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Eigen values of a real orthogonal matrix.
The Next CEO of Stack OverflowMahalanobis Distance using Eigen-Values of the Covariance MatrixWhat is the difference between a $1times 1$ matrix and a scalar?Pick the true options for a $ntimes n$ matrix $A$Identify $E=ain Bbb R:lim a^n A^ntextexists and is different from zero$How to find generalized Eigen vectors of a matrix with Eigen vectors already on diagonal?$T$ has no eigen-valueseigen values inside unit circleHow to determine the eigen values of $A^tA$ where $x^tA A^tx=alpha x^tx ~~~~ forall x in BbbR^m$ holds for some $alpha$Find the eigen values and the associated eigen vectors of the matrix,$~~~~$ $A=beginpmatrix 3~~1~~1\ 2~~4~~2\ 1~~1~~3 endpmatrix$Show that $L_A$ acts on by orthogonal transformation and in particular rotation.
$begingroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigen value of $A$ corresponding to the eigen vector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigen vector $X neq 0.$ Therefore $X^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigen values of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigen values of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
asked 6 hours ago
math maniac.math maniac.
1417
$endgroup$
add a comment |
$begingroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigen value of $A$ corresponding to the eigen vector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigen vector $X neq 0.$ Therefore $X^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigen values of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigen values of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
asked 6 hours ago
math maniac.math maniac.
1417
$endgroup$
add a comment |
$begingroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigen value of $A$ corresponding to the eigen vector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigen vector $X neq 0.$ Therefore $X^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigen values of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigen values of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
asked 6 hours ago
math maniac.math maniac.
1417
$endgroup$
Let $A$ be a real orthogonal matrix. Then $A^text T A = I.$ Let $lambda in Bbb C$ be an eigen value of $A$ corresponding to the eigen vector $X in Bbb C^n.$ Then we have
$$beginalign* X^text T A^text T A X = X^text T X. \ implies (AX)^text T AX & = X^text T X. \ implies (lambda X)^text T lambda X & = X^text T X. \ implies lambda^2 X^text T X & = X^text T X. \ implies (lambda^2 - 1) X^text T X & = 0. endalign*$$
Since $X$ is an eigen vector $X neq 0.$ Therefore $X^2 = X^text T X neq 0.$ Hence we must have $lambda^2 - 1 = 0$ i.e. $lambda^2 = 1.$ So $lambda = pm 1.$
So according to my argument above it follows that eigen values of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigen values of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.
What's going wrong in my argument above. Please help me in this regard.
Thank you very much for your valuable time.
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
asked 6 hours ago
math maniac.math maniac.
1417
asked 6 hours ago
math maniac.math maniac.
1417
asked 6 hours ago
math maniac.math maniac.
1417
asked 6 hours ago
math maniac.math maniac.
1417
asked 6 hours ago
math maniac.math maniac.
1417
1417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
5 hours ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
5 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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votes
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
5 hours ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
5 hours ago
add a comment |
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
5 hours ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
5 hours ago
add a comment |
$begingroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
The mistake is your assumption that $X^TXne0$. Consider a simple example:
$$A=pmatrix0&1\-1&0.$$
It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
$$X=pmatrix1\i.$$
It satisfies $X^TX=0$.
However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
of transpose) will give you the correct conclusion that $|lambda|^2=1$.
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 6 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
5 hours ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
5 hours ago
add a comment |
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
5 hours ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
5 hours ago
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
5 hours ago
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
5 hours ago
$begingroup$
how can Euclidean norm of non zero vector be zero?
$endgroup$
– math maniac.
5 hours ago
1
1
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
5 hours ago
$begingroup$
@mathmaniac. How can $1^2+i^2$ equal zero?
$endgroup$
– Lord Shark the Unknown
5 hours ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
5 hours ago
$begingroup$
I think the Euclidean norm of $X in Bbb C^n$ is $sqrt X^text T overline X text or sqrt overline X^text T X,$ not $sqrt X^text T X.$ Am I right?
$endgroup$
– math maniac.
5 hours ago
1
1
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
5 hours ago
$begingroup$
Which is same as $sqrt X^text HX,$ as you have rightly pointed out.
$endgroup$
– math maniac.
5 hours ago
add a comment |
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