Matrix diagonalisation, and orthogonalization
$begingroup$
Hey I have a a question on how to diagonalize matrices. My lecture focuses on symmetric matrices, but I have examples where I need to calculate $D$ for non- symmetric matrices as well.
I know that I can simply but in the Eigenvalues into $D$ if $D$ exists. But if I would like to calculate $D$ by hand with
$$D= U^{-1}AU$$
I found that i sometimes have to orthogonalize the vectors of U and normalize them and sometimes not.
I cant seem to find a pattern, and I cant find anything online. Maybe I am just completely wrong, this topic is new to me.... so please excuse if the question is stupid ....
Many thanks
linear-algebra
edited Dec 4 '18 at 13:25
Moo
5,57631020
asked Dec 4 '18 at 11:48
LillysLillys
778
$endgroup$
add a comment |
$begingroup$
Hey I have a a question on how to diagonalize matrices. My lecture focuses on symmetric matrices, but I have examples where I need to calculate $D$ for non- symmetric matrices as well.
I know that I can simply but in the Eigenvalues into $D$ if $D$ exists. But if I would like to calculate $D$ by hand with
$$D= U^{-1}AU$$
I found that i sometimes have to orthogonalize the vectors of U and normalize them and sometimes not.
I cant seem to find a pattern, and I cant find anything online. Maybe I am just completely wrong, this topic is new to me.... so please excuse if the question is stupid ....
Many thanks
linear-algebra
edited Dec 4 '18 at 13:25
Moo
5,57631020
asked Dec 4 '18 at 11:48
LillysLillys
778
$endgroup$
add a comment |
$begingroup$
Hey I have a a question on how to diagonalize matrices. My lecture focuses on symmetric matrices, but I have examples where I need to calculate $D$ for non- symmetric matrices as well.
I know that I can simply but in the Eigenvalues into $D$ if $D$ exists. But if I would like to calculate $D$ by hand with
$$D= U^{-1}AU$$
I found that i sometimes have to orthogonalize the vectors of U and normalize them and sometimes not.
I cant seem to find a pattern, and I cant find anything online. Maybe I am just completely wrong, this topic is new to me.... so please excuse if the question is stupid ....
Many thanks
linear-algebra
edited Dec 4 '18 at 13:25
Moo
5,57631020
asked Dec 4 '18 at 11:48
LillysLillys
778
$endgroup$
Hey I have a a question on how to diagonalize matrices. My lecture focuses on symmetric matrices, but I have examples where I need to calculate $D$ for non- symmetric matrices as well.
I know that I can simply but in the Eigenvalues into $D$ if $D$ exists. But if I would like to calculate $D$ by hand with
$$D= U^{-1}AU$$
I found that i sometimes have to orthogonalize the vectors of U and normalize them and sometimes not.
I cant seem to find a pattern, and I cant find anything online. Maybe I am just completely wrong, this topic is new to me.... so please excuse if the question is stupid ....
Many thanks
linear-algebra
linear-algebra
edited Dec 4 '18 at 13:25
Moo
5,57631020
asked Dec 4 '18 at 11:48
LillysLillys
778
edited Dec 4 '18 at 13:25
Moo
5,57631020
asked Dec 4 '18 at 11:48
LillysLillys
778
edited Dec 4 '18 at 13:25
Moo
5,57631020
edited Dec 4 '18 at 13:25
Moo
5,57631020
edited Dec 4 '18 at 13:25
Moo
5,57631020
5,57631020
asked Dec 4 '18 at 11:48
LillysLillys
778
asked Dec 4 '18 at 11:48
LillysLillys
778
asked Dec 4 '18 at 11:48
LillysLillys
778
778
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general we have that if we can find a basis of eigenvectors we can always diagonalize a matrix and
$D$ contains the eigenvalues along the diagonal
$U$ contains the corresponding eigenvectors by columns
such that
$$AU=UD iff U^{-1}AU=D $$
edited Dec 4 '18 at 13:23
Moo
5,57631020
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Thanks for your answer, but where i am uncleared about is when I have to orthogonalize U ....
$endgroup$
– Lillys
Dec 4 '18 at 11:57
$begingroup$
@Lillys Not always we can do that. We can of course when A is symmetric otherwise we cannot do that and in these case we need to take $U^{-1}$ to express $A$ in the form $A=UDU^{-1}$.
$endgroup$
– gimusi
Dec 4 '18 at 12:00
$begingroup$
I hope it is ok if i as so many questions. Ím not sure if I’m using the right words, as i cant find them in any dictonary and ím not studying in English, what i know that wie have to use u^-1, i corrected that above, what i meant with orthogonalize is to treat the vectors in u such that the dot product of each is 0 - meaning they are perpendicular.
$endgroup$
– Lillys
Dec 4 '18 at 12:06
$begingroup$
@Lillys You need to refer to Spectral Theorem: If A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.
$endgroup$
– gimusi
Dec 4 '18 at 12:08
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
In general we have that if we can find a basis of eigenvectors we can always diagonalize a matrix and
$D$ contains the eigenvalues along the diagonal
$U$ contains the corresponding eigenvectors by columns
such that
$$AU=UD iff U^{-1}AU=D $$
edited Dec 4 '18 at 13:23
Moo
5,57631020
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Thanks for your answer, but where i am uncleared about is when I have to orthogonalize U ....
$endgroup$
– Lillys
Dec 4 '18 at 11:57
$begingroup$
@Lillys Not always we can do that. We can of course when A is symmetric otherwise we cannot do that and in these case we need to take $U^{-1}$ to express $A$ in the form $A=UDU^{-1}$.
$endgroup$
– gimusi
Dec 4 '18 at 12:00
$begingroup$
I hope it is ok if i as so many questions. Ím not sure if I’m using the right words, as i cant find them in any dictonary and ím not studying in English, what i know that wie have to use u^-1, i corrected that above, what i meant with orthogonalize is to treat the vectors in u such that the dot product of each is 0 - meaning they are perpendicular.
$endgroup$
– Lillys
Dec 4 '18 at 12:06
$begingroup$
@Lillys You need to refer to Spectral Theorem: If A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.
$endgroup$
– gimusi
Dec 4 '18 at 12:08
add a comment |
$begingroup$
In general we have that if we can find a basis of eigenvectors we can always diagonalize a matrix and
$D$ contains the eigenvalues along the diagonal
$U$ contains the corresponding eigenvectors by columns
such that
$$AU=UD iff U^{-1}AU=D $$
edited Dec 4 '18 at 13:23
Moo
5,57631020
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Thanks for your answer, but where i am uncleared about is when I have to orthogonalize U ....
$endgroup$
– Lillys
Dec 4 '18 at 11:57
$begingroup$
@Lillys Not always we can do that. We can of course when A is symmetric otherwise we cannot do that and in these case we need to take $U^{-1}$ to express $A$ in the form $A=UDU^{-1}$.
$endgroup$
– gimusi
Dec 4 '18 at 12:00
$begingroup$
I hope it is ok if i as so many questions. Ím not sure if I’m using the right words, as i cant find them in any dictonary and ím not studying in English, what i know that wie have to use u^-1, i corrected that above, what i meant with orthogonalize is to treat the vectors in u such that the dot product of each is 0 - meaning they are perpendicular.
$endgroup$
– Lillys
Dec 4 '18 at 12:06
$begingroup$
@Lillys You need to refer to Spectral Theorem: If A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.
$endgroup$
– gimusi
Dec 4 '18 at 12:08
add a comment |
$begingroup$
In general we have that if we can find a basis of eigenvectors we can always diagonalize a matrix and
$D$ contains the eigenvalues along the diagonal
$U$ contains the corresponding eigenvectors by columns
such that
$$AU=UD iff U^{-1}AU=D $$
edited Dec 4 '18 at 13:23
Moo
5,57631020
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
$endgroup$
In general we have that if we can find a basis of eigenvectors we can always diagonalize a matrix and
$D$ contains the eigenvalues along the diagonal
$U$ contains the corresponding eigenvectors by columns
such that
$$AU=UD iff U^{-1}AU=D $$
edited Dec 4 '18 at 13:23
Moo
5,57631020
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
edited Dec 4 '18 at 13:23
Moo
5,57631020
edited Dec 4 '18 at 13:23
Moo
5,57631020
edited Dec 4 '18 at 13:23
Moo
5,57631020
5,57631020
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
answered Dec 4 '18 at 11:54
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
Thanks for your answer, but where i am uncleared about is when I have to orthogonalize U ....
$endgroup$
– Lillys
Dec 4 '18 at 11:57
$begingroup$
@Lillys Not always we can do that. We can of course when A is symmetric otherwise we cannot do that and in these case we need to take $U^{-1}$ to express $A$ in the form $A=UDU^{-1}$.
$endgroup$
– gimusi
Dec 4 '18 at 12:00
$begingroup$
I hope it is ok if i as so many questions. Ím not sure if I’m using the right words, as i cant find them in any dictonary and ím not studying in English, what i know that wie have to use u^-1, i corrected that above, what i meant with orthogonalize is to treat the vectors in u such that the dot product of each is 0 - meaning they are perpendicular.
$endgroup$
– Lillys
Dec 4 '18 at 12:06
$begingroup$
@Lillys You need to refer to Spectral Theorem: If A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.
$endgroup$
– gimusi
Dec 4 '18 at 12:08
add a comment |
$begingroup$
Thanks for your answer, but where i am uncleared about is when I have to orthogonalize U ....
$endgroup$
– Lillys
Dec 4 '18 at 11:57
$begingroup$
@Lillys Not always we can do that. We can of course when A is symmetric otherwise we cannot do that and in these case we need to take $U^{-1}$ to express $A$ in the form $A=UDU^{-1}$.
$endgroup$
– gimusi
Dec 4 '18 at 12:00
$begingroup$
I hope it is ok if i as so many questions. Ím not sure if I’m using the right words, as i cant find them in any dictonary and ím not studying in English, what i know that wie have to use u^-1, i corrected that above, what i meant with orthogonalize is to treat the vectors in u such that the dot product of each is 0 - meaning they are perpendicular.
$endgroup$
– Lillys
Dec 4 '18 at 12:06
$begingroup$
@Lillys You need to refer to Spectral Theorem: If A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.
$endgroup$
– gimusi
Dec 4 '18 at 12:08
$begingroup$
Thanks for your answer, but where i am uncleared about is when I have to orthogonalize U ....
$endgroup$
– Lillys
Dec 4 '18 at 11:57
$begingroup$
Thanks for your answer, but where i am uncleared about is when I have to orthogonalize U ....
$endgroup$
– Lillys
Dec 4 '18 at 11:57
$begingroup$
@Lillys Not always we can do that. We can of course when A is symmetric otherwise we cannot do that and in these case we need to take $U^{-1}$ to express $A$ in the form $A=UDU^{-1}$.
$endgroup$
– gimusi
Dec 4 '18 at 12:00
$begingroup$
@Lillys Not always we can do that. We can of course when A is symmetric otherwise we cannot do that and in these case we need to take $U^{-1}$ to express $A$ in the form $A=UDU^{-1}$.
$endgroup$
– gimusi
Dec 4 '18 at 12:00
$begingroup$
I hope it is ok if i as so many questions. Ím not sure if I’m using the right words, as i cant find them in any dictonary and ím not studying in English, what i know that wie have to use u^-1, i corrected that above, what i meant with orthogonalize is to treat the vectors in u such that the dot product of each is 0 - meaning they are perpendicular.
$endgroup$
– Lillys
Dec 4 '18 at 12:06
$begingroup$
I hope it is ok if i as so many questions. Ím not sure if I’m using the right words, as i cant find them in any dictonary and ím not studying in English, what i know that wie have to use u^-1, i corrected that above, what i meant with orthogonalize is to treat the vectors in u such that the dot product of each is 0 - meaning they are perpendicular.
$endgroup$
– Lillys
Dec 4 '18 at 12:06
$begingroup$
@Lillys You need to refer to Spectral Theorem: If A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.
$endgroup$
– gimusi
Dec 4 '18 at 12:08
$begingroup$
@Lillys You need to refer to Spectral Theorem: If A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.
$endgroup$
– gimusi
Dec 4 '18 at 12:08
add a comment |
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