Mixing
Show that the map $A mapsto AA^{T}$ is continuous. [closed]
0
Let $M$ be the set of all $n times n$ matrices. Show that the map $f : M longrightarrow M$ given by $f(A) = AA^{T}, A in M$ is continuous.
How do I prove it? Please help me in this regard.
Thank you very much.
matrices continuity
asked Jan 1 at 6:43
math maniac.math maniac.
265
closed as off-topic by Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos Jan 1 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
Let $M$ be the set of all $n times n$ matrices. Show that the map $f : M longrightarrow M$ given by $f(A) = AA^{T}, A in M$ is continuous.
How do I prove it? Please help me in this regard.
Thank you very much.
matrices continuity
asked Jan 1 at 6:43
math maniac.math maniac.
265
closed as off-topic by Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos Jan 1 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Hint: write the product as a polynomial and use that fact that polynomials are everywhere continuous.
– Sean Roberson
Jan 1 at 6:46
add a comment |
0
0
0
Let $M$ be the set of all $n times n$ matrices. Show that the map $f : M longrightarrow M$ given by $f(A) = AA^{T}, A in M$ is continuous.
How do I prove it? Please help me in this regard.
Thank you very much.
matrices continuity
asked Jan 1 at 6:43
math maniac.math maniac.
265
Let $M$ be the set of all $n times n$ matrices. Show that the map $f : M longrightarrow M$ given by $f(A) = AA^{T}, A in M$ is continuous.
How do I prove it? Please help me in this regard.
Thank you very much.
matrices continuity
matrices continuity
asked Jan 1 at 6:43
math maniac.math maniac.
265
asked Jan 1 at 6:43
math maniac.math maniac.
265
asked Jan 1 at 6:43
math maniac.math maniac.
265
asked Jan 1 at 6:43
math maniac.math maniac.
265
asked Jan 1 at 6:43
math maniac.math maniac.
265
265
closed as off-topic by Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos Jan 1 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos Jan 1 at 15:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, mrtaurho, Saad, metamorphy, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Hint: write the product as a polynomial and use that fact that polynomials are everywhere continuous.
– Sean Roberson
Jan 1 at 6:46
add a comment |
1
Hint: write the product as a polynomial and use that fact that polynomials are everywhere continuous.
– Sean Roberson
Jan 1 at 6:46
1
1
Hint: write the product as a polynomial and use that fact that polynomials are everywhere continuous.
– Sean Roberson
Jan 1 at 6:46
Hint: write the product as a polynomial and use that fact that polynomials are everywhere continuous.
– Sean Roberson
Jan 1 at 6:46
add a comment |
1 Answer
1
active
oldest
votes
4
Hint: Observe
begin{align}
| AA^T-BB^T| = |AA^T-AB^T+AB^T-BB^T| leq (|A|+|B|)|A-B|
end{align}
then the remaining of the proof follows the same idea as showing $x^2$ is continuous from the definition.
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
You use Euclidean norm. Right?
– math maniac.
Jan 1 at 6:59
You are saying that $|f(X)-f(A)| le (|X|+|A|)|X-A|.$ Now suppose we want to make $|f(X)-f(A)| < epsilon$ then which $delta >0$ should I take such that whenever $|X-A| < delta,$ $|f(X) - f(A) | < epsilon$ ?
– math maniac.
Jan 1 at 7:06
Whenever $|X-A| < delta$ then $|X| < |A| + delta$ by reverse triangle inequality. But then $|f(X)-f(A)| < {delta}^2 + 2|A| delta.$ We may take $0 < delta < 1$. But then ${delta}^2 < delta.$ Then $|f(X)-f(A)| < (2 |A| + 1) delta < epsilon$ if $delta le frac {epsilon} {2|A| + 1}.$
– math maniac.
Jan 1 at 7:16
So we should take $delta = min left {1,frac {epsilon} {2|A| +1} right }$. That will suit our purpose. Am I right?
– math maniac.
Jan 1 at 7:20
We can take any norm. Not necessarily Euclidean norm. That's because any two norms on $Bbb R^4$ are equivalent.
– Dbchatto67
Jan 1 at 7:45
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
Hint: Observe
begin{align}
| AA^T-BB^T| = |AA^T-AB^T+AB^T-BB^T| leq (|A|+|B|)|A-B|
end{align}
then the remaining of the proof follows the same idea as showing $x^2$ is continuous from the definition.
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
You use Euclidean norm. Right?
– math maniac.
Jan 1 at 6:59
You are saying that $|f(X)-f(A)| le (|X|+|A|)|X-A|.$ Now suppose we want to make $|f(X)-f(A)| < epsilon$ then which $delta >0$ should I take such that whenever $|X-A| < delta,$ $|f(X) - f(A) | < epsilon$ ?
– math maniac.
Jan 1 at 7:06
Whenever $|X-A| < delta$ then $|X| < |A| + delta$ by reverse triangle inequality. But then $|f(X)-f(A)| < {delta}^2 + 2|A| delta.$ We may take $0 < delta < 1$. But then ${delta}^2 < delta.$ Then $|f(X)-f(A)| < (2 |A| + 1) delta < epsilon$ if $delta le frac {epsilon} {2|A| + 1}.$
– math maniac.
Jan 1 at 7:16
So we should take $delta = min left {1,frac {epsilon} {2|A| +1} right }$. That will suit our purpose. Am I right?
– math maniac.
Jan 1 at 7:20
We can take any norm. Not necessarily Euclidean norm. That's because any two norms on $Bbb R^4$ are equivalent.
– Dbchatto67
Jan 1 at 7:45
add a comment |
4
Hint: Observe
begin{align}
| AA^T-BB^T| = |AA^T-AB^T+AB^T-BB^T| leq (|A|+|B|)|A-B|
end{align}
then the remaining of the proof follows the same idea as showing $x^2$ is continuous from the definition.
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
You use Euclidean norm. Right?
– math maniac.
Jan 1 at 6:59
You are saying that $|f(X)-f(A)| le (|X|+|A|)|X-A|.$ Now suppose we want to make $|f(X)-f(A)| < epsilon$ then which $delta >0$ should I take such that whenever $|X-A| < delta,$ $|f(X) - f(A) | < epsilon$ ?
– math maniac.
Jan 1 at 7:06
Whenever $|X-A| < delta$ then $|X| < |A| + delta$ by reverse triangle inequality. But then $|f(X)-f(A)| < {delta}^2 + 2|A| delta.$ We may take $0 < delta < 1$. But then ${delta}^2 < delta.$ Then $|f(X)-f(A)| < (2 |A| + 1) delta < epsilon$ if $delta le frac {epsilon} {2|A| + 1}.$
– math maniac.
Jan 1 at 7:16
So we should take $delta = min left {1,frac {epsilon} {2|A| +1} right }$. That will suit our purpose. Am I right?
– math maniac.
Jan 1 at 7:20
We can take any norm. Not necessarily Euclidean norm. That's because any two norms on $Bbb R^4$ are equivalent.
– Dbchatto67
Jan 1 at 7:45
add a comment |
4
4
4
Hint: Observe
begin{align}
| AA^T-BB^T| = |AA^T-AB^T+AB^T-BB^T| leq (|A|+|B|)|A-B|
end{align}
then the remaining of the proof follows the same idea as showing $x^2$ is continuous from the definition.
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
Hint: Observe
begin{align}
| AA^T-BB^T| = |AA^T-AB^T+AB^T-BB^T| leq (|A|+|B|)|A-B|
end{align}
then the remaining of the proof follows the same idea as showing $x^2$ is continuous from the definition.
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
edited Jan 1 at 8:44
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
answered Jan 1 at 6:47
Jacky ChongJacky Chong
17.8k21128
17.8k21128
You use Euclidean norm. Right?
– math maniac.
Jan 1 at 6:59
You are saying that $|f(X)-f(A)| le (|X|+|A|)|X-A|.$ Now suppose we want to make $|f(X)-f(A)| < epsilon$ then which $delta >0$ should I take such that whenever $|X-A| < delta,$ $|f(X) - f(A) | < epsilon$ ?
– math maniac.
Jan 1 at 7:06
Whenever $|X-A| < delta$ then $|X| < |A| + delta$ by reverse triangle inequality. But then $|f(X)-f(A)| < {delta}^2 + 2|A| delta.$ We may take $0 < delta < 1$. But then ${delta}^2 < delta.$ Then $|f(X)-f(A)| < (2 |A| + 1) delta < epsilon$ if $delta le frac {epsilon} {2|A| + 1}.$
– math maniac.
Jan 1 at 7:16
So we should take $delta = min left {1,frac {epsilon} {2|A| +1} right }$. That will suit our purpose. Am I right?
– math maniac.
Jan 1 at 7:20
We can take any norm. Not necessarily Euclidean norm. That's because any two norms on $Bbb R^4$ are equivalent.
– Dbchatto67
Jan 1 at 7:45
add a comment |
You use Euclidean norm. Right?
– math maniac.
Jan 1 at 6:59
You are saying that $|f(X)-f(A)| le (|X|+|A|)|X-A|.$ Now suppose we want to make $|f(X)-f(A)| < epsilon$ then which $delta >0$ should I take such that whenever $|X-A| < delta,$ $|f(X) - f(A) | < epsilon$ ?
– math maniac.
Jan 1 at 7:06
Whenever $|X-A| < delta$ then $|X| < |A| + delta$ by reverse triangle inequality. But then $|f(X)-f(A)| < {delta}^2 + 2|A| delta.$ We may take $0 < delta < 1$. But then ${delta}^2 < delta.$ Then $|f(X)-f(A)| < (2 |A| + 1) delta < epsilon$ if $delta le frac {epsilon} {2|A| + 1}.$
– math maniac.
Jan 1 at 7:16
So we should take $delta = min left {1,frac {epsilon} {2|A| +1} right }$. That will suit our purpose. Am I right?
– math maniac.
Jan 1 at 7:20
We can take any norm. Not necessarily Euclidean norm. That's because any two norms on $Bbb R^4$ are equivalent.
– Dbchatto67
Jan 1 at 7:45
You use Euclidean norm. Right?
– math maniac.
Jan 1 at 6:59
You use Euclidean norm. Right?
– math maniac.
Jan 1 at 6:59
You are saying that $|f(X)-f(A)| le (|X|+|A|)|X-A|.$ Now suppose we want to make $|f(X)-f(A)| < epsilon$ then which $delta >0$ should I take such that whenever $|X-A| < delta,$ $|f(X) - f(A) | < epsilon$ ?
– math maniac.
Jan 1 at 7:06
You are saying that $|f(X)-f(A)| le (|X|+|A|)|X-A|.$ Now suppose we want to make $|f(X)-f(A)| < epsilon$ then which $delta >0$ should I take such that whenever $|X-A| < delta,$ $|f(X) - f(A) | < epsilon$ ?
– math maniac.
Jan 1 at 7:06
Whenever $|X-A| < delta$ then $|X| < |A| + delta$ by reverse triangle inequality. But then $|f(X)-f(A)| < {delta}^2 + 2|A| delta.$ We may take $0 < delta < 1$. But then ${delta}^2 < delta.$ Then $|f(X)-f(A)| < (2 |A| + 1) delta < epsilon$ if $delta le frac {epsilon} {2|A| + 1}.$
– math maniac.
Jan 1 at 7:16
Whenever $|X-A| < delta$ then $|X| < |A| + delta$ by reverse triangle inequality. But then $|f(X)-f(A)| < {delta}^2 + 2|A| delta.$ We may take $0 < delta < 1$. But then ${delta}^2 < delta.$ Then $|f(X)-f(A)| < (2 |A| + 1) delta < epsilon$ if $delta le frac {epsilon} {2|A| + 1}.$
– math maniac.
Jan 1 at 7:16
So we should take $delta = min left {1,frac {epsilon} {2|A| +1} right }$. That will suit our purpose. Am I right?
– math maniac.
Jan 1 at 7:20
So we should take $delta = min left {1,frac {epsilon} {2|A| +1} right }$. That will suit our purpose. Am I right?
– math maniac.
Jan 1 at 7:20
We can take any norm. Not necessarily Euclidean norm. That's because any two norms on $Bbb R^4$ are equivalent.
– Dbchatto67
Jan 1 at 7:45
We can take any norm. Not necessarily Euclidean norm. That's because any two norms on $Bbb R^4$ are equivalent.
– Dbchatto67
Jan 1 at 7:45
add a comment |
1
Hint: write the product as a polynomial and use that fact that polynomials are everywhere continuous.
– Sean Roberson
Jan 1 at 6:46