Possible eigenvalues of a matrix satisfying $A*A = 4A$ [closed]
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What is the eigenvalue or what could the eigenvalue be of a matrix $A$ that if multiplied by itself equals $4A$?
linear-algebra eigenvalues-eigenvectors
edited Jan 22 at 18:44
pwerth
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asked Jan 22 at 18:36
TournaTourna
126
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closed as off-topic by Martin R, caverac, anomaly, MisterRiemann, Dietrich Burde Jan 22 at 19:07
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$begingroup$
What is the eigenvalue or what could the eigenvalue be of a matrix $A$ that if multiplied by itself equals $4A$?
linear-algebra eigenvalues-eigenvectors
edited Jan 22 at 18:44
pwerth
3,243417
asked Jan 22 at 18:36
TournaTourna
126
$endgroup$
closed as off-topic by Martin R, caverac, anomaly, MisterRiemann, Dietrich Burde Jan 22 at 19:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, caverac, anomaly, MisterRiemann, Dietrich Burde
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Hint: for an eigenvector $v$, consider $A^2v=4Av$.
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– Wojowu
Jan 22 at 18:38
add a comment |
$begingroup$
What is the eigenvalue or what could the eigenvalue be of a matrix $A$ that if multiplied by itself equals $4A$?
linear-algebra eigenvalues-eigenvectors
edited Jan 22 at 18:44
pwerth
3,243417
asked Jan 22 at 18:36
TournaTourna
126
$endgroup$
What is the eigenvalue or what could the eigenvalue be of a matrix $A$ that if multiplied by itself equals $4A$?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Jan 22 at 18:44
pwerth
3,243417
asked Jan 22 at 18:36
TournaTourna
126
edited Jan 22 at 18:44
pwerth
3,243417
asked Jan 22 at 18:36
TournaTourna
126
edited Jan 22 at 18:44
pwerth
3,243417
edited Jan 22 at 18:44
pwerth
3,243417
edited Jan 22 at 18:44
pwerth
3,243417
3,243417
asked Jan 22 at 18:36
TournaTourna
126
asked Jan 22 at 18:36
TournaTourna
126
asked Jan 22 at 18:36
TournaTourna
126
126
closed as off-topic by Martin R, caverac, anomaly, MisterRiemann, Dietrich Burde Jan 22 at 19:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, caverac, anomaly, MisterRiemann, Dietrich Burde
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Martin R, caverac, anomaly, MisterRiemann, Dietrich Burde Jan 22 at 19:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, caverac, anomaly, MisterRiemann, Dietrich Burde
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Hint: for an eigenvector $v$, consider $A^2v=4Av$.
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– Wojowu
Jan 22 at 18:38
add a comment |
1
$begingroup$
Hint: for an eigenvector $v$, consider $A^2v=4Av$.
$endgroup$
– Wojowu
Jan 22 at 18:38
1
1
$begingroup$
Hint: for an eigenvector $v$, consider $A^2v=4Av$.
$endgroup$
– Wojowu
Jan 22 at 18:38
$begingroup$
Hint: for an eigenvector $v$, consider $A^2v=4Av$.
$endgroup$
– Wojowu
Jan 22 at 18:38
add a comment |
3 Answers
3
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Suppose $A$ has an eigenvalue $lambda$, so there exists $v$ with $Av=lambda v$. Using the fact that $A^{2}=4A$, we have
$$A^{2}v=(4A)v=4(Av)=4(lambda v)=(4lambda)v$$
so that $v$ is also an eigenvector of $A^{2}$, with eigenvalue $4lambda$. But
$$A^{2}v=A(Av)=A(lambda v)=lambda(Av)=lambda(lambda v)=lambda^{2}v$$
so the eigenvalue of $v$ (with respect to $A^{2}$) is $lambda^{2}$. Since an eigenvector can only correspond to one eigenvalue, we must have $lambda^{2}=4lambda$, so that $lambda = 0$ or $4$.
answered Jan 22 at 18:42
pwerthpwerth
3,243417
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You are given that $A^2-4A=0$. If $v$ is a nonzero eigenvector of $A$, say with eigenvalue $lambda$, then
$$A^2v=A(Av)=A(lambda v)=lambda Av=lambda^2v,$$
and so
$$(A^2-4A)v=A^2v-4Av=lambda^2v-4lambda v=(lambda^2-4lambda)v.$$
But $A^2-4A=0$, so $lambda^2-4lambda=0$. So what are the possible values for $lambda$?
answered Jan 22 at 18:47
ServaesServaes
26k33997
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The minimal polynomial of $A$ is $x^2-4x=x(x-4)$ or $x$ or $x-4$. So the eigenvalues can be $0,4$.
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose $A$ has an eigenvalue $lambda$, so there exists $v$ with $Av=lambda v$. Using the fact that $A^{2}=4A$, we have
$$A^{2}v=(4A)v=4(Av)=4(lambda v)=(4lambda)v$$
so that $v$ is also an eigenvector of $A^{2}$, with eigenvalue $4lambda$. But
$$A^{2}v=A(Av)=A(lambda v)=lambda(Av)=lambda(lambda v)=lambda^{2}v$$
so the eigenvalue of $v$ (with respect to $A^{2}$) is $lambda^{2}$. Since an eigenvector can only correspond to one eigenvalue, we must have $lambda^{2}=4lambda$, so that $lambda = 0$ or $4$.
answered Jan 22 at 18:42
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
Suppose $A$ has an eigenvalue $lambda$, so there exists $v$ with $Av=lambda v$. Using the fact that $A^{2}=4A$, we have
$$A^{2}v=(4A)v=4(Av)=4(lambda v)=(4lambda)v$$
so that $v$ is also an eigenvector of $A^{2}$, with eigenvalue $4lambda$. But
$$A^{2}v=A(Av)=A(lambda v)=lambda(Av)=lambda(lambda v)=lambda^{2}v$$
so the eigenvalue of $v$ (with respect to $A^{2}$) is $lambda^{2}$. Since an eigenvector can only correspond to one eigenvalue, we must have $lambda^{2}=4lambda$, so that $lambda = 0$ or $4$.
answered Jan 22 at 18:42
pwerthpwerth
3,243417
$endgroup$
add a comment |
$begingroup$
Suppose $A$ has an eigenvalue $lambda$, so there exists $v$ with $Av=lambda v$. Using the fact that $A^{2}=4A$, we have
$$A^{2}v=(4A)v=4(Av)=4(lambda v)=(4lambda)v$$
so that $v$ is also an eigenvector of $A^{2}$, with eigenvalue $4lambda$. But
$$A^{2}v=A(Av)=A(lambda v)=lambda(Av)=lambda(lambda v)=lambda^{2}v$$
so the eigenvalue of $v$ (with respect to $A^{2}$) is $lambda^{2}$. Since an eigenvector can only correspond to one eigenvalue, we must have $lambda^{2}=4lambda$, so that $lambda = 0$ or $4$.
answered Jan 22 at 18:42
pwerthpwerth
3,243417
$endgroup$
Suppose $A$ has an eigenvalue $lambda$, so there exists $v$ with $Av=lambda v$. Using the fact that $A^{2}=4A$, we have
$$A^{2}v=(4A)v=4(Av)=4(lambda v)=(4lambda)v$$
so that $v$ is also an eigenvector of $A^{2}$, with eigenvalue $4lambda$. But
$$A^{2}v=A(Av)=A(lambda v)=lambda(Av)=lambda(lambda v)=lambda^{2}v$$
so the eigenvalue of $v$ (with respect to $A^{2}$) is $lambda^{2}$. Since an eigenvector can only correspond to one eigenvalue, we must have $lambda^{2}=4lambda$, so that $lambda = 0$ or $4$.
answered Jan 22 at 18:42
pwerthpwerth
3,243417
answered Jan 22 at 18:42
pwerthpwerth
3,243417
answered Jan 22 at 18:42
pwerthpwerth
3,243417
answered Jan 22 at 18:42
pwerthpwerth
3,243417
3,243417
add a comment |
add a comment |
$begingroup$
You are given that $A^2-4A=0$. If $v$ is a nonzero eigenvector of $A$, say with eigenvalue $lambda$, then
$$A^2v=A(Av)=A(lambda v)=lambda Av=lambda^2v,$$
and so
$$(A^2-4A)v=A^2v-4Av=lambda^2v-4lambda v=(lambda^2-4lambda)v.$$
But $A^2-4A=0$, so $lambda^2-4lambda=0$. So what are the possible values for $lambda$?
answered Jan 22 at 18:47
ServaesServaes
26k33997
$endgroup$
add a comment |
$begingroup$
You are given that $A^2-4A=0$. If $v$ is a nonzero eigenvector of $A$, say with eigenvalue $lambda$, then
$$A^2v=A(Av)=A(lambda v)=lambda Av=lambda^2v,$$
and so
$$(A^2-4A)v=A^2v-4Av=lambda^2v-4lambda v=(lambda^2-4lambda)v.$$
But $A^2-4A=0$, so $lambda^2-4lambda=0$. So what are the possible values for $lambda$?
answered Jan 22 at 18:47
ServaesServaes
26k33997
$endgroup$
add a comment |
$begingroup$
You are given that $A^2-4A=0$. If $v$ is a nonzero eigenvector of $A$, say with eigenvalue $lambda$, then
$$A^2v=A(Av)=A(lambda v)=lambda Av=lambda^2v,$$
and so
$$(A^2-4A)v=A^2v-4Av=lambda^2v-4lambda v=(lambda^2-4lambda)v.$$
But $A^2-4A=0$, so $lambda^2-4lambda=0$. So what are the possible values for $lambda$?
answered Jan 22 at 18:47
ServaesServaes
26k33997
$endgroup$
You are given that $A^2-4A=0$. If $v$ is a nonzero eigenvector of $A$, say with eigenvalue $lambda$, then
$$A^2v=A(Av)=A(lambda v)=lambda Av=lambda^2v,$$
and so
$$(A^2-4A)v=A^2v-4Av=lambda^2v-4lambda v=(lambda^2-4lambda)v.$$
But $A^2-4A=0$, so $lambda^2-4lambda=0$. So what are the possible values for $lambda$?
answered Jan 22 at 18:47
ServaesServaes
26k33997
answered Jan 22 at 18:47
ServaesServaes
26k33997
answered Jan 22 at 18:47
ServaesServaes
26k33997
answered Jan 22 at 18:47
ServaesServaes
26k33997
26k33997
add a comment |
add a comment |
$begingroup$
The minimal polynomial of $A$ is $x^2-4x=x(x-4)$ or $x$ or $x-4$. So the eigenvalues can be $0,4$.
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The minimal polynomial of $A$ is $x^2-4x=x(x-4)$ or $x$ or $x-4$. So the eigenvalues can be $0,4$.
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The minimal polynomial of $A$ is $x^2-4x=x(x-4)$ or $x$ or $x-4$. So the eigenvalues can be $0,4$.
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
$endgroup$
The minimal polynomial of $A$ is $x^2-4x=x(x-4)$ or $x$ or $x-4$. So the eigenvalues can be $0,4$.
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
edited Jan 22 at 18:48
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
answered Jan 22 at 18:47
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
add a comment |
1
$begingroup$
Hint: for an eigenvector $v$, consider $A^2v=4Av$.
$endgroup$
– Wojowu
Jan 22 at 18:38