Mixing
Matrix norm for two matrices simultaneously close to spectral radius
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Suppose $A$ and $B$ have the same spectral radius $rho$. We can find a norm $|| cdot ||_A $ s.t. $||A||_A - epsilon < rho$. We can likewise find a another norm s.t. $||B||_B - epsilon < rho$. Under what conditions can we find a norm $||cdot||_*$ s.t. both $$||A||_* - epsilon < rho $$ and $$||B||_* - epsilon < rho. $$ Having done some digging it would seem one sufficient condition would be for $A$ and $B$ to be normal and another for the matrices to be simultaneously diagonalizable. Is there a necessary and sufficient condition?
linear-algebra matrix-norms
asked Jan 11 at 2:53
JaMcJaMc
112
$endgroup$
add a comment |
$begingroup$
Suppose $A$ and $B$ have the same spectral radius $rho$. We can find a norm $|| cdot ||_A $ s.t. $||A||_A - epsilon < rho$. We can likewise find a another norm s.t. $||B||_B - epsilon < rho$. Under what conditions can we find a norm $||cdot||_*$ s.t. both $$||A||_* - epsilon < rho $$ and $$||B||_* - epsilon < rho. $$ Having done some digging it would seem one sufficient condition would be for $A$ and $B$ to be normal and another for the matrices to be simultaneously diagonalizable. Is there a necessary and sufficient condition?
linear-algebra matrix-norms
asked Jan 11 at 2:53
JaMcJaMc
112
$endgroup$
$begingroup$
It is also sufficient for $A$ and $B$ to be simultaneously upper triangularizable, but if normality is enough then this isn't a necessary condition.
$endgroup$
– Omnomnomnom
Jan 11 at 6:08
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Do you know if there exists a pair $A,B$ where there is no such norm? Off the top of my head, I would think that $$ pmatrix{0&1\0&0}, quad pmatrix{0&0\1&0} $$ might be such a pair
$endgroup$
– Omnomnomnom
Jan 11 at 6:10
1
$begingroup$
Notably, any norm satisfies $|AB| leq |A| , |B|$, so we must have $$ |B| geq frac{|AB|}{|A|} geq frac{rho(AB)}{rho + epsilon} $$ it follows that $rho(AB) leq rho^2$ is a necessary condition, but I don't know if this condition is also sufficient.
$endgroup$
– Omnomnomnom
Jan 11 at 6:18
add a comment |
$begingroup$
Suppose $A$ and $B$ have the same spectral radius $rho$. We can find a norm $|| cdot ||_A $ s.t. $||A||_A - epsilon < rho$. We can likewise find a another norm s.t. $||B||_B - epsilon < rho$. Under what conditions can we find a norm $||cdot||_*$ s.t. both $$||A||_* - epsilon < rho $$ and $$||B||_* - epsilon < rho. $$ Having done some digging it would seem one sufficient condition would be for $A$ and $B$ to be normal and another for the matrices to be simultaneously diagonalizable. Is there a necessary and sufficient condition?
linear-algebra matrix-norms
asked Jan 11 at 2:53
JaMcJaMc
112
$endgroup$
Suppose $A$ and $B$ have the same spectral radius $rho$. We can find a norm $|| cdot ||_A $ s.t. $||A||_A - epsilon < rho$. We can likewise find a another norm s.t. $||B||_B - epsilon < rho$. Under what conditions can we find a norm $||cdot||_*$ s.t. both $$||A||_* - epsilon < rho $$ and $$||B||_* - epsilon < rho. $$ Having done some digging it would seem one sufficient condition would be for $A$ and $B$ to be normal and another for the matrices to be simultaneously diagonalizable. Is there a necessary and sufficient condition?
linear-algebra matrix-norms
linear-algebra matrix-norms
asked Jan 11 at 2:53
JaMcJaMc
112
asked Jan 11 at 2:53
JaMcJaMc
112
asked Jan 11 at 2:53
JaMcJaMc
112
asked Jan 11 at 2:53
JaMcJaMc
112
asked Jan 11 at 2:53
JaMcJaMc
112
112
$begingroup$
It is also sufficient for $A$ and $B$ to be simultaneously upper triangularizable, but if normality is enough then this isn't a necessary condition.
$endgroup$
– Omnomnomnom
Jan 11 at 6:08
$begingroup$
Do you know if there exists a pair $A,B$ where there is no such norm? Off the top of my head, I would think that $$ pmatrix{0&1\0&0}, quad pmatrix{0&0\1&0} $$ might be such a pair
$endgroup$
– Omnomnomnom
Jan 11 at 6:10
1
$begingroup$
Notably, any norm satisfies $|AB| leq |A| , |B|$, so we must have $$ |B| geq frac{|AB|}{|A|} geq frac{rho(AB)}{rho + epsilon} $$ it follows that $rho(AB) leq rho^2$ is a necessary condition, but I don't know if this condition is also sufficient.
$endgroup$
– Omnomnomnom
Jan 11 at 6:18
add a comment |
$begingroup$
It is also sufficient for $A$ and $B$ to be simultaneously upper triangularizable, but if normality is enough then this isn't a necessary condition.
$endgroup$
– Omnomnomnom
Jan 11 at 6:08
$begingroup$
Do you know if there exists a pair $A,B$ where there is no such norm? Off the top of my head, I would think that $$ pmatrix{0&1\0&0}, quad pmatrix{0&0\1&0} $$ might be such a pair
$endgroup$
– Omnomnomnom
Jan 11 at 6:10
1
$begingroup$
Notably, any norm satisfies $|AB| leq |A| , |B|$, so we must have $$ |B| geq frac{|AB|}{|A|} geq frac{rho(AB)}{rho + epsilon} $$ it follows that $rho(AB) leq rho^2$ is a necessary condition, but I don't know if this condition is also sufficient.
$endgroup$
– Omnomnomnom
Jan 11 at 6:18
$begingroup$
It is also sufficient for $A$ and $B$ to be simultaneously upper triangularizable, but if normality is enough then this isn't a necessary condition.
$endgroup$
– Omnomnomnom
Jan 11 at 6:08
$begingroup$
It is also sufficient for $A$ and $B$ to be simultaneously upper triangularizable, but if normality is enough then this isn't a necessary condition.
$endgroup$
– Omnomnomnom
Jan 11 at 6:08
$begingroup$
Do you know if there exists a pair $A,B$ where there is no such norm? Off the top of my head, I would think that $$ pmatrix{0&1\0&0}, quad pmatrix{0&0\1&0} $$ might be such a pair
$endgroup$
– Omnomnomnom
Jan 11 at 6:10
$begingroup$
Do you know if there exists a pair $A,B$ where there is no such norm? Off the top of my head, I would think that $$ pmatrix{0&1\0&0}, quad pmatrix{0&0\1&0} $$ might be such a pair
$endgroup$
– Omnomnomnom
Jan 11 at 6:10
1
1
$begingroup$
Notably, any norm satisfies $|AB| leq |A| , |B|$, so we must have $$ |B| geq frac{|AB|}{|A|} geq frac{rho(AB)}{rho + epsilon} $$ it follows that $rho(AB) leq rho^2$ is a necessary condition, but I don't know if this condition is also sufficient.
$endgroup$
– Omnomnomnom
Jan 11 at 6:18
$begingroup$
Notably, any norm satisfies $|AB| leq |A| , |B|$, so we must have $$ |B| geq frac{|AB|}{|A|} geq frac{rho(AB)}{rho + epsilon} $$ it follows that $rho(AB) leq rho^2$ is a necessary condition, but I don't know if this condition is also sufficient.
$endgroup$
– Omnomnomnom
Jan 11 at 6:18
add a comment |
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$begingroup$
It is also sufficient for $A$ and $B$ to be simultaneously upper triangularizable, but if normality is enough then this isn't a necessary condition.
$endgroup$
– Omnomnomnom
Jan 11 at 6:08
$begingroup$
Do you know if there exists a pair $A,B$ where there is no such norm? Off the top of my head, I would think that $$ pmatrix{0&1\0&0}, quad pmatrix{0&0\1&0} $$ might be such a pair
$endgroup$
– Omnomnomnom
Jan 11 at 6:10
1
$begingroup$
Notably, any norm satisfies $|AB| leq |A| , |B|$, so we must have $$ |B| geq frac{|AB|}{|A|} geq frac{rho(AB)}{rho + epsilon} $$ it follows that $rho(AB) leq rho^2$ is a necessary condition, but I don't know if this condition is also sufficient.
$endgroup$
– Omnomnomnom
Jan 11 at 6:18