If A is invertible and orthogonally diagonalizable, is $A^{-1}$ orthogonally diagonalizable as well?
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I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?
What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?
linear-algebra matrices diagonalization
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
asked May 1 '16 at 23:17
JetJet
12
$endgroup$
add a comment |
$begingroup$
I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?
What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?
linear-algebra matrices diagonalization
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
asked May 1 '16 at 23:17
JetJet
12
$endgroup$
$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25
$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25
1
$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02
add a comment |
$begingroup$
I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?
What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?
linear-algebra matrices diagonalization
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
asked May 1 '16 at 23:17
JetJet
12
$endgroup$
I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?
What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
asked May 1 '16 at 23:17
JetJet
12
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
asked May 1 '16 at 23:17
JetJet
12
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
edited May 2 '16 at 4:16
Martin Sleziak
44.8k9118272
44.8k9118272
asked May 1 '16 at 23:17
JetJet
12
asked May 1 '16 at 23:17
JetJet
12
asked May 1 '16 at 23:17
JetJet
12
12
$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25
$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25
1
$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02
add a comment |
$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25
$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25
1
$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02
$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25
$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25
$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25
$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25
1
1
$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02
$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
$$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
$$begin{pmatrix}
2 & 0 \
0 & 3
end{pmatrix} ^{-1} =
begin{pmatrix}
frac{1}{2} & 0 \
0 & frac{1}{3}
end{pmatrix}$$
Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.
Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
$endgroup$
add a comment |
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$begingroup$
Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
$$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
$$begin{pmatrix}
2 & 0 \
0 & 3
end{pmatrix} ^{-1} =
begin{pmatrix}
frac{1}{2} & 0 \
0 & frac{1}{3}
end{pmatrix}$$
Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.
Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
$endgroup$
add a comment |
$begingroup$
Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
$$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
$$begin{pmatrix}
2 & 0 \
0 & 3
end{pmatrix} ^{-1} =
begin{pmatrix}
frac{1}{2} & 0 \
0 & frac{1}{3}
end{pmatrix}$$
Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.
Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
$endgroup$
add a comment |
$begingroup$
Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
$$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
$$begin{pmatrix}
2 & 0 \
0 & 3
end{pmatrix} ^{-1} =
begin{pmatrix}
frac{1}{2} & 0 \
0 & frac{1}{3}
end{pmatrix}$$
Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.
Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
$endgroup$
Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible,
$$A^{-1} = {left(Q^TDQright)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance
$$begin{pmatrix}
2 & 0 \
0 & 3
end{pmatrix} ^{-1} =
begin{pmatrix}
frac{1}{2} & 0 \
0 & frac{1}{3}
end{pmatrix}$$
Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.
Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
answered May 2 '16 at 5:28
Jimmy XiaoJimmy Xiao
141111
141111
add a comment |
add a comment |
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$begingroup$
Do you need a proof of the claim in the title?
$endgroup$
– user258700
May 1 '16 at 23:25
$begingroup$
Yes @AhmedHussein
$endgroup$
– Jet
May 1 '16 at 23:25
1
$begingroup$
If $A=MDM^{-1}$ where $D$ is diagonal and $M=M^T$, can you write down a formula for $A^{-1}$?
$endgroup$
– Greg Martin
May 2 '16 at 5:02