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Algebra 1 is the second math course in high school and will guide you through among other things expressions, systems of equations, functions, real numbers, inequalities, exponents, polynomials, radical and rational expressions.
This Algebra 1 math course is divided into 12 chapters and each chapter is divided into several lessons. Under each lesson you will find theory, examples and video lessons.
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Section 1.2. Problem 23: The figure shows that $\cos (\alpha)=v_1 /|v|$ and $\sin (\alpha)=$ $v_2 /|v|$. Similarly $\cos (\beta)$ ___is and $\sin (\beta)$ is __. The angle $\theta$ is $\beta-\alpha$. Substitute into the trigonometry formula $\cos (\alpha) \cos \overline{(\beta)+\sin }(\beta) \sin (\alpha)$ for $\cos (\beta-\alpha)$ to find $\cos (\theta)=v \cdot w /|v||w|$.
First blank: $w_1 /|w|$. Second blank: $w_2 /|w|$. Substituting into the trigonometry formula yields
$$
\cos (\beta-\alpha)=\left(w_1 /|w|\right)\left(v_1 /|v|\right)+\left(w_2 /|w|\right)\left(v_2 /|v|\right)=v \cdot w /|v||w| .
$$
If $A=L D U$ and also $A=L_1 D_1 U_1$ with all factors invertible, then $L=L_1$ and $D=D_1$ and $U=U_1$. “The three factors are unique.”
Derive the equation $L_1^{-1} L D=D_1 U_1 U^{-1}$. Are the two sides triangular or diagonal? Deduce $L=L_1$ and $U=U_1$ (they all have diagonal 1’s). Then $D=D_1$.
Notice that $L D U=L_1 D_1 U_1$. Multiply on the left by $L_1^{-1}$ and on the right by $U^{-1}$, getting
$$
L_1^{-1} L D U U^{-1}=L_1^{-1} L_1 D_1 U_1 U^{-1} .
$$
But $U U^{-1}=I$ and $L_1^{-1} L_1=I$. Thus $L_1^{-1} L D=D_1 U_1 U^{-1}$, as desired.
The left side $L_1^{-1} L D$ is lower triangular, and the right side $D_1 U_1 U^{-1}$ is upper triangular. But they’re equal. So they’re both diagonal. Hence $L_1^{-1} L$ and $U_1 U^{-1}$ are diagonal too. But they have diagonal 1’s. So they’re both equal to $I$. Thus $L=L_1$ and $U=U_1$. Also $L_1^{-1} L D=D_1 U_1 U^{-1}$. Thus $D=D_1$.
Suppose $\mathbf{S}$ and $\mathbf{T}$ are two subspaces of a vector space $\mathbf{V}$.
(a) Definition: The sum $\mathbf{S}+\mathbf{T}$ contains all sums $\mathbf{s}+\mathbf{t}$ of a vector $\mathbf{s}$ in $\mathbf{S}$ and a vector $\mathbf{t}$ in T. Show that $\mathbf{S}+\mathbf{T}$ satisfies the requirements (addition and scalar multiplication) for a vector space.
(b) If $\mathbf{S}$ and $\mathbf{T}$ are lines in $\mathbf{R}^m$, what is the difference between $\mathbf{S}+\mathbf{T}$ and $\mathbf{S} \cup \mathbf{T}$ ? That union contains all vectors from $\mathbf{S}$ and $\mathbf{T}$ or both. Explain this statement: The span of $\mathbf{S} \cup \mathbf{T}$ is $\mathbf{S}+\mathbf{T}$. (Section $3.5$ returns to this word “span.”)
(a) Let $\mathbf{s}, \mathbf{s}^{\prime}$ be vectors in $\mathbf{S}$, let $\mathbf{t}, \mathbf{t}^{\prime}$ be vectors in $\mathbf{T}$, and let $c$ be a scalar. Then
$$
(\mathbf{s}+\mathbf{t})+\left(\mathbf{s}^{\prime}+\mathbf{t}^{\prime}\right)=\left(\mathbf{s}+\mathbf{s}^{\prime}\right)+\left(\mathbf{t}+\mathbf{t}^{\prime}\right) \text { and } c(\mathbf{s}+\mathbf{t})=c \mathbf{s}+c \mathbf{t} .
$$
Thus $\mathbf{S}+\mathbf{T}$ is closed under addition and scalar multiplication; in other words, it satisfies the two requirements for a vector space.
(b) If $\mathbf{S}$ and $\mathbf{T}$ are distinct lines, then $\mathbf{S}+\mathbf{T}$ is a plane, whereas $\mathbf{S} \cup \mathbf{T}$ is not even closed under addition. The span of $\mathbf{S} \cup \mathbf{T}$ is the set of all combinations of vectors in this union. In particular, it contains all sums $\mathbf{s}+\mathbf{t}$ of a vector $\mathbf{s}$ in $\mathbf{S}$ and a vector $\mathbf{t}$ in $\mathbf{T}$, and these sums form $\mathbf{S}+\mathbf{T}$. On the other hand, $\mathbf{S}+\mathbf{T}$ contains both $\mathbf{S}$ and $\mathbf{T}$; so it contains $\mathbf{S} \cup \mathbf{T}$. Further, $\mathbf{S}+\mathbf{T}$ is a vector space. So it contains all combinations of vectors in itself; in particular, it contains the span of $\mathbf{S} \cup \mathbf{T}$. Thus the span of $\mathbf{S} \cup \mathbf{T}$ is $\mathbf{S}+\mathbf{T}$.
