5. Consider two basis of R:१-{ () ()B =and (=%3D3.5-with respect to B.If a vector V E FR has coordinatesWhat is its coordindtcs wHh respect tó C?

The given 2 bases of ℝ2 are as follows. B=75, -3-1 and C=3-5, -12 Also, the coordinates of the…

Answered: 2. 5. Consider twO basis of R: B = %3D…

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Inner Product Spaces

Section5.2: Inner Product Spaces

Problem 101E: Consider the vectors u=(6,2,4) and v=(1,2,0) from Example 10. Without using Theorem 5.9, show that...

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5. Consider two basis of R:
१-{ () ()
B =
and (=
%3D
3.
5-
with respect to B.
If a vector V E FR has coordinates
What is its coordindtcs wHh respect tó C?

Expert Solution

Step 1

The given 2 bases of $ℝ^{2}$ are as follows.

$B = \{(\begin{matrix} 7 \\ 5 \end{matrix}), (\begin{matrix} - 3 \\ - 1 \end{matrix})\}$ and $C = \{(\begin{matrix} 3 \\ - 5 \end{matrix}), (\begin{matrix} - 1 \\ 2 \end{matrix})\}$

Also, the coordinates of the vector v with respect to the basis B is given by $(\begin{matrix} 2 \\ 1 \end{matrix})$ .

This implies that the vector $v = 2 (\begin{matrix} 7 \\ 5 \end{matrix}) + 1 (\begin{matrix} - 3 \\ - 1 \end{matrix})$ .

Hence, we have

$\begin{array}{rcl} v & = & 2 (\begin{matrix} 7 \\ 5 \end{matrix}) + 1 (\begin{matrix} - 3 \\ - 1 \end{matrix}) \\ = & (\begin{matrix} 14 \\ 10 \end{matrix}) + (\begin{matrix} - 3 \\ - 1 \end{matrix}) \\ = & (\begin{matrix} 11 \\ 9 \end{matrix}) \end{array}$

Step 2

Let $(\begin{matrix} a \\ b \end{matrix})$ be the coordinates of the vector v with respect to the basis C.

Then,

$\begin{array}{rcl} v & = & a (\begin{matrix} 3 \\ - 5 \end{matrix}) + b (\begin{matrix} - 1 \\ 2 \end{matrix}) \\ = & (\begin{matrix} 3 a \\ - 5 a \end{matrix}) + (\begin{matrix} - b \\ 2 b \end{matrix}) \\ = & (\begin{matrix} 3 a - b \\ - 5 a + 2 b \end{matrix}) \end{array}$

Substituting the vector $\begin{array}{rcl} v & = & (\begin{matrix} 11 \\ 9 \end{matrix}) \end{array}$ in $\begin{array}{rcl} v & = & (\begin{matrix} 3 a - b \\ - 5 a + 2 b \end{matrix}) \end{array}$ , we get $\begin{array}{rcl} (\begin{matrix} 11 \\ 9 \end{matrix}) & = & (\begin{matrix} 3 a - b \\ - 5 a + 2 b \end{matrix}) \end{array}$ .

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