# HG changeset patch # User Rob Beezer # Date 1352941699 28800 # Node ID ad83ec5affedd8a49a34c246bb51f53aae443b91 # Parent 9f3fcd7faefc6b02419c73e23631faf450373454 Line breaks, Chapter VS diff --git a/src/section-S.xml b/src/section-S.xml --- a/src/section-S.xml +++ b/src/section-S.xml @@ -41,25 +41,25 @@ \vect{x}+\vect{y}=\colvector{x_1\\x_2\\x_3}+\colvector{y_1\\y_2\\y_3}= \colvector{x_1+y_1\\x_2+y_2\\x_3+y_3} -and we can test this vector for membership in $W$ as follows, +and we can test this vector for membership in $W$ as follows. Because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$ and because $\vect{y}\in W$ we know $2y_1-5y_2+7y_3=0$. Therefore, 2(x_1+y_1)-5(x_2+y_2)+7(x_3+y_3) - + and by this computation we see that $\vect{x}+\vect{y}\in W$. One property down, nine to go.

-

If $\alpha$ is a scalar and $\vect{x}\in W$, is it always true that $\alpha\vect{x}\in W$? This is what we need to establish . Again, the answer is not as obvious as it was when our set of vectors was all of $\complex{3}$. Let's see. +

If $\alpha$ is a scalar and $\vect{x}\in W$, is it always true that $\alpha\vect{x}\in W$? This is what we need to establish . Again, the answer is not as obvious as it was when our set of vectors was all of $\complex{3}$. Let's see. First, note that because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$. Therefore, \alpha\vect{x}=\alpha\colvector{x_1\\x_2\\x_3}=\colvector{\alpha x_1\\\alpha x_2\\\alpha x_3} -and we can test this vector for membership in $W$ with +and we can test this vector for membership in $W$. First, note that because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$. Therefore, 2(\alpha x_1)-5(\alpha x_2)+7(\alpha x_3) - + and we see that indeed $\alpha\vect{x}\in W$. Always.

@@ -70,11 +70,11 @@ \vect{-x}=\colvector{-x_1\\-x_2\\-x_3} -and we can test this vector for membership in $W$ with +and we can test this vector for membership in $W$. As before, because $\vect{x}\in W$ we know $2x_1-5x_2+7x_3=0$. 2(-x_1)-5(-x_2)+7(-x_3) - + and we now believe that $\vect{-x}\in W$.

@@ -235,7 +235,7 @@ -

We will examine the three requirements of . Recall that $\nsp{A}=\setparts{\vect{x}\in\complex{n}}{A\vect{x}=\zerovector}$.

+

We will examine the three requirements of . Recall that can be formulated as $\nsp{A}=\setparts{\vect{x}\in\complex{n}}{A\vect{x}=\zerovector}$.

First, $\zerovector\in\nsp{A}$, which can be inferred as a consequence of . So $\nsp{A}\neq\emptyset$.

@@ -511,7 +511,8 @@ (\alpha_1+2\alpha_2)x^4+ (-4\alpha_1-3\alpha_2)x^3+ -(5\alpha_1-6\alpha_2)x^2+ +(5\alpha_1-6\alpha_2)x^2\\ + (-\alpha_1+6\alpha_2)x+ (-2\alpha_1+4\alpha_2) @@ -586,9 +587,9 @@ in $W$? To answer this, we want to determine if $\vect{y}$ can be written as a linear combination of the five matrices in $S$. Can we find scalars, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ so that -\begin{bmatrix} + -\end{bmatrix} +\end{bmatrix}\\ \alpha_1 \begin{bmatrix} @@ -675,9 +676,9 @@ in $W$? To answer this, we want to determine if $\vect{x}$ can be written as a linear combination of the five matrices in $S$. Can we find scalars, $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4,\,\alpha_5$ so that -\begin{bmatrix} + -\end{bmatrix} +\end{bmatrix}\\ \alpha_1 \begin{bmatrix} diff --git a/src/section-VS.xml b/src/section-VS.xml --- a/src/section-VS.xml +++ b/src/section-VS.xml @@ -275,9 +275,10 @@ Feeling better? Or worse?

: For each vector, $\vect{u}$, we must locate an additive inverse, $\vect{-u}$. Here it is, $-(x_1,\,x_2)=(-x_1-2,\,-x_2-2)$. As odd as it may look, I hope you are withholding judgment. Check: - -\vect{u}+ (\vect{-u})=(x_1,\,x_2)+(-x_1-2,\,-x_2-2)=(x_1+(-x_1-2)+1,\,-x_2+(x_2-2)+1)=(-1,\,-1)=\zerovector - + + + +

: @@ -294,7 +295,8 @@

: If you have hung on so far, here's where it gets even wilder. In the next two properties we mix and mash the two operations. - + + @@ -308,7 +310,8 @@

: - + +