Question

Consider the following two basis: $\begin{pmatrix} 7\\0\\0\\1 \end{pmatrix}$, $\begin{pmatrix} 3\\0\\1\\0 \end{pmatrix}$, $\begin{pmatrix} -5\\1\\0\\0 \end{pmatrix}$ = {$\vec{a_1}$,$\vec{a_2}$,$\vec{a_3}$} and $B = \begin{pmatrix} 2\\1\\0\\1 \end{pmatrix}$, $\begin{pmatrix} -1\\2\\3\\0 \end{pmatrix}$, $\begin{pmatrix} 1\\-1\\1\\-1 \end{pmatrix}$ = {$\vec{b_1}$,$\vec{b_2}$,$\vec{b_3}$} 1. Show that every vector in A is in Span(B) and every vector in B is in Span(A).

          Consider the following two basis:
$\begin{pmatrix} 7\\0\\0\\1 \end{pmatrix}$, $\begin{pmatrix} 3\\0\\1\\0 \end{pmatrix}$, $\begin{pmatrix} -5\\1\\0\\0 \end{pmatrix}$ = {$\vec{a_1}$,$\vec{a_2}$,$\vec{a_3}$} and $B = \begin{pmatrix} 2\\1\\0\\1 \end{pmatrix}$, $\begin{pmatrix} -1\\2\\3\\0 \end{pmatrix}$, $\begin{pmatrix} 1\\-1\\1\\-1 \end{pmatrix}$ = {$\vec{b_1}$,$\vec{b_2}$,$\vec{b_3}$}
1. Show that every vector in A is in Span(B) and every vector in B is in Span(A).

Consider the following two basis:
< p m a t r i x >, < p m a t r i x >, < p m a t r i x > = a⃗1⃗,a⃗2⃗,a⃗3⃗ and B =
< p m a t r i x >, < p m a t r i x >, < p m a t r i x > = b⃗1⃗,b⃗2⃗,b⃗3⃗
1. Show that every vector in A is in Span(B) and every vector in B is in Span(A).

Added by Vicenta R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

01/25/2023

Step 1

To find proj_(W)(vec(v)), we can use the formula proj_(W)(vec(v)) = (vec(v) dot vec(a_1)/||vec(a_1)||^2) * vec(a_1) + (vec(v) dot vec(a_2)/||vec(a_2)||^2) * vec(a_2) + (vec(v) dot vec(a_3)/||vec(a_3)||^2) * vec(a_3). Similarly, to find perp_(W)(vec(v)), we can Show more…

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Question 1 tells us that A and B are both bases for the same subspaces. Call that subspace W. (a) Find proj_(W)(vec(v)) and perp_(W)(vec(v)). (b) Demonstrate that vec(v)=proj_(W)(vec(v))+perp_(W)(vec(v)). (c) Demonstrate that proj_(W)(vec(v)) and perp_(W)(vec(v)) are orthogonal. (d) Demonstrate that proj_(W)(vec(v)) is in W and perp_(W)(vec(v)) is in W. Consider the following two bases: A={[7],[0],[0],[1]],[[3],[0],[1],[0]],[[-5],[1],[0],[0]]}={vec(a)_(1),vec(a)_(2),vec(a)_(3)} and B={[2],[1],[0],[1]],[[-1],[2],[3],[0]],[[1],[-1],[1],[-1]]}={vec(b)_(1),vec(b)_(2),vec(b)_(3)} Show that every vector in A is in Span(B) and every vector in B is in Span(A). Consider the following two bases: A={a1,a2,a} and B={61,b2,b3} Show that every vector in A is in Span(B) and every vector in B is in Span(A).