependent, e.g. if they contain the zero vector, or if one is a scalar multiple of another. Here's one other situation:
Theorem 2.8: If $m > n$, then any set of $m$ vectors in $\R^n$ is linearly dependent.
Proof: The system is a homogeneous system with $m$ variables and $n$ equations. By Theorem 2.3, a homogeneous system with more variables than equations always has a non-trivial solution.
On whiteboard: An example like Example 2.25 in the text, and a discussion of Theorem 2.7.
.