Mixing
Diagonalizing matrix with fractions [closed]
$begingroup$
I'm revising for an exam in linear algebra, and I've found myself stuck on this one specific exercise.
I'm supposed to decide a matrix $P$ and a diagonal matrix $D$ from my matrix $H$ (which I'll post below, so that $P^{-1}HP = D$.
Normally, I know how to solve tasks like these, but the fractions are what's giving me the issues when trying to get the eigenvalues out of the matrix. I'm just clueless on how to get them, so if any of you could help me out I'd greatly appreciate it.
$$H= begin{pmatrix} frac{3}{2} & - frac{1}{2} &0\ - frac{1}{2} & frac{3}{2}&0 \ 0 &0&1end{pmatrix}$$
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited Jan 17 at 16:11
idriskameni
683319
asked Jan 17 at 15:15
wzndwznd
85
$endgroup$
closed as off-topic by José Carlos Santos, Alexander Gruber♦ Jan 17 at 23:17
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." – José Carlos Santos, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm revising for an exam in linear algebra, and I've found myself stuck on this one specific exercise.
I'm supposed to decide a matrix $P$ and a diagonal matrix $D$ from my matrix $H$ (which I'll post below, so that $P^{-1}HP = D$.
Normally, I know how to solve tasks like these, but the fractions are what's giving me the issues when trying to get the eigenvalues out of the matrix. I'm just clueless on how to get them, so if any of you could help me out I'd greatly appreciate it.
$$H= begin{pmatrix} frac{3}{2} & - frac{1}{2} &0\ - frac{1}{2} & frac{3}{2}&0 \ 0 &0&1end{pmatrix}$$
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited Jan 17 at 16:11
idriskameni
683319
asked Jan 17 at 15:15
wzndwznd
85
$endgroup$
closed as off-topic by José Carlos Santos, Alexander Gruber♦ Jan 17 at 23:17
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." – José Carlos Santos, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
If it's just the fractions that are making it hard for you, then just diagonalize $2H$. Now, all entries are integer.
$endgroup$
– José Carlos Santos
Jan 17 at 15:23
$begingroup$
Whenever I try doing that I get different eigenvalues though. The common problem I have with these sort of tasks are to get the eigenvalues. The rest I got down, but I just keep screwing up on the eigenvalues. Do you possibly think you could show how to get the correct eigenvalues?
$endgroup$
– wznd
Jan 17 at 15:31
$begingroup$
Why? Didn't you write that you know how to solve tasks like these and that you were just having problems with the fractions?
$endgroup$
– José Carlos Santos
Jan 17 at 15:33
$begingroup$
Normally I get through these tasks fine, except for the eigenvalues which is a problem for me. I should probably have mentioned it.
$endgroup$
– wznd
Jan 17 at 15:38
add a comment |
$begingroup$
I'm revising for an exam in linear algebra, and I've found myself stuck on this one specific exercise.
I'm supposed to decide a matrix $P$ and a diagonal matrix $D$ from my matrix $H$ (which I'll post below, so that $P^{-1}HP = D$.
Normally, I know how to solve tasks like these, but the fractions are what's giving me the issues when trying to get the eigenvalues out of the matrix. I'm just clueless on how to get them, so if any of you could help me out I'd greatly appreciate it.
$$H= begin{pmatrix} frac{3}{2} & - frac{1}{2} &0\ - frac{1}{2} & frac{3}{2}&0 \ 0 &0&1end{pmatrix}$$
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited Jan 17 at 16:11
idriskameni
683319
asked Jan 17 at 15:15
wzndwznd
85
$endgroup$
I'm revising for an exam in linear algebra, and I've found myself stuck on this one specific exercise.
I'm supposed to decide a matrix $P$ and a diagonal matrix $D$ from my matrix $H$ (which I'll post below, so that $P^{-1}HP = D$.
Normally, I know how to solve tasks like these, but the fractions are what's giving me the issues when trying to get the eigenvalues out of the matrix. I'm just clueless on how to get them, so if any of you could help me out I'd greatly appreciate it.
$$H= begin{pmatrix} frac{3}{2} & - frac{1}{2} &0\ - frac{1}{2} & frac{3}{2}&0 \ 0 &0&1end{pmatrix}$$
linear-algebra matrices eigenvalues-eigenvectors diagonalization
linear-algebra matrices eigenvalues-eigenvectors diagonalization
edited Jan 17 at 16:11
idriskameni
683319
asked Jan 17 at 15:15
wzndwznd
85
edited Jan 17 at 16:11
idriskameni
683319
asked Jan 17 at 15:15
wzndwznd
85
edited Jan 17 at 16:11
idriskameni
683319
edited Jan 17 at 16:11
idriskameni
683319
edited Jan 17 at 16:11
idriskameni
683319
683319
asked Jan 17 at 15:15
wzndwznd
85
asked Jan 17 at 15:15
wzndwznd
85
asked Jan 17 at 15:15
wzndwznd
85
85
closed as off-topic by José Carlos Santos, Alexander Gruber♦ Jan 17 at 23:17
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." – José Carlos Santos, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Alexander Gruber♦ Jan 17 at 23:17
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." – José Carlos Santos, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
If it's just the fractions that are making it hard for you, then just diagonalize $2H$. Now, all entries are integer.
$endgroup$
– José Carlos Santos
Jan 17 at 15:23
$begingroup$
Whenever I try doing that I get different eigenvalues though. The common problem I have with these sort of tasks are to get the eigenvalues. The rest I got down, but I just keep screwing up on the eigenvalues. Do you possibly think you could show how to get the correct eigenvalues?
$endgroup$
– wznd
Jan 17 at 15:31
$begingroup$
Why? Didn't you write that you know how to solve tasks like these and that you were just having problems with the fractions?
$endgroup$
– José Carlos Santos
Jan 17 at 15:33
$begingroup$
Normally I get through these tasks fine, except for the eigenvalues which is a problem for me. I should probably have mentioned it.
$endgroup$
– wznd
Jan 17 at 15:38
add a comment |
$begingroup$
If it's just the fractions that are making it hard for you, then just diagonalize $2H$. Now, all entries are integer.
$endgroup$
– José Carlos Santos
Jan 17 at 15:23
$begingroup$
Whenever I try doing that I get different eigenvalues though. The common problem I have with these sort of tasks are to get the eigenvalues. The rest I got down, but I just keep screwing up on the eigenvalues. Do you possibly think you could show how to get the correct eigenvalues?
$endgroup$
– wznd
Jan 17 at 15:31
$begingroup$
Why? Didn't you write that you know how to solve tasks like these and that you were just having problems with the fractions?
$endgroup$
– José Carlos Santos
Jan 17 at 15:33
$begingroup$
Normally I get through these tasks fine, except for the eigenvalues which is a problem for me. I should probably have mentioned it.
$endgroup$
– wznd
Jan 17 at 15:38
$begingroup$
If it's just the fractions that are making it hard for you, then just diagonalize $2H$. Now, all entries are integer.
$endgroup$
– José Carlos Santos
Jan 17 at 15:23
$begingroup$
If it's just the fractions that are making it hard for you, then just diagonalize $2H$. Now, all entries are integer.
$endgroup$
– José Carlos Santos
Jan 17 at 15:23
$begingroup$
Whenever I try doing that I get different eigenvalues though. The common problem I have with these sort of tasks are to get the eigenvalues. The rest I got down, but I just keep screwing up on the eigenvalues. Do you possibly think you could show how to get the correct eigenvalues?
$endgroup$
– wznd
Jan 17 at 15:31
$begingroup$
Whenever I try doing that I get different eigenvalues though. The common problem I have with these sort of tasks are to get the eigenvalues. The rest I got down, but I just keep screwing up on the eigenvalues. Do you possibly think you could show how to get the correct eigenvalues?
$endgroup$
– wznd
Jan 17 at 15:31
$begingroup$
Why? Didn't you write that you know how to solve tasks like these and that you were just having problems with the fractions?
$endgroup$
– José Carlos Santos
Jan 17 at 15:33
$begingroup$
Why? Didn't you write that you know how to solve tasks like these and that you were just having problems with the fractions?
$endgroup$
– José Carlos Santos
Jan 17 at 15:33
$begingroup$
Normally I get through these tasks fine, except for the eigenvalues which is a problem for me. I should probably have mentioned it.
$endgroup$
– wznd
Jan 17 at 15:38
$begingroup$
Normally I get through these tasks fine, except for the eigenvalues which is a problem for me. I should probably have mentioned it.
$endgroup$
– wznd
Jan 17 at 15:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that:
$$
|H-lambda I|=
begin{vmatrix}
frac{3}{2}-lambda & -frac{1}{2} & 0 \
-frac{1}{2} & frac{3}{2}-lambda & 0 \
0 & 0 & 1-lambda
end{vmatrix}=(1-lambda)left(left(frac{3}{2}-lambdaright)^2-frac{1}
{4}right)=
(1-lambda)(2-3lambda+lambda^2)=(1-lambda)(lambda-1)(lambda-2)=-(lambda-1)^2(lambda-2)
$$
(I calculated the determinant using the Laplace expansion of the third column). The eigenvalues are hence $1$ and $2$. Can you take it from here?
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
$endgroup$
$begingroup$
Yes, this is exactly the one problem I have with this task, thank you for this. From now on I don't find it very difficult since I just need the eigenvectors to get P etc. What I was having difficulties with/still having a hard time to understand is firstly, where did you ge the 1/4 from? Also, how does (3/2 - λ)^2 - 1/4 become (1-λ)(2-3λ+λ^2). Once again, thank you for helping me out :)
$endgroup$
– wznd
Jan 17 at 16:36
$begingroup$
Using $(a-b)^2=a^2-2ab+b^2$ we get that $left(frac{3}{2}-lambdaright)^2=(3/2)^2-2(3/2)(lambda)+lambda^2=9/4-3lambda+lambda^2$. Subtracting $frac{1}{4}$, we have $frac{9}{4}-frac{1}{4}=frac{8}{4}=2$ so $left(frac{3}{2}-lambdaright)^2-frac{1}{4}=2-3lambda+lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(lambda-1)(lambda-2)$. Multiplying by the $(1-lambda)$ we get from the determinant we get the characteristic polynomial I said earlier.
$endgroup$
– Yuval Gat
Jan 17 at 16:40
$begingroup$
If you have any more questions about this I'll be happy to answer you. If not, please mark the question as solved by accepting my answer ;)
$endgroup$
– Yuval Gat
Jan 17 at 17:56
add a comment |
$begingroup$
If you’re having trouble computing eigenvalues, try looking for eigenvectors instead. You should be able to tell at a glance that $(0,0,1)^T$ is an eigenvector of $H$ since that vector gets mapped to itself—the columns of $H$ are the images of the basis vectors. The corresponding eigenvalue is, of course, $1$.
Now focus on the upper-right $2times2$ submatrix. Both rows have the same elements, so the row sums are equal. Summing the first two rows of $H$ is equivalent to right-multiplying by $(1,1,0)^T$, so there’s another eigenvector that’s linearly independent of the first one, with eigenvalue $frac32-frac12=1$.
You can always get the last eigenvalue “for free” since the sum of the eigenvalues, taking multiplicity into account, is equal to the trace. So, the remaining eigenvalue is $4-1-1=2$. To find a corresponding eigenvector, recall that a real symmetric matrix can be orthogonally diagonalized, so any eigenvector of $2$ has to be orthogonal to both of the eigenvectors of $1$ that you’ve already found. You’re working in $mathbb R^3$, so such a vector can be found via a cross product: $(1,1,0)^Ttimes(0,0,1)^T=(1,-1,0)^T$ (which could also have been found by inspection).
answered Jan 17 at 18:16
amdamd
30.6k21050
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that:
$$
|H-lambda I|=
begin{vmatrix}
frac{3}{2}-lambda & -frac{1}{2} & 0 \
-frac{1}{2} & frac{3}{2}-lambda & 0 \
0 & 0 & 1-lambda
end{vmatrix}=(1-lambda)left(left(frac{3}{2}-lambdaright)^2-frac{1}
{4}right)=
(1-lambda)(2-3lambda+lambda^2)=(1-lambda)(lambda-1)(lambda-2)=-(lambda-1)^2(lambda-2)
$$
(I calculated the determinant using the Laplace expansion of the third column). The eigenvalues are hence $1$ and $2$. Can you take it from here?
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
$endgroup$
$begingroup$
Yes, this is exactly the one problem I have with this task, thank you for this. From now on I don't find it very difficult since I just need the eigenvectors to get P etc. What I was having difficulties with/still having a hard time to understand is firstly, where did you ge the 1/4 from? Also, how does (3/2 - λ)^2 - 1/4 become (1-λ)(2-3λ+λ^2). Once again, thank you for helping me out :)
$endgroup$
– wznd
Jan 17 at 16:36
$begingroup$
Using $(a-b)^2=a^2-2ab+b^2$ we get that $left(frac{3}{2}-lambdaright)^2=(3/2)^2-2(3/2)(lambda)+lambda^2=9/4-3lambda+lambda^2$. Subtracting $frac{1}{4}$, we have $frac{9}{4}-frac{1}{4}=frac{8}{4}=2$ so $left(frac{3}{2}-lambdaright)^2-frac{1}{4}=2-3lambda+lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(lambda-1)(lambda-2)$. Multiplying by the $(1-lambda)$ we get from the determinant we get the characteristic polynomial I said earlier.
$endgroup$
– Yuval Gat
Jan 17 at 16:40
$begingroup$
If you have any more questions about this I'll be happy to answer you. If not, please mark the question as solved by accepting my answer ;)
$endgroup$
– Yuval Gat
Jan 17 at 17:56
add a comment |
$begingroup$
Note that:
$$
|H-lambda I|=
begin{vmatrix}
frac{3}{2}-lambda & -frac{1}{2} & 0 \
-frac{1}{2} & frac{3}{2}-lambda & 0 \
0 & 0 & 1-lambda
end{vmatrix}=(1-lambda)left(left(frac{3}{2}-lambdaright)^2-frac{1}
{4}right)=
(1-lambda)(2-3lambda+lambda^2)=(1-lambda)(lambda-1)(lambda-2)=-(lambda-1)^2(lambda-2)
$$
(I calculated the determinant using the Laplace expansion of the third column). The eigenvalues are hence $1$ and $2$. Can you take it from here?
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
$endgroup$
$begingroup$
Yes, this is exactly the one problem I have with this task, thank you for this. From now on I don't find it very difficult since I just need the eigenvectors to get P etc. What I was having difficulties with/still having a hard time to understand is firstly, where did you ge the 1/4 from? Also, how does (3/2 - λ)^2 - 1/4 become (1-λ)(2-3λ+λ^2). Once again, thank you for helping me out :)
$endgroup$
– wznd
Jan 17 at 16:36
$begingroup$
Using $(a-b)^2=a^2-2ab+b^2$ we get that $left(frac{3}{2}-lambdaright)^2=(3/2)^2-2(3/2)(lambda)+lambda^2=9/4-3lambda+lambda^2$. Subtracting $frac{1}{4}$, we have $frac{9}{4}-frac{1}{4}=frac{8}{4}=2$ so $left(frac{3}{2}-lambdaright)^2-frac{1}{4}=2-3lambda+lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(lambda-1)(lambda-2)$. Multiplying by the $(1-lambda)$ we get from the determinant we get the characteristic polynomial I said earlier.
$endgroup$
– Yuval Gat
Jan 17 at 16:40
$begingroup$
If you have any more questions about this I'll be happy to answer you. If not, please mark the question as solved by accepting my answer ;)
$endgroup$
– Yuval Gat
Jan 17 at 17:56
add a comment |
$begingroup$
Note that:
$$
|H-lambda I|=
begin{vmatrix}
frac{3}{2}-lambda & -frac{1}{2} & 0 \
-frac{1}{2} & frac{3}{2}-lambda & 0 \
0 & 0 & 1-lambda
end{vmatrix}=(1-lambda)left(left(frac{3}{2}-lambdaright)^2-frac{1}
{4}right)=
(1-lambda)(2-3lambda+lambda^2)=(1-lambda)(lambda-1)(lambda-2)=-(lambda-1)^2(lambda-2)
$$
(I calculated the determinant using the Laplace expansion of the third column). The eigenvalues are hence $1$ and $2$. Can you take it from here?
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
$endgroup$
Note that:
$$
|H-lambda I|=
begin{vmatrix}
frac{3}{2}-lambda & -frac{1}{2} & 0 \
-frac{1}{2} & frac{3}{2}-lambda & 0 \
0 & 0 & 1-lambda
end{vmatrix}=(1-lambda)left(left(frac{3}{2}-lambdaright)^2-frac{1}
{4}right)=
(1-lambda)(2-3lambda+lambda^2)=(1-lambda)(lambda-1)(lambda-2)=-(lambda-1)^2(lambda-2)
$$
(I calculated the determinant using the Laplace expansion of the third column). The eigenvalues are hence $1$ and $2$. Can you take it from here?
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
answered Jan 17 at 16:29
Yuval GatYuval Gat
572213
572213
$begingroup$
Yes, this is exactly the one problem I have with this task, thank you for this. From now on I don't find it very difficult since I just need the eigenvectors to get P etc. What I was having difficulties with/still having a hard time to understand is firstly, where did you ge the 1/4 from? Also, how does (3/2 - λ)^2 - 1/4 become (1-λ)(2-3λ+λ^2). Once again, thank you for helping me out :)
$endgroup$
– wznd
Jan 17 at 16:36
$begingroup$
Using $(a-b)^2=a^2-2ab+b^2$ we get that $left(frac{3}{2}-lambdaright)^2=(3/2)^2-2(3/2)(lambda)+lambda^2=9/4-3lambda+lambda^2$. Subtracting $frac{1}{4}$, we have $frac{9}{4}-frac{1}{4}=frac{8}{4}=2$ so $left(frac{3}{2}-lambdaright)^2-frac{1}{4}=2-3lambda+lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(lambda-1)(lambda-2)$. Multiplying by the $(1-lambda)$ we get from the determinant we get the characteristic polynomial I said earlier.
$endgroup$
– Yuval Gat
Jan 17 at 16:40
$begingroup$
If you have any more questions about this I'll be happy to answer you. If not, please mark the question as solved by accepting my answer ;)
$endgroup$
– Yuval Gat
Jan 17 at 17:56
add a comment |
$begingroup$
Yes, this is exactly the one problem I have with this task, thank you for this. From now on I don't find it very difficult since I just need the eigenvectors to get P etc. What I was having difficulties with/still having a hard time to understand is firstly, where did you ge the 1/4 from? Also, how does (3/2 - λ)^2 - 1/4 become (1-λ)(2-3λ+λ^2). Once again, thank you for helping me out :)
$endgroup$
– wznd
Jan 17 at 16:36
$begingroup$
Using $(a-b)^2=a^2-2ab+b^2$ we get that $left(frac{3}{2}-lambdaright)^2=(3/2)^2-2(3/2)(lambda)+lambda^2=9/4-3lambda+lambda^2$. Subtracting $frac{1}{4}$, we have $frac{9}{4}-frac{1}{4}=frac{8}{4}=2$ so $left(frac{3}{2}-lambdaright)^2-frac{1}{4}=2-3lambda+lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(lambda-1)(lambda-2)$. Multiplying by the $(1-lambda)$ we get from the determinant we get the characteristic polynomial I said earlier.
$endgroup$
– Yuval Gat
Jan 17 at 16:40
$begingroup$
If you have any more questions about this I'll be happy to answer you. If not, please mark the question as solved by accepting my answer ;)
$endgroup$
– Yuval Gat
Jan 17 at 17:56
$begingroup$
Yes, this is exactly the one problem I have with this task, thank you for this. From now on I don't find it very difficult since I just need the eigenvectors to get P etc. What I was having difficulties with/still having a hard time to understand is firstly, where did you ge the 1/4 from? Also, how does (3/2 - λ)^2 - 1/4 become (1-λ)(2-3λ+λ^2). Once again, thank you for helping me out :)
$endgroup$
– wznd
Jan 17 at 16:36
$begingroup$
Yes, this is exactly the one problem I have with this task, thank you for this. From now on I don't find it very difficult since I just need the eigenvectors to get P etc. What I was having difficulties with/still having a hard time to understand is firstly, where did you ge the 1/4 from? Also, how does (3/2 - λ)^2 - 1/4 become (1-λ)(2-3λ+λ^2). Once again, thank you for helping me out :)
$endgroup$
– wznd
Jan 17 at 16:36
$begingroup$
Using $(a-b)^2=a^2-2ab+b^2$ we get that $left(frac{3}{2}-lambdaright)^2=(3/2)^2-2(3/2)(lambda)+lambda^2=9/4-3lambda+lambda^2$. Subtracting $frac{1}{4}$, we have $frac{9}{4}-frac{1}{4}=frac{8}{4}=2$ so $left(frac{3}{2}-lambdaright)^2-frac{1}{4}=2-3lambda+lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(lambda-1)(lambda-2)$. Multiplying by the $(1-lambda)$ we get from the determinant we get the characteristic polynomial I said earlier.
$endgroup$
– Yuval Gat
Jan 17 at 16:40
$begingroup$
Using $(a-b)^2=a^2-2ab+b^2$ we get that $left(frac{3}{2}-lambdaright)^2=(3/2)^2-2(3/2)(lambda)+lambda^2=9/4-3lambda+lambda^2$. Subtracting $frac{1}{4}$, we have $frac{9}{4}-frac{1}{4}=frac{8}{4}=2$ so $left(frac{3}{2}-lambdaright)^2-frac{1}{4}=2-3lambda+lambda^2$. Now, note that by finding the roots of that quadratic (using the quadratic formula, for example) we get $1, 2$ and so the polynomial is actually just $(lambda-1)(lambda-2)$. Multiplying by the $(1-lambda)$ we get from the determinant we get the characteristic polynomial I said earlier.
$endgroup$
– Yuval Gat
Jan 17 at 16:40
$begingroup$
If you have any more questions about this I'll be happy to answer you. If not, please mark the question as solved by accepting my answer ;)
$endgroup$
– Yuval Gat
Jan 17 at 17:56
$begingroup$
If you have any more questions about this I'll be happy to answer you. If not, please mark the question as solved by accepting my answer ;)
$endgroup$
– Yuval Gat
Jan 17 at 17:56
add a comment |
$begingroup$
If you’re having trouble computing eigenvalues, try looking for eigenvectors instead. You should be able to tell at a glance that $(0,0,1)^T$ is an eigenvector of $H$ since that vector gets mapped to itself—the columns of $H$ are the images of the basis vectors. The corresponding eigenvalue is, of course, $1$.
Now focus on the upper-right $2times2$ submatrix. Both rows have the same elements, so the row sums are equal. Summing the first two rows of $H$ is equivalent to right-multiplying by $(1,1,0)^T$, so there’s another eigenvector that’s linearly independent of the first one, with eigenvalue $frac32-frac12=1$.
You can always get the last eigenvalue “for free” since the sum of the eigenvalues, taking multiplicity into account, is equal to the trace. So, the remaining eigenvalue is $4-1-1=2$. To find a corresponding eigenvector, recall that a real symmetric matrix can be orthogonally diagonalized, so any eigenvector of $2$ has to be orthogonal to both of the eigenvectors of $1$ that you’ve already found. You’re working in $mathbb R^3$, so such a vector can be found via a cross product: $(1,1,0)^Ttimes(0,0,1)^T=(1,-1,0)^T$ (which could also have been found by inspection).
answered Jan 17 at 18:16
amdamd
30.6k21050
$endgroup$
add a comment |
$begingroup$
If you’re having trouble computing eigenvalues, try looking for eigenvectors instead. You should be able to tell at a glance that $(0,0,1)^T$ is an eigenvector of $H$ since that vector gets mapped to itself—the columns of $H$ are the images of the basis vectors. The corresponding eigenvalue is, of course, $1$.
Now focus on the upper-right $2times2$ submatrix. Both rows have the same elements, so the row sums are equal. Summing the first two rows of $H$ is equivalent to right-multiplying by $(1,1,0)^T$, so there’s another eigenvector that’s linearly independent of the first one, with eigenvalue $frac32-frac12=1$.
You can always get the last eigenvalue “for free” since the sum of the eigenvalues, taking multiplicity into account, is equal to the trace. So, the remaining eigenvalue is $4-1-1=2$. To find a corresponding eigenvector, recall that a real symmetric matrix can be orthogonally diagonalized, so any eigenvector of $2$ has to be orthogonal to both of the eigenvectors of $1$ that you’ve already found. You’re working in $mathbb R^3$, so such a vector can be found via a cross product: $(1,1,0)^Ttimes(0,0,1)^T=(1,-1,0)^T$ (which could also have been found by inspection).
answered Jan 17 at 18:16
amdamd
30.6k21050
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add a comment |
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If you’re having trouble computing eigenvalues, try looking for eigenvectors instead. You should be able to tell at a glance that $(0,0,1)^T$ is an eigenvector of $H$ since that vector gets mapped to itself—the columns of $H$ are the images of the basis vectors. The corresponding eigenvalue is, of course, $1$.
Now focus on the upper-right $2times2$ submatrix. Both rows have the same elements, so the row sums are equal. Summing the first two rows of $H$ is equivalent to right-multiplying by $(1,1,0)^T$, so there’s another eigenvector that’s linearly independent of the first one, with eigenvalue $frac32-frac12=1$.
You can always get the last eigenvalue “for free” since the sum of the eigenvalues, taking multiplicity into account, is equal to the trace. So, the remaining eigenvalue is $4-1-1=2$. To find a corresponding eigenvector, recall that a real symmetric matrix can be orthogonally diagonalized, so any eigenvector of $2$ has to be orthogonal to both of the eigenvectors of $1$ that you’ve already found. You’re working in $mathbb R^3$, so such a vector can be found via a cross product: $(1,1,0)^Ttimes(0,0,1)^T=(1,-1,0)^T$ (which could also have been found by inspection).
answered Jan 17 at 18:16
amdamd
30.6k21050
$endgroup$
If you’re having trouble computing eigenvalues, try looking for eigenvectors instead. You should be able to tell at a glance that $(0,0,1)^T$ is an eigenvector of $H$ since that vector gets mapped to itself—the columns of $H$ are the images of the basis vectors. The corresponding eigenvalue is, of course, $1$.
Now focus on the upper-right $2times2$ submatrix. Both rows have the same elements, so the row sums are equal. Summing the first two rows of $H$ is equivalent to right-multiplying by $(1,1,0)^T$, so there’s another eigenvector that’s linearly independent of the first one, with eigenvalue $frac32-frac12=1$.
You can always get the last eigenvalue “for free” since the sum of the eigenvalues, taking multiplicity into account, is equal to the trace. So, the remaining eigenvalue is $4-1-1=2$. To find a corresponding eigenvector, recall that a real symmetric matrix can be orthogonally diagonalized, so any eigenvector of $2$ has to be orthogonal to both of the eigenvectors of $1$ that you’ve already found. You’re working in $mathbb R^3$, so such a vector can be found via a cross product: $(1,1,0)^Ttimes(0,0,1)^T=(1,-1,0)^T$ (which could also have been found by inspection).
answered Jan 17 at 18:16
amdamd
30.6k21050
answered Jan 17 at 18:16
amdamd
30.6k21050
answered Jan 17 at 18:16
amdamd
30.6k21050
answered Jan 17 at 18:16
amdamd
30.6k21050
30.6k21050
add a comment |
add a comment |
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If it's just the fractions that are making it hard for you, then just diagonalize $2H$. Now, all entries are integer.
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– José Carlos Santos
Jan 17 at 15:23
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Whenever I try doing that I get different eigenvalues though. The common problem I have with these sort of tasks are to get the eigenvalues. The rest I got down, but I just keep screwing up on the eigenvalues. Do you possibly think you could show how to get the correct eigenvalues?
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– wznd
Jan 17 at 15:31
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Why? Didn't you write that you know how to solve tasks like these and that you were just having problems with the fractions?
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– José Carlos Santos
Jan 17 at 15:33
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Normally I get through these tasks fine, except for the eigenvalues which is a problem for me. I should probably have mentioned it.
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– wznd
Jan 17 at 15:38