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How to find the coordinate vector respect to a matrix basis?
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I know how to find the coordinate vector respect to a vector matrix. However, in my textbook, I see the following problem, which asks to find the coordinate vector of M respect to a matrix basis:
$M=begin{bmatrix}1 & 1\0& 1end{bmatrix}in Mat(2,2,mathbb{Q})$
$mathcal{B} _ { 1 } = left{ A = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right] , B = left[ begin{array} { c c } { 0 } & { 1 } \ { 0 } & { 0 } end{array} right] , C = left[ begin{array} { c c } { 0 } & { 0 } \ { 1 } & { 0 } end{array} right] , D = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { - 1 } end{array} right] right}$
How can I proceed? Can I explicit the matrix components as column vectors?
linear-algebra matrices
asked Jan 12 at 12:55
KevinKevin
13311
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add a comment |
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I know how to find the coordinate vector respect to a vector matrix. However, in my textbook, I see the following problem, which asks to find the coordinate vector of M respect to a matrix basis:
$M=begin{bmatrix}1 & 1\0& 1end{bmatrix}in Mat(2,2,mathbb{Q})$
$mathcal{B} _ { 1 } = left{ A = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right] , B = left[ begin{array} { c c } { 0 } & { 1 } \ { 0 } & { 0 } end{array} right] , C = left[ begin{array} { c c } { 0 } & { 0 } \ { 1 } & { 0 } end{array} right] , D = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { - 1 } end{array} right] right}$
How can I proceed? Can I explicit the matrix components as column vectors?
linear-algebra matrices
asked Jan 12 at 12:55
KevinKevin
13311
$endgroup$
add a comment |
$begingroup$
I know how to find the coordinate vector respect to a vector matrix. However, in my textbook, I see the following problem, which asks to find the coordinate vector of M respect to a matrix basis:
$M=begin{bmatrix}1 & 1\0& 1end{bmatrix}in Mat(2,2,mathbb{Q})$
$mathcal{B} _ { 1 } = left{ A = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right] , B = left[ begin{array} { c c } { 0 } & { 1 } \ { 0 } & { 0 } end{array} right] , C = left[ begin{array} { c c } { 0 } & { 0 } \ { 1 } & { 0 } end{array} right] , D = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { - 1 } end{array} right] right}$
How can I proceed? Can I explicit the matrix components as column vectors?
linear-algebra matrices
asked Jan 12 at 12:55
KevinKevin
13311
$endgroup$
I know how to find the coordinate vector respect to a vector matrix. However, in my textbook, I see the following problem, which asks to find the coordinate vector of M respect to a matrix basis:
$M=begin{bmatrix}1 & 1\0& 1end{bmatrix}in Mat(2,2,mathbb{Q})$
$mathcal{B} _ { 1 } = left{ A = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { 1 } end{array} right] , B = left[ begin{array} { c c } { 0 } & { 1 } \ { 0 } & { 0 } end{array} right] , C = left[ begin{array} { c c } { 0 } & { 0 } \ { 1 } & { 0 } end{array} right] , D = left[ begin{array} { c c } { 1 } & { 0 } \ { 0 } & { - 1 } end{array} right] right}$
How can I proceed? Can I explicit the matrix components as column vectors?
linear-algebra matrices
linear-algebra matrices
asked Jan 12 at 12:55
KevinKevin
13311
asked Jan 12 at 12:55
KevinKevin
13311
asked Jan 12 at 12:55
KevinKevin
13311
asked Jan 12 at 12:55
KevinKevin
13311
asked Jan 12 at 12:55
KevinKevin
13311
13311
add a comment |
add a comment |
1 Answer
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Start by setting up the equation you want to solve:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = abegin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix} + bbegin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix} + cbegin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix} + d begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix}.$$
Next, simplify:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = begin{bmatrix} a + d & b \ c & a - d end{bmatrix}.$$
Then turn into a system of equations:
begin{align*}
1 &= a + d \
1 &= b \
0 &= c \
1 &= a - d.
end{align*}
Finally, solve!
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
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General approach is OK but $(cdot)_{21}=-c$, not $c$.
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– obareey
Jan 12 at 17:41
$begingroup$
@obareey Actually, I just put down the wrong matrix, but thanks for letting me know.
$endgroup$
– Theo Bendit
Jan 12 at 22:45
add a comment |
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1 Answer
1
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
Start by setting up the equation you want to solve:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = abegin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix} + bbegin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix} + cbegin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix} + d begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix}.$$
Next, simplify:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = begin{bmatrix} a + d & b \ c & a - d end{bmatrix}.$$
Then turn into a system of equations:
begin{align*}
1 &= a + d \
1 &= b \
0 &= c \
1 &= a - d.
end{align*}
Finally, solve!
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
$endgroup$
$begingroup$
General approach is OK but $(cdot)_{21}=-c$, not $c$.
$endgroup$
– obareey
Jan 12 at 17:41
$begingroup$
@obareey Actually, I just put down the wrong matrix, but thanks for letting me know.
$endgroup$
– Theo Bendit
Jan 12 at 22:45
add a comment |
$begingroup$
Start by setting up the equation you want to solve:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = abegin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix} + bbegin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix} + cbegin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix} + d begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix}.$$
Next, simplify:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = begin{bmatrix} a + d & b \ c & a - d end{bmatrix}.$$
Then turn into a system of equations:
begin{align*}
1 &= a + d \
1 &= b \
0 &= c \
1 &= a - d.
end{align*}
Finally, solve!
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
$endgroup$
$begingroup$
General approach is OK but $(cdot)_{21}=-c$, not $c$.
$endgroup$
– obareey
Jan 12 at 17:41
$begingroup$
@obareey Actually, I just put down the wrong matrix, but thanks for letting me know.
$endgroup$
– Theo Bendit
Jan 12 at 22:45
add a comment |
$begingroup$
Start by setting up the equation you want to solve:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = abegin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix} + bbegin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix} + cbegin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix} + d begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix}.$$
Next, simplify:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = begin{bmatrix} a + d & b \ c & a - d end{bmatrix}.$$
Then turn into a system of equations:
begin{align*}
1 &= a + d \
1 &= b \
0 &= c \
1 &= a - d.
end{align*}
Finally, solve!
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
$endgroup$
Start by setting up the equation you want to solve:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = abegin{bmatrix} 1 & 0 \ 0 & 1 end{bmatrix} + bbegin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix} + cbegin{bmatrix} 0 & 0 \ 1 & 0 end{bmatrix} + d begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix}.$$
Next, simplify:
$$begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} = begin{bmatrix} a + d & b \ c & a - d end{bmatrix}.$$
Then turn into a system of equations:
begin{align*}
1 &= a + d \
1 &= b \
0 &= c \
1 &= a - d.
end{align*}
Finally, solve!
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
edited Jan 12 at 22:44
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
answered Jan 12 at 14:06
Theo BenditTheo Bendit
18.1k12152
18.1k12152
$begingroup$
General approach is OK but $(cdot)_{21}=-c$, not $c$.
$endgroup$
– obareey
Jan 12 at 17:41
$begingroup$
@obareey Actually, I just put down the wrong matrix, but thanks for letting me know.
$endgroup$
– Theo Bendit
Jan 12 at 22:45
add a comment |
$begingroup$
General approach is OK but $(cdot)_{21}=-c$, not $c$.
$endgroup$
– obareey
Jan 12 at 17:41
$begingroup$
@obareey Actually, I just put down the wrong matrix, but thanks for letting me know.
$endgroup$
– Theo Bendit
Jan 12 at 22:45
$begingroup$
General approach is OK but $(cdot)_{21}=-c$, not $c$.
$endgroup$
– obareey
Jan 12 at 17:41
$begingroup$
General approach is OK but $(cdot)_{21}=-c$, not $c$.
$endgroup$
– obareey
Jan 12 at 17:41
$begingroup$
@obareey Actually, I just put down the wrong matrix, but thanks for letting me know.
$endgroup$
– Theo Bendit
Jan 12 at 22:45
$begingroup$
@obareey Actually, I just put down the wrong matrix, but thanks for letting me know.
$endgroup$
– Theo Bendit
Jan 12 at 22:45
add a comment |
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