Chapter 1

Chapter 1


A linear equation in the variables $x_1, x_2, ... , x_n$ is an equation that can be written in the form

\begin{equation} a_1 x_1 + a_2 x_2 + ... + a_n x_n = b \end{equation}

A system of linear equations is a collection of linear equations involving the same variables.

A solution of the system is a list of numbers that makes each equation true.

The set of all solutions is the solution set.

Two systems are equivalent if they have the same solution set.

Linear Combination

Let $\bar{v}_1 , \bar{v}_2, ... , \bar{v}_p$ be vectors in $\mathbb{R}^n$.
Let $c_1, c_2, ..., c_p$ be scalars.

\begin{align} y=c_1\bar{v}_1 + c_2\bar{v}_2 + ... + c_p \bar{v}_p \end{align}

is the linear combination of $\bar{v}_1 , \bar{v}_2, ... , \bar{v}_p$


Let $\bar{v}_1 , \bar{v}_2, ... , \bar{v}_p \in \mathbb{R}^n$. The **span** of $\bar{v}_1 , \bar{v}_2, ... , \bar{v}_p$ is the set of all linear combinations of$\bar{v}_1 , \bar{v}_2, ... , \bar{v}_p$

Linear Independence

The vectors $\bar{v}_1,...,\bar{v}_p$ in $\mathbb{R}^n$ are linearly independent if $x_1 \bar{v}_1 + ... + x_p \bar{v}_p = 0$ has only the trivial solution.


Theorem 1

Each matrix is row equivalent to one and only one reduced echelon matrix.

Theorem 2

A linear system is consistent if and only if the right most column of the augmented matrix is not a pivot. Furthermore, if a linear system is consistent then the solution set is unique if there are no free variables and infinite if there is at least one free variable.

Theorem 3

If $A$ is and $m \times n$ matrix with columns $\bar{a}_1, \bar{a}_2, ... ,\bar{a}_n$ and if $\bar{b}$ is in $\mathbb{R}^m$, the matrix equation

\begin{align} A \bar{x} = \bar{b} \end{align}

has the same solution set as the vector equations

\begin{align} x_1\bar{a}_1 + x_2\bar{a}_2+ ... + x_n \bar{a}_n = \bar{b} \end{align}

which also has the same solution set as the augmented matrix

\begin{align} [\bar{a}_1 \text{ } \bar{a}_2 ... \bar{a}_n \text{ } \bar{b} ] \end{align}

Theorem 4

Let $A$ be an $m \times n$ matrix. Then the following statements are logically equivalent. That is, for a particular $A$, either they are all true statements or they are all false.

  • For each $\vec{b}$ in $\mathbb{R}^m$, the equation $A\vec{x} = \vec{b}$ has a solution.
  • Each $\vec{b}$ in $\mathbb{R}^m$ is a linear combination of the columns of $A$.
  • The columns of $A$ span $\mathbb{R}^m$.
  • A has a pivot position in every row.

Theorem 5

If $A$ is an $m \times n$ matrix, $\vec{u}$ and $\vec{v}$ are vectors in $\mathbb{R}^n$, and $c$ is a scalar, then:

  • $A(\vec{u}+\vec{v}) = A\vec{u}+A\vec{v}$ ;
  • $A(c\vec{u}) = c(A\vec{u})$.

Theorem 6

Suppose the equation $A\vec{x} = \vec{b}$ is consistent for some given $\vec{b}$ and let $\vec{p}$ be a solution. Then the solution set of $A\vec{x} = \vec{b}$ is the set of all vectors of the form $\vec{w} = \vec{p}+\vec{v}_h$, where $\vec{v}_h$ is any solution of the homogeneous equation $A\vec{x} = 0$.

Theorem 7

An indexed set $S = {\vec{v}_1, ..., \vec{v}_p}$ of two or more vectors is linearly dependent if and only if at least one of the vectors in $S$ is a linear combination of the others. In fact, if $S$ is linearly dependent and $\vec{v}_1\not= 0$, then some $\vec{v}_j$ (with $j > 1$) is a linear combination of the preceding vectors, $\vec{v}_1, ..., \vec{v}_{j-1}$.

Theorem 8

If a set contains more vectors than there are entries in each vector, then the set is linearly dependent. That is, any set ${[\vec{v}_1, ...,\vec{v}_p]}$ in $\mathbb{R}^n$ is linearly dependent if $p > n$.

Theorem 9

If a set $S = {[\vec{v}_1, ...,\vec{v}_p]}$ in $\mathbb{R}^n$ contains the zero vector, then the set is linearly dependent.

Let $S$ be a set of vectors such that $\vec{0}$ is an element of $S$.
$S =${$\vec{v}_1, \vec{v}_2, ..., \vec{v}_m, \vec{0}$}
$A = [\vec{v}_1 \vec{v}_2 ... \vec{0}]$
Consider $A\vec{x}=\vec{0}$, the column of $\vec{0}$ cannot be a pivot, therefore, it must be free. Consequently, the columns of $A$ are linearly dependent. $S$ is linearly dependent.

Linear Transformation

A transformation (or mapping) T is linear if:
(i) T(u + v) = T(u) + T(v) for all u,v, in the domain of T
(ii) T(cu) = cT(u) for all scalars c and all u in the domain of T.

If $T$ is a linear transformation, then
for all vectors $\vec{u}$, $\vec{v}$ in the domain of $T$ and all scalars $c$, $d$.

Parametric Vector Form

Parametric Vector form is the explicit description of the plane as the set spanned by u and v.
Parametric form:

x=s*u+t*v (s,t,$\mathbb{R}$)

Homework Problems

Section 1.4
31) Let $A$ be a $3 \times 2$ matrix. Explain why the equation $A \vec{x} = \vec{b}$ cannot be consistent for all $\vec{b}$ in $\mathbb{R}^3$.
The matrix $A$ has more rows than it has columns. This makes it impossible for each row to contain a pivot position. As per Theorem 4, this means that $A \vec{x} = \vec{b}$ cannot be consistent for all $\vec{b}$ in $\mathbb{R}^3$. (i.e. There is not a solution for every $\vec{b}$ in $\mathbb{R}^3$ using the equation $A \vec{x} = \vec{b}$.)

Section 1.6
6) We are given $Al_2O_3+C=Al + CO_2$. From here, we assign the matrix $\vec{X}=\begin{bmatrix}Al\\O\\C\end{bmatrix}$ to the two sides of the equation to get the matrix $\begin{bmatrix}2\\3\\0\end{bmatrix}X_1+\begin{bmatrix}0\\0\\1\end{bmatrix}X_2=\begin{bmatrix}1\\0\\0\end{bmatrix}X_3+\begin{bmatrix}0\\2\\1\end{bmatrix}X_4$.
We move the $X's$ to one side to get: $\begin{bmatrix}2\\3\\0\end{bmatrix}X_1+\begin{bmatrix}0\\0\\1\end{bmatrix}X_2+\begin{bmatrix}-1\\0\\0\end{bmatrix}X_3+\begin{bmatrix}0\\-2\\-1\end{bmatrix}X_4=\begin{bmatrix}0\\0\\0\end{bmatrix}$
This gives us the augmented matrix:
$\begin{bmatrix}�1362�2&0&{-1}&0$0\\3$0$0${-2}&0\\0&1&0&{-1}&0�1363�\end{bmatrix}$ which row reduces to $\begin{bmatrix}�1364�2&0&{-1}&0$0\\0$1$0${-1}&0\\0&0&{\frac{3}{2}}&{-2}&0�1365�\end{bmatrix}$
From here, we know that

We assign:

Therefore, the balanced equation is:
$2Al_2O_3+3C=4Al + 3CO_2$.

Section 1.8
25) Given $\vec{v} \neq 0$ and $\vec{p}$ in $\mathbb{R}^n$, the line through $\vec{p}$ in direction $\vec{v}$ has the parametric equation $\vec{x}=\vec{p}+t\vec{v}$. Show that linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ maps this line onto another line or a single point.
First, just attempt to apply the transformation to the equation, yielding $T(\vec{x})=T(\vec{p}+t\vec{v})$. Notice that since our transformation is linear, our definitions apply and the following manipulations take place:
Notice this equation holds the form of a point if $t=0$, and a parametric line otherwise. $\blacksquare$

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