eigenvalues and eigenvectors of Diagonalisable matrices
$begingroup$
Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.
Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.
As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.
And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.
Could someone tell me if I am on the right way?
linear-algebra eigenvalues-eigenvectors
asked Jan 6 at 10:23
KaiKai
686
$endgroup$
add a comment |
$begingroup$
Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.
Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.
As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.
And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.
Could someone tell me if I am on the right way?
linear-algebra eigenvalues-eigenvectors
asked Jan 6 at 10:23
KaiKai
686
$endgroup$
2
$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26
$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28
add a comment |
$begingroup$
Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.
Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.
As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.
And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.
Could someone tell me if I am on the right way?
linear-algebra eigenvalues-eigenvectors
asked Jan 6 at 10:23
KaiKai
686
$endgroup$
Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.
Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.
As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.
And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.
Could someone tell me if I am on the right way?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Jan 6 at 10:23
KaiKai
686
asked Jan 6 at 10:23
KaiKai
686
asked Jan 6 at 10:23
KaiKai
686
asked Jan 6 at 10:23
KaiKai
686
asked Jan 6 at 10:23
KaiKai
686
686
2
$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26
$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28
add a comment |
2
$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26
$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28
2
2
$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26
$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26
$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28
$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $u_i$ is an eigenvector of $D$:
$$
Du_i=lambda_iu_i
$$
than, from:
$$
A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$
we have:
$$
A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
$$
that gives the eigenvectors for the same eigenvalues.
Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
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add a comment |
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1 Answer
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votes
$begingroup$
If $u_i$ is an eigenvector of $D$:
$$
Du_i=lambda_iu_i
$$
than, from:
$$
A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$
we have:
$$
A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
$$
that gives the eigenvectors for the same eigenvalues.
Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
$endgroup$
add a comment |
$begingroup$
If $u_i$ is an eigenvector of $D$:
$$
Du_i=lambda_iu_i
$$
than, from:
$$
A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$
we have:
$$
A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
$$
that gives the eigenvectors for the same eigenvalues.
Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
$endgroup$
add a comment |
$begingroup$
If $u_i$ is an eigenvector of $D$:
$$
Du_i=lambda_iu_i
$$
than, from:
$$
A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$
we have:
$$
A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
$$
that gives the eigenvectors for the same eigenvalues.
Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
$endgroup$
If $u_i$ is an eigenvector of $D$:
$$
Du_i=lambda_iu_i
$$
than, from:
$$
A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$
we have:
$$
A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
$$
that gives the eigenvectors for the same eigenvalues.
Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
answered Jan 6 at 10:33
Emilio NovatiEmilio Novati
52.2k43574
52.2k43574
add a comment |
add a comment |
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$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26
$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28