Mixing
Questions on diagonalization
$begingroup$
- We have "$lambda in K$ is an eigenvalue of $A in M_n(K)$ $iff chi_A(lambda)=det(lambda I_n-A)=0$", and this analogue: "Consider an $n$-dimensional $K$-vector space $V$ and $f:Vto V$ a linear operator on $V$. $lambda in K$ is an eigenvalue of $fiff chi_f(lambda)=det(lambda mathbf{1}-f)=0$".
The proof of the second statements uses $0=det(lambda I_n-A)=det(lambda mathbf{1} -f)$ (with $A$ the matrix representation of $f$), but I can't really see where the equality come from.
$Ain M_n(K)$ is diagonalizable if and only if there exists a matrix $Pin GL_n(K)$ such that $P^{-1}AP$ is a diagonal matrix.
This is the proof:
$Rightarrow$:Consider the standard basis $mathcal{B}$ for $K^n$. Suppose that $A$ (and therefor $L_A$) is diagonalizable. Then there is a basis $mathcal{B'}={b_1,dots,b_n}$ consisting of eigenvectors of $A$. Define $P$, the transition matrix from $mathcal{B}$ to $mathcal{B'}$; Then $L_{P^{-1}AP}$ maps $b_i$ to $lambda_i b_i$. Conclusion: $P^{-1}AP = operatorname{diag}(lambda_1dots,lambda_n)$.
$Leftarrow$: Suppose there is a $P in GL_n(K)$ such that $P^{-1}AP$ is diagonal, and consider the basis $mathcal{B'} = {Pe_1,dots,Pe_n}$ of $K^n$. It is clear that the matrix of $L_A$ $(A_{mathcal{B},mathcal{B}})$ will be the diagonal matrix $P^{-1}AP$.
I don't understand the sentences that are put in bold. Can someone help me see the logic behind it.
Thanks in advance.
linear-algebra
asked Jan 11 at 8:58
ZacharyZachary
1559
$endgroup$
add a comment |
$begingroup$
- We have "$lambda in K$ is an eigenvalue of $A in M_n(K)$ $iff chi_A(lambda)=det(lambda I_n-A)=0$", and this analogue: "Consider an $n$-dimensional $K$-vector space $V$ and $f:Vto V$ a linear operator on $V$. $lambda in K$ is an eigenvalue of $fiff chi_f(lambda)=det(lambda mathbf{1}-f)=0$".
The proof of the second statements uses $0=det(lambda I_n-A)=det(lambda mathbf{1} -f)$ (with $A$ the matrix representation of $f$), but I can't really see where the equality come from.
$Ain M_n(K)$ is diagonalizable if and only if there exists a matrix $Pin GL_n(K)$ such that $P^{-1}AP$ is a diagonal matrix.
This is the proof:
$Rightarrow$:Consider the standard basis $mathcal{B}$ for $K^n$. Suppose that $A$ (and therefor $L_A$) is diagonalizable. Then there is a basis $mathcal{B'}={b_1,dots,b_n}$ consisting of eigenvectors of $A$. Define $P$, the transition matrix from $mathcal{B}$ to $mathcal{B'}$; Then $L_{P^{-1}AP}$ maps $b_i$ to $lambda_i b_i$. Conclusion: $P^{-1}AP = operatorname{diag}(lambda_1dots,lambda_n)$.
$Leftarrow$: Suppose there is a $P in GL_n(K)$ such that $P^{-1}AP$ is diagonal, and consider the basis $mathcal{B'} = {Pe_1,dots,Pe_n}$ of $K^n$. It is clear that the matrix of $L_A$ $(A_{mathcal{B},mathcal{B}})$ will be the diagonal matrix $P^{-1}AP$.
I don't understand the sentences that are put in bold. Can someone help me see the logic behind it.
Thanks in advance.
linear-algebra
asked Jan 11 at 8:58
ZacharyZachary
1559
$endgroup$
add a comment |
$begingroup$
- We have "$lambda in K$ is an eigenvalue of $A in M_n(K)$ $iff chi_A(lambda)=det(lambda I_n-A)=0$", and this analogue: "Consider an $n$-dimensional $K$-vector space $V$ and $f:Vto V$ a linear operator on $V$. $lambda in K$ is an eigenvalue of $fiff chi_f(lambda)=det(lambda mathbf{1}-f)=0$".
The proof of the second statements uses $0=det(lambda I_n-A)=det(lambda mathbf{1} -f)$ (with $A$ the matrix representation of $f$), but I can't really see where the equality come from.
$Ain M_n(K)$ is diagonalizable if and only if there exists a matrix $Pin GL_n(K)$ such that $P^{-1}AP$ is a diagonal matrix.
This is the proof:
$Rightarrow$:Consider the standard basis $mathcal{B}$ for $K^n$. Suppose that $A$ (and therefor $L_A$) is diagonalizable. Then there is a basis $mathcal{B'}={b_1,dots,b_n}$ consisting of eigenvectors of $A$. Define $P$, the transition matrix from $mathcal{B}$ to $mathcal{B'}$; Then $L_{P^{-1}AP}$ maps $b_i$ to $lambda_i b_i$. Conclusion: $P^{-1}AP = operatorname{diag}(lambda_1dots,lambda_n)$.
$Leftarrow$: Suppose there is a $P in GL_n(K)$ such that $P^{-1}AP$ is diagonal, and consider the basis $mathcal{B'} = {Pe_1,dots,Pe_n}$ of $K^n$. It is clear that the matrix of $L_A$ $(A_{mathcal{B},mathcal{B}})$ will be the diagonal matrix $P^{-1}AP$.
I don't understand the sentences that are put in bold. Can someone help me see the logic behind it.
Thanks in advance.
linear-algebra
asked Jan 11 at 8:58
ZacharyZachary
1559
$endgroup$
- We have "$lambda in K$ is an eigenvalue of $A in M_n(K)$ $iff chi_A(lambda)=det(lambda I_n-A)=0$", and this analogue: "Consider an $n$-dimensional $K$-vector space $V$ and $f:Vto V$ a linear operator on $V$. $lambda in K$ is an eigenvalue of $fiff chi_f(lambda)=det(lambda mathbf{1}-f)=0$".
The proof of the second statements uses $0=det(lambda I_n-A)=det(lambda mathbf{1} -f)$ (with $A$ the matrix representation of $f$), but I can't really see where the equality come from.
$Ain M_n(K)$ is diagonalizable if and only if there exists a matrix $Pin GL_n(K)$ such that $P^{-1}AP$ is a diagonal matrix.
This is the proof:
$Rightarrow$:Consider the standard basis $mathcal{B}$ for $K^n$. Suppose that $A$ (and therefor $L_A$) is diagonalizable. Then there is a basis $mathcal{B'}={b_1,dots,b_n}$ consisting of eigenvectors of $A$. Define $P$, the transition matrix from $mathcal{B}$ to $mathcal{B'}$; Then $L_{P^{-1}AP}$ maps $b_i$ to $lambda_i b_i$. Conclusion: $P^{-1}AP = operatorname{diag}(lambda_1dots,lambda_n)$.
$Leftarrow$: Suppose there is a $P in GL_n(K)$ such that $P^{-1}AP$ is diagonal, and consider the basis $mathcal{B'} = {Pe_1,dots,Pe_n}$ of $K^n$. It is clear that the matrix of $L_A$ $(A_{mathcal{B},mathcal{B}})$ will be the diagonal matrix $P^{-1}AP$.
I don't understand the sentences that are put in bold. Can someone help me see the logic behind it.
Thanks in advance.
linear-algebra
linear-algebra
asked Jan 11 at 8:58
ZacharyZachary
1559
asked Jan 11 at 8:58
ZacharyZachary
1559
asked Jan 11 at 8:58
ZacharyZachary
1559
asked Jan 11 at 8:58
ZacharyZachary
1559
asked Jan 11 at 8:58
ZacharyZachary
1559
1559
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
- Since $lambda$ is an eigenvalue of $A$, there is a non-zero vector $v$ such that $Av=lambda v$. But$$Av=lambda viff(lambdaoperatorname{Id}-A).v=0impliesdet(lambdaoperatorname{Id}-A)=0.$$
- If $mathcal{B}={e_1,ldots,e_n}$, then the entries of the $i$th column of $P$ are the coordinates of $b_i$ with respect to $mathcal B$. So, $(AP).e_i=A.b_i=lambda_ib_i$. On the other hand, $P^{-1}.b_i=e_i$, and so $(P^{-1}AP).e_i=lambda_ie_i$. And if $P^{-1}AP$ is diagonal, then $(P^{-1}AP)e_i=lambda_ie_i$. But$$(P^{-1}AP)e_i=lambda_ie_iiff A.(Pe_i)=lambda_i(Pe_i).$$So, yes, the matrix $L_A$ with respect to ${Pe_1,ldots,Pe_n}$ will be diagonal.
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Thank you so much! Just two quick questions: 1) I see why $det(lambda I_n - A) = 0$, but why is $det(lambda I_n - A) = det(lambda mathbf{1} - f)$? 2) You say that $(P^{-1} A P)cdot e_i = lambda_i e_i $, does this also hold for the basis elements $b_i$ in $mathcal{B'}$?
$endgroup$
– Zachary
Jan 11 at 9:29
$begingroup$
1) $mathbf{1}$ means $operatorname{Id}$. 2) There are no $b_i$'s in this part of the proof.
$endgroup$
– José Carlos Santos
Jan 11 at 9:44
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%2f3069623%2fquestions-on-diagonalization%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
$begingroup$
- Since $lambda$ is an eigenvalue of $A$, there is a non-zero vector $v$ such that $Av=lambda v$. But$$Av=lambda viff(lambdaoperatorname{Id}-A).v=0impliesdet(lambdaoperatorname{Id}-A)=0.$$
- If $mathcal{B}={e_1,ldots,e_n}$, then the entries of the $i$th column of $P$ are the coordinates of $b_i$ with respect to $mathcal B$. So, $(AP).e_i=A.b_i=lambda_ib_i$. On the other hand, $P^{-1}.b_i=e_i$, and so $(P^{-1}AP).e_i=lambda_ie_i$. And if $P^{-1}AP$ is diagonal, then $(P^{-1}AP)e_i=lambda_ie_i$. But$$(P^{-1}AP)e_i=lambda_ie_iiff A.(Pe_i)=lambda_i(Pe_i).$$So, yes, the matrix $L_A$ with respect to ${Pe_1,ldots,Pe_n}$ will be diagonal.
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Thank you so much! Just two quick questions: 1) I see why $det(lambda I_n - A) = 0$, but why is $det(lambda I_n - A) = det(lambda mathbf{1} - f)$? 2) You say that $(P^{-1} A P)cdot e_i = lambda_i e_i $, does this also hold for the basis elements $b_i$ in $mathcal{B'}$?
$endgroup$
– Zachary
Jan 11 at 9:29
$begingroup$
1) $mathbf{1}$ means $operatorname{Id}$. 2) There are no $b_i$'s in this part of the proof.
$endgroup$
– José Carlos Santos
Jan 11 at 9:44
add a comment |
$begingroup$
- Since $lambda$ is an eigenvalue of $A$, there is a non-zero vector $v$ such that $Av=lambda v$. But$$Av=lambda viff(lambdaoperatorname{Id}-A).v=0impliesdet(lambdaoperatorname{Id}-A)=0.$$
- If $mathcal{B}={e_1,ldots,e_n}$, then the entries of the $i$th column of $P$ are the coordinates of $b_i$ with respect to $mathcal B$. So, $(AP).e_i=A.b_i=lambda_ib_i$. On the other hand, $P^{-1}.b_i=e_i$, and so $(P^{-1}AP).e_i=lambda_ie_i$. And if $P^{-1}AP$ is diagonal, then $(P^{-1}AP)e_i=lambda_ie_i$. But$$(P^{-1}AP)e_i=lambda_ie_iiff A.(Pe_i)=lambda_i(Pe_i).$$So, yes, the matrix $L_A$ with respect to ${Pe_1,ldots,Pe_n}$ will be diagonal.
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Thank you so much! Just two quick questions: 1) I see why $det(lambda I_n - A) = 0$, but why is $det(lambda I_n - A) = det(lambda mathbf{1} - f)$? 2) You say that $(P^{-1} A P)cdot e_i = lambda_i e_i $, does this also hold for the basis elements $b_i$ in $mathcal{B'}$?
$endgroup$
– Zachary
Jan 11 at 9:29
$begingroup$
1) $mathbf{1}$ means $operatorname{Id}$. 2) There are no $b_i$'s in this part of the proof.
$endgroup$
– José Carlos Santos
Jan 11 at 9:44
add a comment |
$begingroup$
- Since $lambda$ is an eigenvalue of $A$, there is a non-zero vector $v$ such that $Av=lambda v$. But$$Av=lambda viff(lambdaoperatorname{Id}-A).v=0impliesdet(lambdaoperatorname{Id}-A)=0.$$
- If $mathcal{B}={e_1,ldots,e_n}$, then the entries of the $i$th column of $P$ are the coordinates of $b_i$ with respect to $mathcal B$. So, $(AP).e_i=A.b_i=lambda_ib_i$. On the other hand, $P^{-1}.b_i=e_i$, and so $(P^{-1}AP).e_i=lambda_ie_i$. And if $P^{-1}AP$ is diagonal, then $(P^{-1}AP)e_i=lambda_ie_i$. But$$(P^{-1}AP)e_i=lambda_ie_iiff A.(Pe_i)=lambda_i(Pe_i).$$So, yes, the matrix $L_A$ with respect to ${Pe_1,ldots,Pe_n}$ will be diagonal.
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
- Since $lambda$ is an eigenvalue of $A$, there is a non-zero vector $v$ such that $Av=lambda v$. But$$Av=lambda viff(lambdaoperatorname{Id}-A).v=0impliesdet(lambdaoperatorname{Id}-A)=0.$$
- If $mathcal{B}={e_1,ldots,e_n}$, then the entries of the $i$th column of $P$ are the coordinates of $b_i$ with respect to $mathcal B$. So, $(AP).e_i=A.b_i=lambda_ib_i$. On the other hand, $P^{-1}.b_i=e_i$, and so $(P^{-1}AP).e_i=lambda_ie_i$. And if $P^{-1}AP$ is diagonal, then $(P^{-1}AP)e_i=lambda_ie_i$. But$$(P^{-1}AP)e_i=lambda_ie_iiff A.(Pe_i)=lambda_i(Pe_i).$$So, yes, the matrix $L_A$ with respect to ${Pe_1,ldots,Pe_n}$ will be diagonal.
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 11 at 9:13
José Carlos SantosJosé Carlos Santos
160k22126232
160k22126232
$begingroup$
Thank you so much! Just two quick questions: 1) I see why $det(lambda I_n - A) = 0$, but why is $det(lambda I_n - A) = det(lambda mathbf{1} - f)$? 2) You say that $(P^{-1} A P)cdot e_i = lambda_i e_i $, does this also hold for the basis elements $b_i$ in $mathcal{B'}$?
$endgroup$
– Zachary
Jan 11 at 9:29
$begingroup$
1) $mathbf{1}$ means $operatorname{Id}$. 2) There are no $b_i$'s in this part of the proof.
$endgroup$
– José Carlos Santos
Jan 11 at 9:44
add a comment |
$begingroup$
Thank you so much! Just two quick questions: 1) I see why $det(lambda I_n - A) = 0$, but why is $det(lambda I_n - A) = det(lambda mathbf{1} - f)$? 2) You say that $(P^{-1} A P)cdot e_i = lambda_i e_i $, does this also hold for the basis elements $b_i$ in $mathcal{B'}$?
$endgroup$
– Zachary
Jan 11 at 9:29
$begingroup$
1) $mathbf{1}$ means $operatorname{Id}$. 2) There are no $b_i$'s in this part of the proof.
$endgroup$
– José Carlos Santos
Jan 11 at 9:44
$begingroup$
Thank you so much! Just two quick questions: 1) I see why $det(lambda I_n - A) = 0$, but why is $det(lambda I_n - A) = det(lambda mathbf{1} - f)$? 2) You say that $(P^{-1} A P)cdot e_i = lambda_i e_i $, does this also hold for the basis elements $b_i$ in $mathcal{B'}$?
$endgroup$
– Zachary
Jan 11 at 9:29
$begingroup$
Thank you so much! Just two quick questions: 1) I see why $det(lambda I_n - A) = 0$, but why is $det(lambda I_n - A) = det(lambda mathbf{1} - f)$? 2) You say that $(P^{-1} A P)cdot e_i = lambda_i e_i $, does this also hold for the basis elements $b_i$ in $mathcal{B'}$?
$endgroup$
– Zachary
Jan 11 at 9:29
$begingroup$
1) $mathbf{1}$ means $operatorname{Id}$. 2) There are no $b_i$'s in this part of the proof.
$endgroup$
– José Carlos Santos
Jan 11 at 9:44
$begingroup$
1) $mathbf{1}$ means $operatorname{Id}$. 2) There are no $b_i$'s in this part of the proof.
$endgroup$
– José Carlos Santos
Jan 11 at 9:44
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%2f3069623%2fquestions-on-diagonalization%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
