Coordinate vector with Respect to Basis (Matrices/Linear Algebra)
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
85
add a comment |
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
85
add a comment |
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
85
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
85
asked Mar 21 '15 at 4:30
user4567587
85
asked Mar 21 '15 at 4:30
user4567587
85
asked Mar 21 '15 at 4:30
user4567587
85
asked Mar 21 '15 at 4:30
user4567587
85
85
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
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 |
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1199255%2fcoordinate-vector-with-respect-to-basis-matrices-linear-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
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 |
$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
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 |
$(a_1-a_2)(1,0,0)+(a_2-a_3)(1,1,0)+a_3(1,1,1)=(a_1,a_2,a_3)$
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 |
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 |
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 |
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1199255%2fcoordinate-vector-with-respect-to-basis-matrices-linear-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
