Coordinate vector with Respect to Basis (Matrices/Linear Algebra)
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I'm studying for a linear algebra test and I'm really stuck on changing basis or finding transformation vectors with respect to other basis and all of that. I've read so many guides and done so many problems, but the concepts won't stick. I feel like this link:
https://math.dartmouth.edu/archive/m24w07/public_html/Lecture12.pdf
is really good, but I don't understand the material. Everything makes sense up to slide five, where they write:
Suppose now that we choose as a basis for V = R^3 the set B = { (1,0,0), (1,1,0), (1,1,1,)}.
Then the coordinate vector of an element (a1, a2, a3) is:
(a1 - a2)
(a2 - a3)
( a3 )
How did they get that coordinate vector?
linear-algebra matrices matrix-equations
asked Mar 21 '15 at 4:30
user4567587
84
add a comment |
up vote
0
down vote
favorite
I'm studying for a linear algebra test and I'm really stuck on changing basis or finding transformation vectors with respect to other basis and all of that. I've read so many guides and done so many problems, but the concepts won't stick. I feel like this link:
https://math.dartmouth.edu/archive/m24w07/public_html/Lecture12.pdf
is really good, but I don't understand the material. Everything makes sense up to slide five, where they write:
Suppose now that we choose as a basis for V = R^3 the set B = { (1,0,0), (1,1,0), (1,1,1,)}.
Then the coordinate vector of an element (a1, a2, a3) is:
(a1 - a2)
(a2 - a3)
( a3 )
How did they get that coordinate vector?
linear-algebra matrices matrix-equations
asked Mar 21 '15 at 4:30
user4567587
84
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm studying for a linear algebra test and I'm really stuck on changing basis or finding transformation vectors with respect to other basis and all of that. I've read so many guides and done so many problems, but the concepts won't stick. I feel like this link:
https://math.dartmouth.edu/archive/m24w07/public_html/Lecture12.pdf
is really good, but I don't understand the material. Everything makes sense up to slide five, where they write:
Suppose now that we choose as a basis for V = R^3 the set B = { (1,0,0), (1,1,0), (1,1,1,)}.
Then the coordinate vector of an element (a1, a2, a3) is:
(a1 - a2)
(a2 - a3)
( a3 )
How did they get that coordinate vector?
linear-algebra matrices matrix-equations
asked Mar 21 '15 at 4:30
user4567587
84
I'm studying for a linear algebra test and I'm really stuck on changing basis or finding transformation vectors with respect to other basis and all of that. I've read so many guides and done so many problems, but the concepts won't stick. I feel like this link:
https://math.dartmouth.edu/archive/m24w07/public_html/Lecture12.pdf
is really good, but I don't understand the material. Everything makes sense up to slide five, where they write:
Suppose now that we choose as a basis for V = R^3 the set B = { (1,0,0), (1,1,0), (1,1,1,)}.
Then the coordinate vector of an element (a1, a2, a3) is:
(a1 - a2)
(a2 - a3)
( a3 )
How did they get that coordinate vector?
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
asked Mar 21 '15 at 4:30
user4567587
84
asked Mar 21 '15 at 4:30
user4567587
84
asked Mar 21 '15 at 4:30
user4567587
84
asked Mar 21 '15 at 4:30
user4567587
84
asked Mar 21 '15 at 4:30
user4567587
84
84
add a comment |
add a comment |
2 Answers
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$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
Edit: how you would actually figure this out is to solve
$c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(a_1,a_2,a_3)$
Edit 2: to answer your question, how they came up with the matrix for $T$. You have three things you know. First, $Tb_1=T[1,0,0]^t=(3,1,1)$. Second, $Tb_2=T[0,1,0]^t=(4,1,1)$. Third, $Tb_3=T[0,0,1]^t=(4,2,0)$. Then you have 9 equations, 9 unknowns. Simply checking here should be good enough.
I never would have seen that! Thank you!! One more question (I was hoping the answer to the last would lead to it, but it didn't for me), how did they find the matrix representation of T with respect to that basis? They have (these are columns) [ 3, 1, 1 ], [ 4, 1, 1 ], [ 4, 2, 0 ]
– user4567587
Mar 21 '15 at 4:59
Edited my post.
– user223391
Mar 21 '15 at 5:11
I'm sorry, but I really don't understand. I tried reviewing my notes and got nowhere.
– user4567587
Mar 21 '15 at 6:05
Simply check that $T[1,0,0]^t=(3,1,1)$, etc. with the given matrix.
– user223391
Mar 21 '15 at 7:33
add a comment |
up vote
0
down vote
There's a couple ways to do this:
You know that in the new basis a vector $v$ has coordinates:
$[b_1,b_2,b_3] stackrel{text{def}}{=} b_1(1,0,0) + b_2(1,1,0) + b_3(1,1,1)$.
So a matrix that turns "new to old" is:
$P = begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
This sends $[0,1,0] = 0(1,0,0) + 1(1,1,0) + 0(1,1,1)$ to $(1,1,0)$, for example.
But that's the opposite of what we want, we want "old to new", so we actually need to find $P^{-1}$.
We have some shortcuts we can take here:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1&a&b\0&1&c\0&0&1end{bmatrix}=begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}$
is what we need to solve (convince yourself $P^{-1}$ HAS to be upper-triangular with $1$'s on the diagonal). This leads to:
$a+1 = 0\b+c+1 = 0\c+1 = 0$
which implies $a = -1,b = 0,c = -1$. Hence:
$P^{-1} = begin{bmatrix}1&-1&0\0&1&-1\0&0&1end{bmatrix}$, and we see:
$P^{-1}(a_1,a_2,a_3)^t = [a_1-a_2,a_2-a_3,a_3]^t$, as your notes claim.
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
add a comment |
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2 Answers
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2 Answers
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up vote
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$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
Edit: how you would actually figure this out is to solve
$c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(a_1,a_2,a_3)$
Edit 2: to answer your question, how they came up with the matrix for $T$. You have three things you know. First, $Tb_1=T[1,0,0]^t=(3,1,1)$. Second, $Tb_2=T[0,1,0]^t=(4,1,1)$. Third, $Tb_3=T[0,0,1]^t=(4,2,0)$. Then you have 9 equations, 9 unknowns. Simply checking here should be good enough.
I never would have seen that! Thank you!! One more question (I was hoping the answer to the last would lead to it, but it didn't for me), how did they find the matrix representation of T with respect to that basis? They have (these are columns) [ 3, 1, 1 ], [ 4, 1, 1 ], [ 4, 2, 0 ]
– user4567587
Mar 21 '15 at 4:59
Edited my post.
– user223391
Mar 21 '15 at 5:11
I'm sorry, but I really don't understand. I tried reviewing my notes and got nowhere.
– user4567587
Mar 21 '15 at 6:05
Simply check that $T[1,0,0]^t=(3,1,1)$, etc. with the given matrix.
– user223391
Mar 21 '15 at 7:33
add a comment |
up vote
0
down vote
$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
Edit: how you would actually figure this out is to solve
$c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(a_1,a_2,a_3)$
Edit 2: to answer your question, how they came up with the matrix for $T$. You have three things you know. First, $Tb_1=T[1,0,0]^t=(3,1,1)$. Second, $Tb_2=T[0,1,0]^t=(4,1,1)$. Third, $Tb_3=T[0,0,1]^t=(4,2,0)$. Then you have 9 equations, 9 unknowns. Simply checking here should be good enough.
I never would have seen that! Thank you!! One more question (I was hoping the answer to the last would lead to it, but it didn't for me), how did they find the matrix representation of T with respect to that basis? They have (these are columns) [ 3, 1, 1 ], [ 4, 1, 1 ], [ 4, 2, 0 ]
– user4567587
Mar 21 '15 at 4:59
Edited my post.
– user223391
Mar 21 '15 at 5:11
I'm sorry, but I really don't understand. I tried reviewing my notes and got nowhere.
– user4567587
Mar 21 '15 at 6:05
Simply check that $T[1,0,0]^t=(3,1,1)$, etc. with the given matrix.
– user223391
Mar 21 '15 at 7:33
add a comment |
up vote
0
down vote
up vote
0
down vote
$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
Edit: how you would actually figure this out is to solve
$c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(a_1,a_2,a_3)$
Edit 2: to answer your question, how they came up with the matrix for $T$. You have three things you know. First, $Tb_1=T[1,0,0]^t=(3,1,1)$. Second, $Tb_2=T[0,1,0]^t=(4,1,1)$. Third, $Tb_3=T[0,0,1]^t=(4,2,0)$. Then you have 9 equations, 9 unknowns. Simply checking here should be good enough.
$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
Edit: how you would actually figure this out is to solve
$c_1(1,0,0)+c_2(1,1,0)+c_3(1,1,1)=(a_1,a_2,a_3)$
Edit 2: to answer your question, how they came up with the matrix for $T$. You have three things you know. First, $Tb_1=T[1,0,0]^t=(3,1,1)$. Second, $Tb_2=T[0,1,0]^t=(4,1,1)$. Third, $Tb_3=T[0,0,1]^t=(4,2,0)$. Then you have 9 equations, 9 unknowns. Simply checking here should be good enough.
edited Mar 21 '15 at 5:10
answered Mar 21 '15 at 4:35
user223391
I never would have seen that! Thank you!! One more question (I was hoping the answer to the last would lead to it, but it didn't for me), how did they find the matrix representation of T with respect to that basis? They have (these are columns) [ 3, 1, 1 ], [ 4, 1, 1 ], [ 4, 2, 0 ]
– user4567587
Mar 21 '15 at 4:59
Edited my post.
– user223391
Mar 21 '15 at 5:11
I'm sorry, but I really don't understand. I tried reviewing my notes and got nowhere.
– user4567587
Mar 21 '15 at 6:05
Simply check that $T[1,0,0]^t=(3,1,1)$, etc. with the given matrix.
– user223391
Mar 21 '15 at 7:33
add a comment |
I never would have seen that! Thank you!! One more question (I was hoping the answer to the last would lead to it, but it didn't for me), how did they find the matrix representation of T with respect to that basis? They have (these are columns) [ 3, 1, 1 ], [ 4, 1, 1 ], [ 4, 2, 0 ]
– user4567587
Mar 21 '15 at 4:59
Edited my post.
– user223391
Mar 21 '15 at 5:11
I'm sorry, but I really don't understand. I tried reviewing my notes and got nowhere.
– user4567587
Mar 21 '15 at 6:05
Simply check that $T[1,0,0]^t=(3,1,1)$, etc. with the given matrix.
– user223391
Mar 21 '15 at 7:33
I never would have seen that! Thank you!! One more question (I was hoping the answer to the last would lead to it, but it didn't for me), how did they find the matrix representation of T with respect to that basis? They have (these are columns) [ 3, 1, 1 ], [ 4, 1, 1 ], [ 4, 2, 0 ]
– user4567587
Mar 21 '15 at 4:59
I never would have seen that! Thank you!! One more question (I was hoping the answer to the last would lead to it, but it didn't for me), how did they find the matrix representation of T with respect to that basis? They have (these are columns) [ 3, 1, 1 ], [ 4, 1, 1 ], [ 4, 2, 0 ]
– user4567587
Mar 21 '15 at 4:59
Edited my post.
– user223391
Mar 21 '15 at 5:11
Edited my post.
– user223391
Mar 21 '15 at 5:11
I'm sorry, but I really don't understand. I tried reviewing my notes and got nowhere.
– user4567587
Mar 21 '15 at 6:05
I'm sorry, but I really don't understand. I tried reviewing my notes and got nowhere.
– user4567587
Mar 21 '15 at 6:05
Simply check that $T[1,0,0]^t=(3,1,1)$, etc. with the given matrix.
– user223391
Mar 21 '15 at 7:33
Simply check that $T[1,0,0]^t=(3,1,1)$, etc. with the given matrix.
– user223391
Mar 21 '15 at 7:33
add a comment |
up vote
0
down vote
There's a couple ways to do this:
You know that in the new basis a vector $v$ has coordinates:
$[b_1,b_2,b_3] stackrel{text{def}}{=} b_1(1,0,0) + b_2(1,1,0) + b_3(1,1,1)$.
So a matrix that turns "new to old" is:
$P = begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
This sends $[0,1,0] = 0(1,0,0) + 1(1,1,0) + 0(1,1,1)$ to $(1,1,0)$, for example.
But that's the opposite of what we want, we want "old to new", so we actually need to find $P^{-1}$.
We have some shortcuts we can take here:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1&a&b\0&1&c\0&0&1end{bmatrix}=begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}$
is what we need to solve (convince yourself $P^{-1}$ HAS to be upper-triangular with $1$'s on the diagonal). This leads to:
$a+1 = 0\b+c+1 = 0\c+1 = 0$
which implies $a = -1,b = 0,c = -1$. Hence:
$P^{-1} = begin{bmatrix}1&-1&0\0&1&-1\0&0&1end{bmatrix}$, and we see:
$P^{-1}(a_1,a_2,a_3)^t = [a_1-a_2,a_2-a_3,a_3]^t$, as your notes claim.
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
add a comment |
up vote
0
down vote
There's a couple ways to do this:
You know that in the new basis a vector $v$ has coordinates:
$[b_1,b_2,b_3] stackrel{text{def}}{=} b_1(1,0,0) + b_2(1,1,0) + b_3(1,1,1)$.
So a matrix that turns "new to old" is:
$P = begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
This sends $[0,1,0] = 0(1,0,0) + 1(1,1,0) + 0(1,1,1)$ to $(1,1,0)$, for example.
But that's the opposite of what we want, we want "old to new", so we actually need to find $P^{-1}$.
We have some shortcuts we can take here:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1&a&b\0&1&c\0&0&1end{bmatrix}=begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}$
is what we need to solve (convince yourself $P^{-1}$ HAS to be upper-triangular with $1$'s on the diagonal). This leads to:
$a+1 = 0\b+c+1 = 0\c+1 = 0$
which implies $a = -1,b = 0,c = -1$. Hence:
$P^{-1} = begin{bmatrix}1&-1&0\0&1&-1\0&0&1end{bmatrix}$, and we see:
$P^{-1}(a_1,a_2,a_3)^t = [a_1-a_2,a_2-a_3,a_3]^t$, as your notes claim.
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
add a comment |
up vote
0
down vote
up vote
0
down vote
There's a couple ways to do this:
You know that in the new basis a vector $v$ has coordinates:
$[b_1,b_2,b_3] stackrel{text{def}}{=} b_1(1,0,0) + b_2(1,1,0) + b_3(1,1,1)$.
So a matrix that turns "new to old" is:
$P = begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
This sends $[0,1,0] = 0(1,0,0) + 1(1,1,0) + 0(1,1,1)$ to $(1,1,0)$, for example.
But that's the opposite of what we want, we want "old to new", so we actually need to find $P^{-1}$.
We have some shortcuts we can take here:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1&a&b\0&1&c\0&0&1end{bmatrix}=begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}$
is what we need to solve (convince yourself $P^{-1}$ HAS to be upper-triangular with $1$'s on the diagonal). This leads to:
$a+1 = 0\b+c+1 = 0\c+1 = 0$
which implies $a = -1,b = 0,c = -1$. Hence:
$P^{-1} = begin{bmatrix}1&-1&0\0&1&-1\0&0&1end{bmatrix}$, and we see:
$P^{-1}(a_1,a_2,a_3)^t = [a_1-a_2,a_2-a_3,a_3]^t$, as your notes claim.
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
There's a couple ways to do this:
You know that in the new basis a vector $v$ has coordinates:
$[b_1,b_2,b_3] stackrel{text{def}}{=} b_1(1,0,0) + b_2(1,1,0) + b_3(1,1,1)$.
So a matrix that turns "new to old" is:
$P = begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}$
This sends $[0,1,0] = 0(1,0,0) + 1(1,1,0) + 0(1,1,1)$ to $(1,1,0)$, for example.
But that's the opposite of what we want, we want "old to new", so we actually need to find $P^{-1}$.
We have some shortcuts we can take here:
$begin{bmatrix}1&1&1\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1&a&b\0&1&c\0&0&1end{bmatrix}=begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix}$
is what we need to solve (convince yourself $P^{-1}$ HAS to be upper-triangular with $1$'s on the diagonal). This leads to:
$a+1 = 0\b+c+1 = 0\c+1 = 0$
which implies $a = -1,b = 0,c = -1$. Hence:
$P^{-1} = begin{bmatrix}1&-1&0\0&1&-1\0&0&1end{bmatrix}$, and we see:
$P^{-1}(a_1,a_2,a_3)^t = [a_1-a_2,a_2-a_3,a_3]^t$, as your notes claim.
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
answered Mar 21 '15 at 10:38
David Wheeler
12.6k11630
12.6k11630
add a comment |
add a comment |
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