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19.4F_Mid_Builder_$\tkcth{8.5.}\tkco{9}$__$\tkco{Mock 3}$
Related Topics
Wize University Linear Algebra Textbook > Vector Spaces and Subspaces
Coordinates
4 Activities
Taken together,
v
⃗
1
=
[
1
1
]
\vv_1 = \colvec{1}{1}
v
1
=
[
1
1
]
and
v
⃗
2
=
[
1
−
1
]
\vv_2 = \colvec{1}{-1}
v
2
=
[
1
−
1
]
constitute a basis for
R
2
\mathbb{R}^2
R
2
. Determine the coordinates
[
x
⃗
]
v
=
[
x
v
1
x
v
2
]
[ \vx ]_{v} = \colvec{\bcth{x_{v_1}}}{\bct{x_{v_2}}}
[
x
]
v
=
[
x
v
1
x
v
2
]
of the vector
x
⃗
=
[
2
1
]
\vx = \colvec{2}{1}
x
=
[
2
1
]
relative to this basis.
x
v
1
=
\bcth{x_{v_1} = }
x
v
1
=
x
v
2
=
\bct{x_{v_2} = }
x
v
2
=
I don't know
Check Submission
More Coordinates Questions:
Coordinates
Practice: Coordinates
Find the coordinates of the polynomial
p
(
x
)
=
5
+
x
−
3
x
2
−
3
x
3
p(x)=5+x-3x^2-3x^3
p
(
x
)
=
5
+
x
−
3
x
2
−
3
x
3
relative to the following basis of
P
3
P_3
P
3
:
B
=
{
(
1
+
x
)
,
(
x
+
x
2
)
,
(
x
2
+
x
3
)
,
(
1
−
x
3
)
}
B = \{(1+x),\ \ (x+x^2),\ \ (x^2+x^3),\ \ (1-x^3)\}
B
=
{(
1
+
x
)
,
(
x
+
x
2
)
,
(
x
2
+
x
3
)
,
(
1
−
x
3
)}
Practice: Coordinates
Find the coordinates of the polynomial
p
(
x
)
=
5
+
x
−
3
x
2
−
3
x
3
p(x)=5+x-3x^2-3x^3
p
(
x
)
=
5
+
x
−
3
x
2
−
3
x
3
relative to the following basis of
P
3
P_3
P
3
:
{
(
1
+
x
)
,
(
x
+
x
2
)
,
(
x
2
+
x
3
)
,
(
1
−
x
3
)
}
\{(1+x),(x+x^2),(x^2+x^3),(1-x^3)\}
{(
1
+
x
)
,
(
x
+
x
2
)
,
(
x
2
+
x
3
)
,
(
1
−
x
3
)}
Verify if the following vectors form a basis for
R
3
\mathbb{R}^3
R
3
:
{
[
1
1
2
]
,
[
1
−
2
1
]
,
[
2
−
1
1
]
}
\left\{ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix},\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \right\}
⎩
⎨
⎧
1
1
2
,
1
−
2
1
,
2
−
1
1
⎭
⎬
⎫
Concept Clarifier: Finding Coordinate Vectors
The following is a basis for
R
2
\mathbb{R}^2
R
2
.
{
(
1
1
)
,
(
−
2
0
)
}
\left \{ \left ( \begin{array}{c} 1 \\ 1 \end{array} \right ), \left ( \begin{array}{c} -2 \\0 \end{array} \right ) \right \}
{
(
1
1
)
,
(
−
2
0
)
}
If
v
⃗
=
(
2
,
3
)
\vec{v} = (2, 3)
v
=
(
2
,
3
)
is a coordinate vector in this basis, what is
v
⃗
\vec{v}
v
in coordinates with respect to the basis
19.4F_Mid_Builder_$\tkcth{8.4.16}$_$\tkcth{mock 1 ✓}$
Find the coordinates
(
c
1
,
c
2
)
(c_1,\, c_2)
(
c
1
,
c
2
)
for the point
P
=
(
8
,
11
)
P = (8,11)
P
=
(
8
,
11
)
in terms of the nonstandard basis vectors
e
⃗
1
=
<
1
,
1
>
\ve{1} = \rowvec{1}{1}
e
1
=
⟨
1
,
1
⟩
and
e
⃗
2
=
<
6
,
9
>
\ve{2} = \rowvec{6}{9}
e
2
=
⟨
6
,
9
⟩
.
Coordinates
Find the coordinates of the polynomial
p
(
x
)
=
2
−
x
+
x
2
p(x)=2-x+x^2
p
(
x
)
=
2
−
x
+
x
2
relative to the basis
{
1
+
x
,
1
+
x
2
,
x
+
x
2
}
\{1+x,1+x^2,x+x^2\}
{
1
+
x
,
1
+
x
2
,
x
+
x
2
}
of
P
2
P_2
P
2
19.4F_133_8.2_Mock_F1_$\tkco{eg6}$_$\key{Final}$_Builder_$\tkcth{8.2.}\tkcf{6}$_
The set
{
[
2
1
]
}
\left\{ \colvec{2}{1} \right\}
{
[
2
1
]
}
is a basis for the subspace
T
T
T
, where
T
⊆
R
2
T \subseteq \mathbb{R}^2
T
⊆
R
2
.
Give the position vector
r
⃗
P
\vr_{_{P}}
r
P
(in the standard basis) corresponding to the point in
T
T
T
whose coordinate is given by
P
T
=
(
3
)
P_{T} = (3)
P
T
=
(
3
)
.
If
r
⃗
=
[
x
y
]
\vr = \colvec{\bco{x}}{\bct{y}}
r
=
[
x
y
]
, then
Coordinates
Practice: Coordinates
Find the coordinates of the polynomial
p
(
x
)
=
5
+
x
−
3
x
2
−
3
x
3
p(x)=5+x-3x^2-3x^3
p
(
x
)
=
5
+
x
−
3
x
2
−
3
x
3
relative to the following basis of
P
3
P_3
P
3
:
B
=
{
(
1
+
x
)
,
(
x
+
x
2
)
,
(
x
2
+
x
3
)
,
(
1
−
x
3
)
}
B = \{(1+x),\ \ (x+x^2),\ \ (x^2+x^3),\ \ (1-x^3)\}
B
=
{(
1
+
x
)
,
(
x
+
x
2
)
,
(
x
2
+
x
3
)
,
(
1
−
x
3
)}
Coordinates
Practice: Coordinates
Let
B
1
=
{
[
−
1
2
0
]
,
[
3
−
5
2
]
,
[
4
−
7
3
]
}
B_1= \left\{ \begin{bmatrix} -1\\ 2 \\0\\ \end{bmatrix} , \begin{bmatrix} 3\\ -5\\ 2\\ \end{bmatrix} , \begin{bmatrix} 4\\ -7\\ 3\\ \end{bmatrix} \right\}
B
1
=
⎩
⎨
⎧
−
1
2
0
,
3
−
5
2
,
4
−
7
3
⎭
⎬
⎫
and
B
2
=
{
[
2
−
3
1
]
,
[
1
1
−
1
]
,
[
−
1
−
3
0
]
}
B_2= \left\{ \begin{bmatrix} 2\\ -3 \\1\\ \end{bmatrix} , \begin{bmatrix} 1\\ 1\\ -1\\ \end{bmatrix} , \begin{bmatrix} -1\\ -3\\ 0\\ \end{bmatrix} \right\}
B
2
=
⎩
⎨
⎧
2
−
3
1
,
1
1
−
1
,
−
1
−
3
0
⎭
⎬
⎫
be two bases of
R
3
\mathbb{R}^3
R
3
.
Suppose
[
x
⃗
]
B
1
=
[
−
4
8
−
7
]
[\vec x]_{\small B_1} = \begin{bmatrix} -4\\ 8\\ -7\\ \end{bmatrix}
[
x
]
B
1
=
−
4
8
−
7
.
Example: Change of Basis
Example: Change of Basis
Suppose we have 2
ordered bases
:
A
=
(
(
1
,
−
1
)
,
(
−
2
,
1
)
)
,
B
=
(
(
1
,
1
)
,
(
1
,
2
)
)
A=((1,-1),(-2,1)),B=((1,1),(1,2))
A
=
((
1
,
−
1
)
,
(
−
2
,
1
))
,
B
=
((
1
,
1
)
,
(
1
,
2
))
.
Compute the
change of basis
matrix
C
A
→
B
C_{A\rightarrow B}
C
A
→
B
, and if
[
v
⃗
]
A
=
(
−
3
,
−
4
)
[\vec v]_A=(-3,-4)
[
v
]
A
=
(
−
3
,
−
4
)
, then find
[
v
⃗
]
B
[\vec v]_B
[
v
]
B
.
Practice Question: Change of Basis
Suppose we have an
ordered basis
B
=
(
(
1
,
1
,
1
)
,
(
2
,
0
,
−
1
)
,
(
−
1
,
1
,
0
)
)
B=((1,1,1),(2,0,-1),(-1,1,0))
B
=
((
1
,
1
,
1
)
,
(
2
,
0
,
−
1
)
,
(
−
1
,
1
,
0
))
. Let
S
=
(
(
1
,
0
,
0
)
,
(
0
,
1
,
0
)
,
(
0
,
0
,
1
)
)
S=((1,0,0),(0,1,0),(0,0,1))
S
=
((
1
,
0
,
0
)
,
(
0
,
1
,
0
)
,
(
0
,
0
,
1
))
be the standard
ordered basis
. Compute the
change of basis
matrix
C
S
→
B
C_{S\rightarrow B}
C
S
→
B
, and use it to find
(
2
,
−
1
,
2
)
B
(2,-1,2)_B
(
2
,
−
1
,
2
)
B
(i.e. find
[
v
⃗
]
B
[\vec v]_B
[
v
]
B
, if
[
v
⃗
]
S
=
(
2
,
−
1
,
2
)
[\vec v]_S=(2,-1,2)
[
v
]
S
=
(
2
,
−
1
,
2
)
).
Let
{
1
−
x
,
x
,
1
+
x
2
,
x
−
x
3
,
x
4
}
\{ 1 - x, x, 1 + x^2, x - x^3, x^4 \}
{
1
−
x
,
x
,
1
+
x
2
,
x
−
x
3
,
x
4
}
be the basis for the vector space of polynomials with degree at most 4. What is the coordinate vector for
v
v
v
in this basis?
v
=
4
−
x
2
+
x
3
−
2
x
4
v = 4 - x^2 + x^3 - 2x^4
v
=
4
−
x
2
+
x
3
−
2
x
4
Consider the vector
v
=
(
1
,
2
,
3
)
v = (1, 2, 3)
v
=
(
1
,
2
,
3
)
with respect to the standard unit vector basis. What is the coordinate vector for
v
v
v
in the following basis:
{
(
1
1
0
)
,
(
0
1
0
)
,
(
−
1
0
−
2
)
}
\left \{ \left ( \begin{array}{c} 1 \\ 1\\ 0 \end{array} \right ), \left ( \begin{array}{c} 0 \\1 \\ 0 \end{array} \right ) , \left ( \begin{array}{c} -1 \\ 0 \\ -2 \end{array} \right ) \right \}
⎩
⎨
⎧
1
1
0
,
0
1
0
,
−
1
0
−
2
⎭
⎬
⎫