Similar Matrices Confer Equivalent Linear Transformations
$begingroup$
Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.
Let's say I have the following matrices:
$$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.
If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
$$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$
I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?
Thanks for help and sorry if an elementary question.
linear-algebra matrices linear-transformations
asked Jan 4 at 5:15
AndrewAndrew
346211
$endgroup$
add a comment |
$begingroup$
Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.
Let's say I have the following matrices:
$$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.
If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
$$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$
I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?
Thanks for help and sorry if an elementary question.
linear-algebra matrices linear-transformations
asked Jan 4 at 5:15
AndrewAndrew
346211
$endgroup$
add a comment |
$begingroup$
Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.
Let's say I have the following matrices:
$$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.
If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
$$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$
I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?
Thanks for help and sorry if an elementary question.
linear-algebra matrices linear-transformations
asked Jan 4 at 5:15
AndrewAndrew
346211
$endgroup$
Apologies - I know there are some similar questions on this site, but I couldn't quite apply what they are saying here.
Let's say I have the following matrices:
$$A=begin{bmatrix}-1&6\-2&6end{bmatrix}; B=begin{bmatrix}1&2\-1&4end{bmatrix}$$
I conclude that they are similar (without finding $P$ such that $PAP^{-1} = B$) because both have eigenvalues 2,3 and thus will be similar to $D=begin{bmatrix}2&0\0&3end{bmatrix}$.
If we let $x=begin{bmatrix}2\3 end{bmatrix}$, we have:
$$Ax=begin{bmatrix}16\14end{bmatrix}; Bx=begin{bmatrix}8\10end{bmatrix}$$
I was told, however, that if two matrices are similar then they induce the same linear transformation. How can they do this if they map the same vector to a different vector? What is meant by 'same' here?
Thanks for help and sorry if an elementary question.
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Jan 4 at 5:15
AndrewAndrew
346211
asked Jan 4 at 5:15
AndrewAndrew
346211
asked Jan 4 at 5:15
AndrewAndrew
346211
asked Jan 4 at 5:15
AndrewAndrew
346211
asked Jan 4 at 5:15
AndrewAndrew
346211
346211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When you write
$$x = begin{bmatrix} 2\3end{bmatrix}$$
what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that
$$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$
but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that
$$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$
Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$
$$A[x]_{alpha} = B[x]_{beta}.$$
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
$endgroup$
$begingroup$
Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
$endgroup$
– Andrew
Jan 4 at 9:20
$begingroup$
You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
$endgroup$
– Santana Afton
Jan 4 at 14:41
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%2f3061335%2fsimilar-matrices-confer-equivalent-linear-transformations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you write
$$x = begin{bmatrix} 2\3end{bmatrix}$$
what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that
$$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$
but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that
$$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$
Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$
$$A[x]_{alpha} = B[x]_{beta}.$$
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
$endgroup$
$begingroup$
Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
$endgroup$
– Andrew
Jan 4 at 9:20
$begingroup$
You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
$endgroup$
– Santana Afton
Jan 4 at 14:41
add a comment |
$begingroup$
When you write
$$x = begin{bmatrix} 2\3end{bmatrix}$$
what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that
$$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$
but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that
$$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$
Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$
$$A[x]_{alpha} = B[x]_{beta}.$$
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
$endgroup$
$begingroup$
Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
$endgroup$
– Andrew
Jan 4 at 9:20
$begingroup$
You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
$endgroup$
– Santana Afton
Jan 4 at 14:41
add a comment |
$begingroup$
When you write
$$x = begin{bmatrix} 2\3end{bmatrix}$$
what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that
$$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$
but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that
$$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$
Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$
$$A[x]_{alpha} = B[x]_{beta}.$$
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
$endgroup$
When you write
$$x = begin{bmatrix} 2\3end{bmatrix}$$
what you are really referencing is the linear combination $2v_1 + 3v_2$ where $v_1, v_2$ is a basis for $mathbb{R}^2$. Usually, we assume that
$$v_1 = begin{bmatrix} 1\0end{bmatrix}qquad v_2 = begin{bmatrix} 0\1end{bmatrix}, $$
but we don’t need to. Given a vector $x$ and a basis $alpha = {alpha_1, alpha_2}$, if $x = aalpha_1 + balpha_2$, I’ll be explicit and write that
$$[x]_{alpha} = begin{bmatrix} a\bend{bmatrix}. $$
Your statement, that similar matrices induce the same linear transformation, is true assuming you choose the correct basis. That is, there are bases $alpha$ and $beta$ such that given any $xinmathbb{R}^2$
$$A[x]_{alpha} = B[x]_{beta}.$$
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
edited Jan 4 at 8:24
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
answered Jan 4 at 5:47
Santana AftonSantana Afton
2,9132629
2,9132629
$begingroup$
Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
$endgroup$
– Andrew
Jan 4 at 9:20
$begingroup$
You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
$endgroup$
– Santana Afton
Jan 4 at 14:41
add a comment |
$begingroup$
Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
$endgroup$
– Andrew
Jan 4 at 9:20
$begingroup$
You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
$endgroup$
– Santana Afton
Jan 4 at 14:41
$begingroup$
Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
$endgroup$
– Andrew
Jan 4 at 9:20
$begingroup$
Thanks for your answer. Sorry if I it sounds stupid, but I still don’t understand how your explanation above means that the two similar matrices induce the SAME liner transformation. You can always write any vector as the linear combination of basis vectors, but not all invertible matrices are similar to each other.
$endgroup$
– Andrew
Jan 4 at 9:20
$begingroup$
You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
$endgroup$
– Santana Afton
Jan 4 at 14:41
$begingroup$
You can always write any vector as a linear combination of basis vectors, but you can’t always find two bases so that $A[x]_{alpha} = B[x]_{beta}$. For two non-similar matrices, this won’t be possible.
$endgroup$
– Santana Afton
Jan 4 at 14:41
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.
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%2f3061335%2fsimilar-matrices-confer-equivalent-linear-transformations%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
