Changing the base of a transformed vector
Let $V$ be an $n$-dimensional vector space, let $W$ be an $m$-dimensional vector space, and let $T$ be any linear transformation from $V$ to $W$ . To associate a matrix with $T$, choose (ordered) bases $B$ and $C$ for $V$ and $W$, respectively.
Given any $x$ in $V$ , the coordinate vector $[x]_B$ is in $mathbb R^n$, and the coordinate vector of its image, $[T(x)]_C$ , is in $mathbb R^m$, as shown in Figure 1:
The connection between $[x]_B$ and $[T(x)]_C$ is easy to find.
Let ${b_1,cdots,b_n}=B$ be the basis for $V$.
If $x=r_1b_1+cdots+r_nb_n$, then
$$[x]_B=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$$
and
$$T(x)=T(r_1b_1+cdots+r_nb_n)=r_1T(b_1)+cdots+r_nT(b_n)$$ because $T$ is linear.
The coordinate mapping from $W$ to $mathbb R^m$ is linear, which leads to
$$[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$$
Since base-$C$ coordinate vectors are in $mathbb R^m$, the vector equation above can be written as a matrix equation, namely $[T(x)]_C = M[x]_B$ where
$$M=begin{bmatrix} [T(b_1)]_C & [T(b_2)]_C & cdots & [T(b_n)]_C end{bmatrix}$$
My question: I'm confused on the jump from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$ to $[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$.
Since the basis $B={b_1,cdots,b_n}$, I see how when $T$ maps it to $W$, the basis $C={T(b_1),cdots,T(b_n)}$. Then from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$, the $r$s can be arranged as $[x]_C=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$ and $M$ as $M=begin{bmatrix} [T(b_1)] & [T(b_2)] & cdots & [T(b_n)] end{bmatrix}$. To get, $x=M[x]_C$, but I don't see how $M$ is in $mathbb{R}^m$.
linear-algebra change-of-basis
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
asked Apr 19 '17 at 4:20
stumped
573724
add a comment |
Let $V$ be an $n$-dimensional vector space, let $W$ be an $m$-dimensional vector space, and let $T$ be any linear transformation from $V$ to $W$ . To associate a matrix with $T$, choose (ordered) bases $B$ and $C$ for $V$ and $W$, respectively.
Given any $x$ in $V$ , the coordinate vector $[x]_B$ is in $mathbb R^n$, and the coordinate vector of its image, $[T(x)]_C$ , is in $mathbb R^m$, as shown in Figure 1:
The connection between $[x]_B$ and $[T(x)]_C$ is easy to find.
Let ${b_1,cdots,b_n}=B$ be the basis for $V$.
If $x=r_1b_1+cdots+r_nb_n$, then
$$[x]_B=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$$
and
$$T(x)=T(r_1b_1+cdots+r_nb_n)=r_1T(b_1)+cdots+r_nT(b_n)$$ because $T$ is linear.
The coordinate mapping from $W$ to $mathbb R^m$ is linear, which leads to
$$[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$$
Since base-$C$ coordinate vectors are in $mathbb R^m$, the vector equation above can be written as a matrix equation, namely $[T(x)]_C = M[x]_B$ where
$$M=begin{bmatrix} [T(b_1)]_C & [T(b_2)]_C & cdots & [T(b_n)]_C end{bmatrix}$$
My question: I'm confused on the jump from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$ to $[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$.
Since the basis $B={b_1,cdots,b_n}$, I see how when $T$ maps it to $W$, the basis $C={T(b_1),cdots,T(b_n)}$. Then from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$, the $r$s can be arranged as $[x]_C=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$ and $M$ as $M=begin{bmatrix} [T(b_1)] & [T(b_2)] & cdots & [T(b_n)] end{bmatrix}$. To get, $x=M[x]_C$, but I don't see how $M$ is in $mathbb{R}^m$.
linear-algebra change-of-basis
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
asked Apr 19 '17 at 4:20
stumped
573724
I am not sure what your confusion is but it seems like you are assuming that ${T(b_1),cdots, T(b_n)}$ forms a basis of $W$ since ${b_1, cdots, b_n}$ forms a basis of $V$. This is not necessarily true: Linear transformations do not preserve linear independence.
– PJK
Apr 19 '17 at 7:06
add a comment |
Let $V$ be an $n$-dimensional vector space, let $W$ be an $m$-dimensional vector space, and let $T$ be any linear transformation from $V$ to $W$ . To associate a matrix with $T$, choose (ordered) bases $B$ and $C$ for $V$ and $W$, respectively.
Given any $x$ in $V$ , the coordinate vector $[x]_B$ is in $mathbb R^n$, and the coordinate vector of its image, $[T(x)]_C$ , is in $mathbb R^m$, as shown in Figure 1:
The connection between $[x]_B$ and $[T(x)]_C$ is easy to find.
Let ${b_1,cdots,b_n}=B$ be the basis for $V$.
If $x=r_1b_1+cdots+r_nb_n$, then
$$[x]_B=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$$
and
$$T(x)=T(r_1b_1+cdots+r_nb_n)=r_1T(b_1)+cdots+r_nT(b_n)$$ because $T$ is linear.
The coordinate mapping from $W$ to $mathbb R^m$ is linear, which leads to
$$[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$$
Since base-$C$ coordinate vectors are in $mathbb R^m$, the vector equation above can be written as a matrix equation, namely $[T(x)]_C = M[x]_B$ where
$$M=begin{bmatrix} [T(b_1)]_C & [T(b_2)]_C & cdots & [T(b_n)]_C end{bmatrix}$$
My question: I'm confused on the jump from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$ to $[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$.
Since the basis $B={b_1,cdots,b_n}$, I see how when $T$ maps it to $W$, the basis $C={T(b_1),cdots,T(b_n)}$. Then from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$, the $r$s can be arranged as $[x]_C=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$ and $M$ as $M=begin{bmatrix} [T(b_1)] & [T(b_2)] & cdots & [T(b_n)] end{bmatrix}$. To get, $x=M[x]_C$, but I don't see how $M$ is in $mathbb{R}^m$.
linear-algebra change-of-basis
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
asked Apr 19 '17 at 4:20
stumped
573724
Let $V$ be an $n$-dimensional vector space, let $W$ be an $m$-dimensional vector space, and let $T$ be any linear transformation from $V$ to $W$ . To associate a matrix with $T$, choose (ordered) bases $B$ and $C$ for $V$ and $W$, respectively.
Given any $x$ in $V$ , the coordinate vector $[x]_B$ is in $mathbb R^n$, and the coordinate vector of its image, $[T(x)]_C$ , is in $mathbb R^m$, as shown in Figure 1:
The connection between $[x]_B$ and $[T(x)]_C$ is easy to find.
Let ${b_1,cdots,b_n}=B$ be the basis for $V$.
If $x=r_1b_1+cdots+r_nb_n$, then
$$[x]_B=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$$
and
$$T(x)=T(r_1b_1+cdots+r_nb_n)=r_1T(b_1)+cdots+r_nT(b_n)$$ because $T$ is linear.
The coordinate mapping from $W$ to $mathbb R^m$ is linear, which leads to
$$[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$$
Since base-$C$ coordinate vectors are in $mathbb R^m$, the vector equation above can be written as a matrix equation, namely $[T(x)]_C = M[x]_B$ where
$$M=begin{bmatrix} [T(b_1)]_C & [T(b_2)]_C & cdots & [T(b_n)]_C end{bmatrix}$$
My question: I'm confused on the jump from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$ to $[T(x)]_C=r_1[T(b_1)]_C+cdots+r_n[T(b_n)]_C$.
Since the basis $B={b_1,cdots,b_n}$, I see how when $T$ maps it to $W$, the basis $C={T(b_1),cdots,T(b_n)}$. Then from $T(x)=r_1T(b_1)+cdots+r_nT(b_n)$, the $r$s can be arranged as $[x]_C=begin{bmatrix}r_1\ vdots \r_nend{bmatrix}$ and $M$ as $M=begin{bmatrix} [T(b_1)] & [T(b_2)] & cdots & [T(b_n)] end{bmatrix}$. To get, $x=M[x]_C$, but I don't see how $M$ is in $mathbb{R}^m$.
linear-algebra change-of-basis
linear-algebra change-of-basis
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
asked Apr 19 '17 at 4:20
stumped
573724
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
asked Apr 19 '17 at 4:20
stumped
573724
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
edited Dec 9 at 7:58
Chase Ryan Taylor
4,37521530
4,37521530
asked Apr 19 '17 at 4:20
stumped
573724
asked Apr 19 '17 at 4:20
stumped
573724
asked Apr 19 '17 at 4:20
stumped
573724
573724
I am not sure what your confusion is but it seems like you are assuming that ${T(b_1),cdots, T(b_n)}$ forms a basis of $W$ since ${b_1, cdots, b_n}$ forms a basis of $V$. This is not necessarily true: Linear transformations do not preserve linear independence.
– PJK
Apr 19 '17 at 7:06
add a comment |
I am not sure what your confusion is but it seems like you are assuming that ${T(b_1),cdots, T(b_n)}$ forms a basis of $W$ since ${b_1, cdots, b_n}$ forms a basis of $V$. This is not necessarily true: Linear transformations do not preserve linear independence.
– PJK
Apr 19 '17 at 7:06
I am not sure what your confusion is but it seems like you are assuming that ${T(b_1),cdots, T(b_n)}$ forms a basis of $W$ since ${b_1, cdots, b_n}$ forms a basis of $V$. This is not necessarily true: Linear transformations do not preserve linear independence.
– PJK
Apr 19 '17 at 7:06
I am not sure what your confusion is but it seems like you are assuming that ${T(b_1),cdots, T(b_n)}$ forms a basis of $W$ since ${b_1, cdots, b_n}$ forms a basis of $V$. This is not necessarily true: Linear transformations do not preserve linear independence.
– PJK
Apr 19 '17 at 7:06
add a comment |
active
oldest
votes
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%2f2241272%2fchanging-the-base-of-a-transformed-vector%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2241272%2fchanging-the-base-of-a-transformed-vector%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
I am not sure what your confusion is but it seems like you are assuming that ${T(b_1),cdots, T(b_n)}$ forms a basis of $W$ since ${b_1, cdots, b_n}$ forms a basis of $V$. This is not necessarily true: Linear transformations do not preserve linear independence.
– PJK
Apr 19 '17 at 7:06