Mixing
Vector norm with a full rank matrix
0
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Let $|| . || $ be a vector norm on $R^m$ and assume $A$ is m by n. Show that if rank(A) = n, then $|| x ||_A$ = $|| Ax ||$ is a vector norm on $R^n$.
I know the 1, 2 and p-norm of a vector but don't understand how to go about here?
linear-algebra
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
$endgroup$
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show 1 more comment
0
$begingroup$
Let $|| . || $ be a vector norm on $R^m$ and assume $A$ is m by n. Show that if rank(A) = n, then $|| x ||_A$ = $|| Ax ||$ is a vector norm on $R^n$.
I know the 1, 2 and p-norm of a vector but don't understand how to go about here?
linear-algebra
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
$endgroup$
$begingroup$
Didn't you forget to add some hypothesis concerning $m$ and $n$?
$endgroup$
– José Carlos Santos
Jan 30 at 13:09
$begingroup$
You need to check the axioms of the norm. Everything should be trivial except the axiom that the norm of a vector is zero iff it is the zero vector. Here is the part where you have to use the fact that the matrix has full rank.
$endgroup$
– Mark
Jan 30 at 13:11
$begingroup$
@JoséCarlosSantos, I have checked there isn't any thing like that expect we know A is m by n.
$endgroup$
– Usman Ashraf
Jan 30 at 13:14
$begingroup$
@Mark, I know the axioms of norm. But, how to use them with a martix and how can ||x||_A = ||Ax||?
$endgroup$
– Usman Ashraf
Jan 30 at 13:16
$begingroup$
$||x||_A=||Ax||$ is just the definition of the norm. You need to use the fact that $||.||$ is a norm. For example why does the triangle inequality hold? We have $||x+y||_A=||A(x+y)||=||Ax+Ay||leq||Ax||+||Ay||=||x||_A+||y||_A$. Now in a similar way you can check the other axioms.
$endgroup$
– Mark
Jan 30 at 13:56
|
show 1 more comment
0
0
0
$begingroup$
Let $|| . || $ be a vector norm on $R^m$ and assume $A$ is m by n. Show that if rank(A) = n, then $|| x ||_A$ = $|| Ax ||$ is a vector norm on $R^n$.
I know the 1, 2 and p-norm of a vector but don't understand how to go about here?
linear-algebra
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
$endgroup$
Let $|| . || $ be a vector norm on $R^m$ and assume $A$ is m by n. Show that if rank(A) = n, then $|| x ||_A$ = $|| Ax ||$ is a vector norm on $R^n$.
I know the 1, 2 and p-norm of a vector but don't understand how to go about here?
linear-algebra
linear-algebra
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
edited Jan 30 at 13:13
Usman Ashraf
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
asked Jan 30 at 13:08
Usman AshrafUsman Ashraf
13
13
$begingroup$
Didn't you forget to add some hypothesis concerning $m$ and $n$?
$endgroup$
– José Carlos Santos
Jan 30 at 13:09
$begingroup$
You need to check the axioms of the norm. Everything should be trivial except the axiom that the norm of a vector is zero iff it is the zero vector. Here is the part where you have to use the fact that the matrix has full rank.
$endgroup$
– Mark
Jan 30 at 13:11
$begingroup$
@JoséCarlosSantos, I have checked there isn't any thing like that expect we know A is m by n.
$endgroup$
– Usman Ashraf
Jan 30 at 13:14
$begingroup$
@Mark, I know the axioms of norm. But, how to use them with a martix and how can ||x||_A = ||Ax||?
$endgroup$
– Usman Ashraf
Jan 30 at 13:16
$begingroup$
$||x||_A=||Ax||$ is just the definition of the norm. You need to use the fact that $||.||$ is a norm. For example why does the triangle inequality hold? We have $||x+y||_A=||A(x+y)||=||Ax+Ay||leq||Ax||+||Ay||=||x||_A+||y||_A$. Now in a similar way you can check the other axioms.
$endgroup$
– Mark
Jan 30 at 13:56
|
show 1 more comment
$begingroup$
Didn't you forget to add some hypothesis concerning $m$ and $n$?
$endgroup$
– José Carlos Santos
Jan 30 at 13:09
$begingroup$
You need to check the axioms of the norm. Everything should be trivial except the axiom that the norm of a vector is zero iff it is the zero vector. Here is the part where you have to use the fact that the matrix has full rank.
$endgroup$
– Mark
Jan 30 at 13:11
$begingroup$
@JoséCarlosSantos, I have checked there isn't any thing like that expect we know A is m by n.
$endgroup$
– Usman Ashraf
Jan 30 at 13:14
$begingroup$
@Mark, I know the axioms of norm. But, how to use them with a martix and how can ||x||_A = ||Ax||?
$endgroup$
– Usman Ashraf
Jan 30 at 13:16
$begingroup$
$||x||_A=||Ax||$ is just the definition of the norm. You need to use the fact that $||.||$ is a norm. For example why does the triangle inequality hold? We have $||x+y||_A=||A(x+y)||=||Ax+Ay||leq||Ax||+||Ay||=||x||_A+||y||_A$. Now in a similar way you can check the other axioms.
$endgroup$
– Mark
Jan 30 at 13:56
$begingroup$
Didn't you forget to add some hypothesis concerning $m$ and $n$?
$endgroup$
– José Carlos Santos
Jan 30 at 13:09
$begingroup$
Didn't you forget to add some hypothesis concerning $m$ and $n$?
$endgroup$
– José Carlos Santos
Jan 30 at 13:09
$begingroup$
You need to check the axioms of the norm. Everything should be trivial except the axiom that the norm of a vector is zero iff it is the zero vector. Here is the part where you have to use the fact that the matrix has full rank.
$endgroup$
– Mark
Jan 30 at 13:11
$begingroup$
You need to check the axioms of the norm. Everything should be trivial except the axiom that the norm of a vector is zero iff it is the zero vector. Here is the part where you have to use the fact that the matrix has full rank.
$endgroup$
– Mark
Jan 30 at 13:11
$begingroup$
@JoséCarlosSantos, I have checked there isn't any thing like that expect we know A is m by n.
$endgroup$
– Usman Ashraf
Jan 30 at 13:14
$begingroup$
@JoséCarlosSantos, I have checked there isn't any thing like that expect we know A is m by n.
$endgroup$
– Usman Ashraf
Jan 30 at 13:14
$begingroup$
@Mark, I know the axioms of norm. But, how to use them with a martix and how can ||x||_A = ||Ax||?
$endgroup$
– Usman Ashraf
Jan 30 at 13:16
$begingroup$
@Mark, I know the axioms of norm. But, how to use them with a martix and how can ||x||_A = ||Ax||?
$endgroup$
– Usman Ashraf
Jan 30 at 13:16
$begingroup$
$||x||_A=||Ax||$ is just the definition of the norm. You need to use the fact that $||.||$ is a norm. For example why does the triangle inequality hold? We have $||x+y||_A=||A(x+y)||=||Ax+Ay||leq||Ax||+||Ay||=||x||_A+||y||_A$. Now in a similar way you can check the other axioms.
$endgroup$
– Mark
Jan 30 at 13:56
$begingroup$
$||x||_A=||Ax||$ is just the definition of the norm. You need to use the fact that $||.||$ is a norm. For example why does the triangle inequality hold? We have $||x+y||_A=||A(x+y)||=||Ax+Ay||leq||Ax||+||Ay||=||x||_A+||y||_A$. Now in a similar way you can check the other axioms.
$endgroup$
– Mark
Jan 30 at 13:56
|
show 1 more comment
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$begingroup$
Didn't you forget to add some hypothesis concerning $m$ and $n$?
$endgroup$
– José Carlos Santos
Jan 30 at 13:09
$begingroup$
You need to check the axioms of the norm. Everything should be trivial except the axiom that the norm of a vector is zero iff it is the zero vector. Here is the part where you have to use the fact that the matrix has full rank.
$endgroup$
– Mark
Jan 30 at 13:11
$begingroup$
@JoséCarlosSantos, I have checked there isn't any thing like that expect we know A is m by n.
$endgroup$
– Usman Ashraf
Jan 30 at 13:14
$begingroup$
@Mark, I know the axioms of norm. But, how to use them with a martix and how can ||x||_A = ||Ax||?
$endgroup$
– Usman Ashraf
Jan 30 at 13:16
$begingroup$
$||x||_A=||Ax||$ is just the definition of the norm. You need to use the fact that $||.||$ is a norm. For example why does the triangle inequality hold? We have $||x+y||_A=||A(x+y)||=||Ax+Ay||leq||Ax||+||Ay||=||x||_A+||y||_A$. Now in a similar way you can check the other axioms.
$endgroup$
– Mark
Jan 30 at 13:56