Mixing
Zero linear transformation in limits with Euclidean norm
$begingroup$
This does not seem too difficult, but I've been stuck here for a while. Could someone give me a hint?
Question: Let $A$ be a linear transformation on $R^n$. $v$ is a vector in $R^n$. Prove that if $frac{||Av||}{||v||}rightarrow0$ as $vrightarrow 0$, then $A=0$. $||cdot||$ denotes Euclidean norm.
My try: proof by contradiction. Suppose at least one element in $A$, namely $A_{pq}$, is nonzero. Ideally, the contradiction would be the limit of the above fraction is nonzero. The problem is the numerator always approaches zero as $vrightarrow 0$. Is there any other approach?
limits multivariable-calculus linear-transformations norm
asked Jan 28 at 16:49
James WangJames Wang
50228
$endgroup$
add a comment |
$begingroup$
This does not seem too difficult, but I've been stuck here for a while. Could someone give me a hint?
Question: Let $A$ be a linear transformation on $R^n$. $v$ is a vector in $R^n$. Prove that if $frac{||Av||}{||v||}rightarrow0$ as $vrightarrow 0$, then $A=0$. $||cdot||$ denotes Euclidean norm.
My try: proof by contradiction. Suppose at least one element in $A$, namely $A_{pq}$, is nonzero. Ideally, the contradiction would be the limit of the above fraction is nonzero. The problem is the numerator always approaches zero as $vrightarrow 0$. Is there any other approach?
limits multivariable-calculus linear-transformations norm
asked Jan 28 at 16:49
James WangJames Wang
50228
$endgroup$
add a comment |
$begingroup$
This does not seem too difficult, but I've been stuck here for a while. Could someone give me a hint?
Question: Let $A$ be a linear transformation on $R^n$. $v$ is a vector in $R^n$. Prove that if $frac{||Av||}{||v||}rightarrow0$ as $vrightarrow 0$, then $A=0$. $||cdot||$ denotes Euclidean norm.
My try: proof by contradiction. Suppose at least one element in $A$, namely $A_{pq}$, is nonzero. Ideally, the contradiction would be the limit of the above fraction is nonzero. The problem is the numerator always approaches zero as $vrightarrow 0$. Is there any other approach?
limits multivariable-calculus linear-transformations norm
asked Jan 28 at 16:49
James WangJames Wang
50228
$endgroup$
This does not seem too difficult, but I've been stuck here for a while. Could someone give me a hint?
Question: Let $A$ be a linear transformation on $R^n$. $v$ is a vector in $R^n$. Prove that if $frac{||Av||}{||v||}rightarrow0$ as $vrightarrow 0$, then $A=0$. $||cdot||$ denotes Euclidean norm.
My try: proof by contradiction. Suppose at least one element in $A$, namely $A_{pq}$, is nonzero. Ideally, the contradiction would be the limit of the above fraction is nonzero. The problem is the numerator always approaches zero as $vrightarrow 0$. Is there any other approach?
limits multivariable-calculus linear-transformations norm
limits multivariable-calculus linear-transformations norm
asked Jan 28 at 16:49
James WangJames Wang
50228
asked Jan 28 at 16:49
James WangJames Wang
50228
asked Jan 28 at 16:49
James WangJames Wang
50228
asked Jan 28 at 16:49
James WangJames Wang
50228
asked Jan 28 at 16:49
James WangJames Wang
50228
50228
add a comment |
add a comment |
1 Answer
1
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For any $v$ with $||v||ne0$,
$$
||Av||=||v||,frac{||Av||}{||v||}=||v||,frac{||lambda Av||}{||lambda v||}=||v||,frac{||A(lambda v)||}{||lambda v||}
to 0
$$
as $lambdato0$. Hence $Av=0$.
answered Jan 28 at 17:11
VladimirVladimir
5,413618
$endgroup$
$begingroup$
How did you take care of $||lambda v||$ in the denominator when $vrightarrow 0$?
$endgroup$
– James Wang
Jan 28 at 19:02
1
$begingroup$
$v$ is constant. Instead, $lambdato0$ and hence $lambda vto0$ and hence $||A(lambda v)||/||lambda v||to 0$ and hence $||Av||=||v||,||A(lambda v)||/||lambda v||to 0$ as $lambdato0$. But $Av$ is independent of $lambda$, so this can only happen if $||Av||=0$,
$endgroup$
– Vladimir
Jan 28 at 20:14
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
For any $v$ with $||v||ne0$,
$$
||Av||=||v||,frac{||Av||}{||v||}=||v||,frac{||lambda Av||}{||lambda v||}=||v||,frac{||A(lambda v)||}{||lambda v||}
to 0
$$
as $lambdato0$. Hence $Av=0$.
answered Jan 28 at 17:11
VladimirVladimir
5,413618
$endgroup$
$begingroup$
How did you take care of $||lambda v||$ in the denominator when $vrightarrow 0$?
$endgroup$
– James Wang
Jan 28 at 19:02
1
$begingroup$
$v$ is constant. Instead, $lambdato0$ and hence $lambda vto0$ and hence $||A(lambda v)||/||lambda v||to 0$ and hence $||Av||=||v||,||A(lambda v)||/||lambda v||to 0$ as $lambdato0$. But $Av$ is independent of $lambda$, so this can only happen if $||Av||=0$,
$endgroup$
– Vladimir
Jan 28 at 20:14
add a comment |
$begingroup$
For any $v$ with $||v||ne0$,
$$
||Av||=||v||,frac{||Av||}{||v||}=||v||,frac{||lambda Av||}{||lambda v||}=||v||,frac{||A(lambda v)||}{||lambda v||}
to 0
$$
as $lambdato0$. Hence $Av=0$.
answered Jan 28 at 17:11
VladimirVladimir
5,413618
$endgroup$
$begingroup$
How did you take care of $||lambda v||$ in the denominator when $vrightarrow 0$?
$endgroup$
– James Wang
Jan 28 at 19:02
1
$begingroup$
$v$ is constant. Instead, $lambdato0$ and hence $lambda vto0$ and hence $||A(lambda v)||/||lambda v||to 0$ and hence $||Av||=||v||,||A(lambda v)||/||lambda v||to 0$ as $lambdato0$. But $Av$ is independent of $lambda$, so this can only happen if $||Av||=0$,
$endgroup$
– Vladimir
Jan 28 at 20:14
add a comment |
$begingroup$
For any $v$ with $||v||ne0$,
$$
||Av||=||v||,frac{||Av||}{||v||}=||v||,frac{||lambda Av||}{||lambda v||}=||v||,frac{||A(lambda v)||}{||lambda v||}
to 0
$$
as $lambdato0$. Hence $Av=0$.
answered Jan 28 at 17:11
VladimirVladimir
5,413618
$endgroup$
For any $v$ with $||v||ne0$,
$$
||Av||=||v||,frac{||Av||}{||v||}=||v||,frac{||lambda Av||}{||lambda v||}=||v||,frac{||A(lambda v)||}{||lambda v||}
to 0
$$
as $lambdato0$. Hence $Av=0$.
answered Jan 28 at 17:11
VladimirVladimir
5,413618
answered Jan 28 at 17:11
VladimirVladimir
5,413618
answered Jan 28 at 17:11
VladimirVladimir
5,413618
answered Jan 28 at 17:11
VladimirVladimir
5,413618
5,413618
$begingroup$
How did you take care of $||lambda v||$ in the denominator when $vrightarrow 0$?
$endgroup$
– James Wang
Jan 28 at 19:02
1
$begingroup$
$v$ is constant. Instead, $lambdato0$ and hence $lambda vto0$ and hence $||A(lambda v)||/||lambda v||to 0$ and hence $||Av||=||v||,||A(lambda v)||/||lambda v||to 0$ as $lambdato0$. But $Av$ is independent of $lambda$, so this can only happen if $||Av||=0$,
$endgroup$
– Vladimir
Jan 28 at 20:14
add a comment |
$begingroup$
How did you take care of $||lambda v||$ in the denominator when $vrightarrow 0$?
$endgroup$
– James Wang
Jan 28 at 19:02
1
$begingroup$
$v$ is constant. Instead, $lambdato0$ and hence $lambda vto0$ and hence $||A(lambda v)||/||lambda v||to 0$ and hence $||Av||=||v||,||A(lambda v)||/||lambda v||to 0$ as $lambdato0$. But $Av$ is independent of $lambda$, so this can only happen if $||Av||=0$,
$endgroup$
– Vladimir
Jan 28 at 20:14
$begingroup$
How did you take care of $||lambda v||$ in the denominator when $vrightarrow 0$?
$endgroup$
– James Wang
Jan 28 at 19:02
$begingroup$
How did you take care of $||lambda v||$ in the denominator when $vrightarrow 0$?
$endgroup$
– James Wang
Jan 28 at 19:02
1
1
$begingroup$
$v$ is constant. Instead, $lambdato0$ and hence $lambda vto0$ and hence $||A(lambda v)||/||lambda v||to 0$ and hence $||Av||=||v||,||A(lambda v)||/||lambda v||to 0$ as $lambdato0$. But $Av$ is independent of $lambda$, so this can only happen if $||Av||=0$,
$endgroup$
– Vladimir
Jan 28 at 20:14
$begingroup$
$v$ is constant. Instead, $lambdato0$ and hence $lambda vto0$ and hence $||A(lambda v)||/||lambda v||to 0$ and hence $||Av||=||v||,||A(lambda v)||/||lambda v||to 0$ as $lambdato0$. But $Av$ is independent of $lambda$, so this can only happen if $||Av||=0$,
$endgroup$
– Vladimir
Jan 28 at 20:14
add a comment |
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