Mixing
Show that $dim E_lambda (A^TA) = dim E_lambda (AA^T)$.
$begingroup$
Let $lambda in mathbb R^*$, $n ge 2$.
Let $A in M_n(mathbb R)$, $f$ the linear map defined by $A$.
Let $E_lambda (A^TA) = ker (A^TA - I_n)$ and $E_lambda (A^TA) = ker (AA^T - I_n)$.
Show that $dim E_lambda (A^TA) = dim E_lambda (AA^T)$.
I started by showing that $dim E_lambda (A^TA) le dim E_lambda (AA^T)$:
Let $tilde{f}$ be the restriction of $f$ on $E_lambda (A^TA)$.
We can prove that $tilde{f}$ is injective:
Let $x, x'$ such that $tilde{f}(x) = tilde{f}(x')$, then
$$Ax = Ax'$$
thus $$A^TAx = A^TAx'$$
$i.e$ $$lambda x = lambda x'$$
as $lambda ne 0$, $x=x'$ wich shows that $tilde{f}$ is injective.
We can also prove that $operatorname{Im} tilde{f} subset E_lambda (AA^T)$:
Let $y in operatorname{Im} tilde{f}$, there exists $x in E_lambda (A^TA)$ such that $y = Ax$, hence $$AA^T y = AA^T Ax = lambda Ax = lambda y.$$
Now, using Rank-Nullity theorem, we can write $$dim E_lambda (A^TA) = dim operatorname{Im} tilde{f} le dim E_lambda (AA^T).$$
To finish, I would have to prove that $$dim E_lambda (A^TA) ge dim E_lambda (AA^T).$$
I can't figure out how to do it, any help would be greatly appreciated.
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
$endgroup$
add a comment |
$begingroup$
Let $lambda in mathbb R^*$, $n ge 2$.
Let $A in M_n(mathbb R)$, $f$ the linear map defined by $A$.
Let $E_lambda (A^TA) = ker (A^TA - I_n)$ and $E_lambda (A^TA) = ker (AA^T - I_n)$.
Show that $dim E_lambda (A^TA) = dim E_lambda (AA^T)$.
I started by showing that $dim E_lambda (A^TA) le dim E_lambda (AA^T)$:
Let $tilde{f}$ be the restriction of $f$ on $E_lambda (A^TA)$.
We can prove that $tilde{f}$ is injective:
Let $x, x'$ such that $tilde{f}(x) = tilde{f}(x')$, then
$$Ax = Ax'$$
thus $$A^TAx = A^TAx'$$
$i.e$ $$lambda x = lambda x'$$
as $lambda ne 0$, $x=x'$ wich shows that $tilde{f}$ is injective.
We can also prove that $operatorname{Im} tilde{f} subset E_lambda (AA^T)$:
Let $y in operatorname{Im} tilde{f}$, there exists $x in E_lambda (A^TA)$ such that $y = Ax$, hence $$AA^T y = AA^T Ax = lambda Ax = lambda y.$$
Now, using Rank-Nullity theorem, we can write $$dim E_lambda (A^TA) = dim operatorname{Im} tilde{f} le dim E_lambda (AA^T).$$
To finish, I would have to prove that $$dim E_lambda (A^TA) ge dim E_lambda (AA^T).$$
I can't figure out how to do it, any help would be greatly appreciated.
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
$endgroup$
1
$begingroup$
If you proved $dim E_lambda (A^TA) le dim E_lambda (AA^T)$ you can simply swap $A$ and $A^T$ to get $dim E_lambda (AA^T) le dim E_lambda (A^TA)$...
$endgroup$
– Gabriel Romon
Jan 27 at 10:32
$begingroup$
Beware, $E_{lambda}(A^TA)=ker(A^TA-lambda I)$.
$endgroup$
– loup blanc
Jan 27 at 16:06
$begingroup$
I see that you have not corrected your statement (line 3)... On the other hand, if you have to assume that $lambdanot= 0$, then it's certainly because $A$ is not assumed to be square (otherwise the equality stands also for $lambda=0$). OK, I will not answer your questions any more.
$endgroup$
– loup blanc
Jan 31 at 0:03
add a comment |
$begingroup$
Let $lambda in mathbb R^*$, $n ge 2$.
Let $A in M_n(mathbb R)$, $f$ the linear map defined by $A$.
Let $E_lambda (A^TA) = ker (A^TA - I_n)$ and $E_lambda (A^TA) = ker (AA^T - I_n)$.
Show that $dim E_lambda (A^TA) = dim E_lambda (AA^T)$.
I started by showing that $dim E_lambda (A^TA) le dim E_lambda (AA^T)$:
Let $tilde{f}$ be the restriction of $f$ on $E_lambda (A^TA)$.
We can prove that $tilde{f}$ is injective:
Let $x, x'$ such that $tilde{f}(x) = tilde{f}(x')$, then
$$Ax = Ax'$$
thus $$A^TAx = A^TAx'$$
$i.e$ $$lambda x = lambda x'$$
as $lambda ne 0$, $x=x'$ wich shows that $tilde{f}$ is injective.
We can also prove that $operatorname{Im} tilde{f} subset E_lambda (AA^T)$:
Let $y in operatorname{Im} tilde{f}$, there exists $x in E_lambda (A^TA)$ such that $y = Ax$, hence $$AA^T y = AA^T Ax = lambda Ax = lambda y.$$
Now, using Rank-Nullity theorem, we can write $$dim E_lambda (A^TA) = dim operatorname{Im} tilde{f} le dim E_lambda (AA^T).$$
To finish, I would have to prove that $$dim E_lambda (A^TA) ge dim E_lambda (AA^T).$$
I can't figure out how to do it, any help would be greatly appreciated.
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
$endgroup$
Let $lambda in mathbb R^*$, $n ge 2$.
Let $A in M_n(mathbb R)$, $f$ the linear map defined by $A$.
Let $E_lambda (A^TA) = ker (A^TA - I_n)$ and $E_lambda (A^TA) = ker (AA^T - I_n)$.
Show that $dim E_lambda (A^TA) = dim E_lambda (AA^T)$.
I started by showing that $dim E_lambda (A^TA) le dim E_lambda (AA^T)$:
Let $tilde{f}$ be the restriction of $f$ on $E_lambda (A^TA)$.
We can prove that $tilde{f}$ is injective:
Let $x, x'$ such that $tilde{f}(x) = tilde{f}(x')$, then
$$Ax = Ax'$$
thus $$A^TAx = A^TAx'$$
$i.e$ $$lambda x = lambda x'$$
as $lambda ne 0$, $x=x'$ wich shows that $tilde{f}$ is injective.
We can also prove that $operatorname{Im} tilde{f} subset E_lambda (AA^T)$:
Let $y in operatorname{Im} tilde{f}$, there exists $x in E_lambda (A^TA)$ such that $y = Ax$, hence $$AA^T y = AA^T Ax = lambda Ax = lambda y.$$
Now, using Rank-Nullity theorem, we can write $$dim E_lambda (A^TA) = dim operatorname{Im} tilde{f} le dim E_lambda (AA^T).$$
To finish, I would have to prove that $$dim E_lambda (A^TA) ge dim E_lambda (AA^T).$$
I can't figure out how to do it, any help would be greatly appreciated.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
edited Jan 27 at 10:22
Euler Pythagoras
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
asked Jan 27 at 9:48
Euler PythagorasEuler Pythagoras
53512
53512
1
$begingroup$
If you proved $dim E_lambda (A^TA) le dim E_lambda (AA^T)$ you can simply swap $A$ and $A^T$ to get $dim E_lambda (AA^T) le dim E_lambda (A^TA)$...
$endgroup$
– Gabriel Romon
Jan 27 at 10:32
$begingroup$
Beware, $E_{lambda}(A^TA)=ker(A^TA-lambda I)$.
$endgroup$
– loup blanc
Jan 27 at 16:06
$begingroup$
I see that you have not corrected your statement (line 3)... On the other hand, if you have to assume that $lambdanot= 0$, then it's certainly because $A$ is not assumed to be square (otherwise the equality stands also for $lambda=0$). OK, I will not answer your questions any more.
$endgroup$
– loup blanc
Jan 31 at 0:03
add a comment |
1
$begingroup$
If you proved $dim E_lambda (A^TA) le dim E_lambda (AA^T)$ you can simply swap $A$ and $A^T$ to get $dim E_lambda (AA^T) le dim E_lambda (A^TA)$...
$endgroup$
– Gabriel Romon
Jan 27 at 10:32
$begingroup$
Beware, $E_{lambda}(A^TA)=ker(A^TA-lambda I)$.
$endgroup$
– loup blanc
Jan 27 at 16:06
$begingroup$
I see that you have not corrected your statement (line 3)... On the other hand, if you have to assume that $lambdanot= 0$, then it's certainly because $A$ is not assumed to be square (otherwise the equality stands also for $lambda=0$). OK, I will not answer your questions any more.
$endgroup$
– loup blanc
Jan 31 at 0:03
1
1
$begingroup$
If you proved $dim E_lambda (A^TA) le dim E_lambda (AA^T)$ you can simply swap $A$ and $A^T$ to get $dim E_lambda (AA^T) le dim E_lambda (A^TA)$...
$endgroup$
– Gabriel Romon
Jan 27 at 10:32
$begingroup$
If you proved $dim E_lambda (A^TA) le dim E_lambda (AA^T)$ you can simply swap $A$ and $A^T$ to get $dim E_lambda (AA^T) le dim E_lambda (A^TA)$...
$endgroup$
– Gabriel Romon
Jan 27 at 10:32
$begingroup$
Beware, $E_{lambda}(A^TA)=ker(A^TA-lambda I)$.
$endgroup$
– loup blanc
Jan 27 at 16:06
$begingroup$
Beware, $E_{lambda}(A^TA)=ker(A^TA-lambda I)$.
$endgroup$
– loup blanc
Jan 27 at 16:06
$begingroup$
I see that you have not corrected your statement (line 3)... On the other hand, if you have to assume that $lambdanot= 0$, then it's certainly because $A$ is not assumed to be square (otherwise the equality stands also for $lambda=0$). OK, I will not answer your questions any more.
$endgroup$
– loup blanc
Jan 31 at 0:03
$begingroup$
I see that you have not corrected your statement (line 3)... On the other hand, if you have to assume that $lambdanot= 0$, then it's certainly because $A$ is not assumed to be square (otherwise the equality stands also for $lambda=0$). OK, I will not answer your questions any more.
$endgroup$
– loup blanc
Jan 31 at 0:03
add a comment |
1 Answer
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$begingroup$
$AA^T$ and $A^TA$ are real symmetric, then diagonalizable over $mathbb{R}$. Moreover $spectrum(AA^T)=spectrum(A^TA)$ (equality of lists).
Thus $AA^T$ and $A^TA$ are similar over $mathbb{R}$.
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
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$begingroup$
$AA^T$ and $A^TA$ are real symmetric, then diagonalizable over $mathbb{R}$. Moreover $spectrum(AA^T)=spectrum(A^TA)$ (equality of lists).
Thus $AA^T$ and $A^TA$ are similar over $mathbb{R}$.
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
$AA^T$ and $A^TA$ are real symmetric, then diagonalizable over $mathbb{R}$. Moreover $spectrum(AA^T)=spectrum(A^TA)$ (equality of lists).
Thus $AA^T$ and $A^TA$ are similar over $mathbb{R}$.
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
$endgroup$
add a comment |
$begingroup$
$AA^T$ and $A^TA$ are real symmetric, then diagonalizable over $mathbb{R}$. Moreover $spectrum(AA^T)=spectrum(A^TA)$ (equality of lists).
Thus $AA^T$ and $A^TA$ are similar over $mathbb{R}$.
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
$endgroup$
$AA^T$ and $A^TA$ are real symmetric, then diagonalizable over $mathbb{R}$. Moreover $spectrum(AA^T)=spectrum(A^TA)$ (equality of lists).
Thus $AA^T$ and $A^TA$ are similar over $mathbb{R}$.
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
answered Jan 27 at 16:04
loup blancloup blanc
24.1k21851
24.1k21851
add a comment |
add a comment |
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$begingroup$
If you proved $dim E_lambda (A^TA) le dim E_lambda (AA^T)$ you can simply swap $A$ and $A^T$ to get $dim E_lambda (AA^T) le dim E_lambda (A^TA)$...
$endgroup$
– Gabriel Romon
Jan 27 at 10:32
$begingroup$
Beware, $E_{lambda}(A^TA)=ker(A^TA-lambda I)$.
$endgroup$
– loup blanc
Jan 27 at 16:06
$begingroup$
I see that you have not corrected your statement (line 3)... On the other hand, if you have to assume that $lambdanot= 0$, then it's certainly because $A$ is not assumed to be square (otherwise the equality stands also for $lambda=0$). OK, I will not answer your questions any more.
$endgroup$
– loup blanc
Jan 31 at 0:03