How can I find the eigenvalues of this 3x3 matrix [closed]Understanding direct sum of matricesEigenvalues and Eigenvectors of Large MatrixHow can I find a matrix $bf B$, with positive eigenvalues, such that its square $bf B^2$ is another matrix $bf A$?Sum of eigenvalues of a symmetric matrixAre the eigenvalues always the diagonal entries of a triangular matrix?Matrix polynomials/eigenvaluesFinding a symmetric 3x3 matrix from 2 eigenvectors and 2 eigenvaluesthe largest real part of the eigenvalues of a matrix$2times 2$ matrix similar to a matrix of its eigenvalues' real and imaginary partsFinding Eigenvalues with Variable Matrix
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How can I find the eigenvalues of this 3x3 matrix [closed]
Understanding direct sum of matricesEigenvalues and Eigenvectors of Large MatrixHow can I find a matrix $bf B$, with positive eigenvalues, such that its square $bf B^2$ is another matrix $bf A$?Sum of eigenvalues of a symmetric matrixAre the eigenvalues always the diagonal entries of a triangular matrix?Matrix polynomials/eigenvaluesFinding a symmetric 3x3 matrix from 2 eigenvectors and 2 eigenvaluesthe largest real part of the eigenvalues of a matrix$2times 2$ matrix similar to a matrix of its eigenvalues' real and imaginary partsFinding Eigenvalues with Variable Matrix
$begingroup$
I have a test in Algebra in a few days and a problem with eigenvalues will be there. I tried many ways to solve it and none helped me. How could you find the eigenvalues of the following matrix?
$$A=beginpmatrix8 &-6 &2\-6 &7 &-4\2 &-4 &3endpmatrix$$
I've tried add/substract rows and columns and the way this pdf shows https://www.scss.tcd.ie/~dahyotr/CS1BA1/SolutionEigen.pdf
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 28 at 14:10
ZoeZoe
82
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by 5xum, Yanko, Morgan Rodgers, max_zorn, Strants Mar 28 at 20:26
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." – 5xum, Yanko, Morgan Rodgers, max_zorn, Strants
add a comment |
$begingroup$
I have a test in Algebra in a few days and a problem with eigenvalues will be there. I tried many ways to solve it and none helped me. How could you find the eigenvalues of the following matrix?
$$A=beginpmatrix8 &-6 &2\-6 &7 &-4\2 &-4 &3endpmatrix$$
I've tried add/substract rows and columns and the way this pdf shows https://www.scss.tcd.ie/~dahyotr/CS1BA1/SolutionEigen.pdf
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 28 at 14:10
ZoeZoe
82
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by 5xum, Yanko, Morgan Rodgers, max_zorn, Strants Mar 28 at 20:26
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." – 5xum, Yanko, Morgan Rodgers, max_zorn, Strants
1
$begingroup$
The way to calculate the eigenvalues in such cases is pretty straightforward. What have you tried?
$endgroup$
– Rebellos
Mar 28 at 14:11
$begingroup$
Do you know about characteristic polynomials?
$endgroup$
– blub
Mar 28 at 14:12
3
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Mar 28 at 14:12
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Mar 28 at 14:12
$begingroup$
One shortcut here: since this matrix is symmetric, there is an orthogonal basis of eigenvectors, so once you’ve found eigenvectors for two distinct eigenvalues, you can compute a third with a cross product.
$endgroup$
– amd
Mar 28 at 19:37
add a comment |
$begingroup$
I have a test in Algebra in a few days and a problem with eigenvalues will be there. I tried many ways to solve it and none helped me. How could you find the eigenvalues of the following matrix?
$$A=beginpmatrix8 &-6 &2\-6 &7 &-4\2 &-4 &3endpmatrix$$
I've tried add/substract rows and columns and the way this pdf shows https://www.scss.tcd.ie/~dahyotr/CS1BA1/SolutionEigen.pdf
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 28 at 14:10
ZoeZoe
82
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a test in Algebra in a few days and a problem with eigenvalues will be there. I tried many ways to solve it and none helped me. How could you find the eigenvalues of the following matrix?
$$A=beginpmatrix8 &-6 &2\-6 &7 &-4\2 &-4 &3endpmatrix$$
I've tried add/substract rows and columns and the way this pdf shows https://www.scss.tcd.ie/~dahyotr/CS1BA1/SolutionEigen.pdf
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 28 at 14:10
ZoeZoe
82
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:10
ZoeZoe
82
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 14:14
Zoe
asked Mar 28 at 14:10
ZoeZoe
82
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:10
ZoeZoe
82
asked Mar 28 at 14:10
ZoeZoe
82
82
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Zoe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by 5xum, Yanko, Morgan Rodgers, max_zorn, Strants Mar 28 at 20:26
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." – 5xum, Yanko, Morgan Rodgers, max_zorn, Strants
closed as off-topic by 5xum, Yanko, Morgan Rodgers, max_zorn, Strants Mar 28 at 20:26
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." – 5xum, Yanko, Morgan Rodgers, max_zorn, Strants
1
$begingroup$
The way to calculate the eigenvalues in such cases is pretty straightforward. What have you tried?
$endgroup$
– Rebellos
Mar 28 at 14:11
$begingroup$
Do you know about characteristic polynomials?
$endgroup$
– blub
Mar 28 at 14:12
3
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Mar 28 at 14:12
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Mar 28 at 14:12
$begingroup$
One shortcut here: since this matrix is symmetric, there is an orthogonal basis of eigenvectors, so once you’ve found eigenvectors for two distinct eigenvalues, you can compute a third with a cross product.
$endgroup$
– amd
Mar 28 at 19:37
add a comment |
1
$begingroup$
The way to calculate the eigenvalues in such cases is pretty straightforward. What have you tried?
$endgroup$
– Rebellos
Mar 28 at 14:11
$begingroup$
Do you know about characteristic polynomials?
$endgroup$
– blub
Mar 28 at 14:12
3
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Mar 28 at 14:12
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Mar 28 at 14:12
$begingroup$
One shortcut here: since this matrix is symmetric, there is an orthogonal basis of eigenvectors, so once you’ve found eigenvectors for two distinct eigenvalues, you can compute a third with a cross product.
$endgroup$
– amd
Mar 28 at 19:37
1
1
$begingroup$
The way to calculate the eigenvalues in such cases is pretty straightforward. What have you tried?
$endgroup$
– Rebellos
Mar 28 at 14:11
$begingroup$
The way to calculate the eigenvalues in such cases is pretty straightforward. What have you tried?
$endgroup$
– Rebellos
Mar 28 at 14:11
$begingroup$
Do you know about characteristic polynomials?
$endgroup$
– blub
Mar 28 at 14:12
$begingroup$
Do you know about characteristic polynomials?
$endgroup$
– blub
Mar 28 at 14:12
3
3
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Mar 28 at 14:12
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Mar 28 at 14:12
1
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Mar 28 at 14:12
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Mar 28 at 14:12
$begingroup$
One shortcut here: since this matrix is symmetric, there is an orthogonal basis of eigenvectors, so once you’ve found eigenvectors for two distinct eigenvalues, you can compute a third with a cross product.
$endgroup$
– amd
Mar 28 at 19:37
$begingroup$
One shortcut here: since this matrix is symmetric, there is an orthogonal basis of eigenvectors, so once you’ve found eigenvectors for two distinct eigenvalues, you can compute a third with a cross product.
$endgroup$
– amd
Mar 28 at 19:37
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I suppose you have some (course) notes on the topic? You find the eigenvalues by solving:
$$detleft(A-lambda I_nright)=0$$
where $A$ is the $ntimes n$-matrix given and $I_n$ is the identity matrix. In your case, this means solving:
$$beginvmatrix
8-lambda & -6 & 2 \ -6& 7-lambda & -4 \ 2 & -4 & 3-lambda
endvmatrix=0$$
I've tried add/substract rows and columns and the way this pdf shows
Do you know how to calculate a ($3 times 3$) determinant?
You can use properties of determinants to try and simplify the calculations, but you can also go for the straightforward calculation by expanding along a column or row of your choice: that will lead to a 3rd degree polynomial equation in $lambda$.
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
$endgroup$
$begingroup$
A down-vote seems a bit harsh?
$endgroup$
– StackTD
Mar 28 at 14:20
1
$begingroup$
I concur: +1 from me.
$endgroup$
– Moritz
Mar 28 at 14:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suppose you have some (course) notes on the topic? You find the eigenvalues by solving:
$$detleft(A-lambda I_nright)=0$$
where $A$ is the $ntimes n$-matrix given and $I_n$ is the identity matrix. In your case, this means solving:
$$beginvmatrix
8-lambda & -6 & 2 \ -6& 7-lambda & -4 \ 2 & -4 & 3-lambda
endvmatrix=0$$
I've tried add/substract rows and columns and the way this pdf shows
Do you know how to calculate a ($3 times 3$) determinant?
You can use properties of determinants to try and simplify the calculations, but you can also go for the straightforward calculation by expanding along a column or row of your choice: that will lead to a 3rd degree polynomial equation in $lambda$.
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
$endgroup$
$begingroup$
A down-vote seems a bit harsh?
$endgroup$
– StackTD
Mar 28 at 14:20
1
$begingroup$
I concur: +1 from me.
$endgroup$
– Moritz
Mar 28 at 14:30
add a comment |
$begingroup$
I suppose you have some (course) notes on the topic? You find the eigenvalues by solving:
$$detleft(A-lambda I_nright)=0$$
where $A$ is the $ntimes n$-matrix given and $I_n$ is the identity matrix. In your case, this means solving:
$$beginvmatrix
8-lambda & -6 & 2 \ -6& 7-lambda & -4 \ 2 & -4 & 3-lambda
endvmatrix=0$$
I've tried add/substract rows and columns and the way this pdf shows
Do you know how to calculate a ($3 times 3$) determinant?
You can use properties of determinants to try and simplify the calculations, but you can also go for the straightforward calculation by expanding along a column or row of your choice: that will lead to a 3rd degree polynomial equation in $lambda$.
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
$endgroup$
$begingroup$
A down-vote seems a bit harsh?
$endgroup$
– StackTD
Mar 28 at 14:20
1
$begingroup$
I concur: +1 from me.
$endgroup$
– Moritz
Mar 28 at 14:30
add a comment |
$begingroup$
I suppose you have some (course) notes on the topic? You find the eigenvalues by solving:
$$detleft(A-lambda I_nright)=0$$
where $A$ is the $ntimes n$-matrix given and $I_n$ is the identity matrix. In your case, this means solving:
$$beginvmatrix
8-lambda & -6 & 2 \ -6& 7-lambda & -4 \ 2 & -4 & 3-lambda
endvmatrix=0$$
I've tried add/substract rows and columns and the way this pdf shows
Do you know how to calculate a ($3 times 3$) determinant?
You can use properties of determinants to try and simplify the calculations, but you can also go for the straightforward calculation by expanding along a column or row of your choice: that will lead to a 3rd degree polynomial equation in $lambda$.
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
$endgroup$
I suppose you have some (course) notes on the topic? You find the eigenvalues by solving:
$$detleft(A-lambda I_nright)=0$$
where $A$ is the $ntimes n$-matrix given and $I_n$ is the identity matrix. In your case, this means solving:
$$beginvmatrix
8-lambda & -6 & 2 \ -6& 7-lambda & -4 \ 2 & -4 & 3-lambda
endvmatrix=0$$
I've tried add/substract rows and columns and the way this pdf shows
Do you know how to calculate a ($3 times 3$) determinant?
You can use properties of determinants to try and simplify the calculations, but you can also go for the straightforward calculation by expanding along a column or row of your choice: that will lead to a 3rd degree polynomial equation in $lambda$.
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
edited Mar 28 at 14:19
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
answered Mar 28 at 14:13
StackTDStackTD
24.3k2254
24.3k2254
$begingroup$
A down-vote seems a bit harsh?
$endgroup$
– StackTD
Mar 28 at 14:20
1
$begingroup$
I concur: +1 from me.
$endgroup$
– Moritz
Mar 28 at 14:30
add a comment |
$begingroup$
A down-vote seems a bit harsh?
$endgroup$
– StackTD
Mar 28 at 14:20
1
$begingroup$
I concur: +1 from me.
$endgroup$
– Moritz
Mar 28 at 14:30
$begingroup$
A down-vote seems a bit harsh?
$endgroup$
– StackTD
Mar 28 at 14:20
$begingroup$
A down-vote seems a bit harsh?
$endgroup$
– StackTD
Mar 28 at 14:20
1
1
$begingroup$
I concur: +1 from me.
$endgroup$
– Moritz
Mar 28 at 14:30
$begingroup$
I concur: +1 from me.
$endgroup$
– Moritz
Mar 28 at 14:30
add a comment |
1
$begingroup$
The way to calculate the eigenvalues in such cases is pretty straightforward. What have you tried?
$endgroup$
– Rebellos
Mar 28 at 14:11
$begingroup$
Do you know about characteristic polynomials?
$endgroup$
– blub
Mar 28 at 14:12
3
$begingroup$
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
$endgroup$
– 5xum
Mar 28 at 14:12
1
$begingroup$
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
$endgroup$
– 5xum
Mar 28 at 14:12
$begingroup$
One shortcut here: since this matrix is symmetric, there is an orthogonal basis of eigenvectors, so once you’ve found eigenvectors for two distinct eigenvalues, you can compute a third with a cross product.
$endgroup$
– amd
Mar 28 at 19:37