Is $A$ a linear transformation on $mathbbC^n$, where $A(u) = | u| u$?

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1

Is $A$ a linear transformation or not on $mathbbC^n$, where
$A(u) = |u| u$ and $|u| $ is the length of the vector $u$.

I know the definition of linear transformations: $T$ is linear if it satisfies

$$
T(mathbf x+mathbf y) = T(mathbf x) + T(mathbf y)\
T(cmathbf x) = cT(mathbf x)
$$

But here how can I apply it? I am confused. . .

linear-algebra

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edited Aug 6 at 6:48

Brahadeesh

3,74931449

asked Aug 6 at 4:19

Messi fifa

1718

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take two arbitrary vectors and try to check the first rule.
  – Anik Bhowmick
  Aug 6 at 4:35
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: consider $mathbfx neq 0$ and $c = 2$.
  – Theo Bendit
  Aug 6 at 4:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

Is $A$ a linear transformation or not on $mathbbC^n$, where
$A(u) = |u| u$ and $|u| $ is the length of the vector $u$.

I know the definition of linear transformations: $T$ is linear if it satisfies

$$
T(mathbf x+mathbf y) = T(mathbf x) + T(mathbf y)\
T(cmathbf x) = cT(mathbf x)
$$

But here how can I apply it? I am confused. . .

linear-algebra

share|cite|improve this question

edited Aug 6 at 6:48

Brahadeesh

3,74931449

asked Aug 6 at 4:19

Messi fifa

1718

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take two arbitrary vectors and try to check the first rule.
  – Anik Bhowmick
  Aug 6 at 4:35
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: consider $mathbfx neq 0$ and $c = 2$.
  – Theo Bendit
  Aug 6 at 4:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Is $A$ a linear transformation or not on $mathbbC^n$, where
$A(u) = |u| u$ and $|u| $ is the length of the vector $u$.

I know the definition of linear transformations: $T$ is linear if it satisfies

$$
T(mathbf x+mathbf y) = T(mathbf x) + T(mathbf y)\
T(cmathbf x) = cT(mathbf x)
$$

But here how can I apply it? I am confused. . .

linear-algebra

share|cite|improve this question

edited Aug 6 at 6:48

Brahadeesh

3,74931449

asked Aug 6 at 4:19

Messi fifa

1718

Is $A$ a linear transformation or not on $mathbbC^n$, where
$A(u) = |u| u$ and $|u| $ is the length of the vector $u$.

I know the definition of linear transformations: $T$ is linear if it satisfies

$$
T(mathbf x+mathbf y) = T(mathbf x) + T(mathbf y)\
T(cmathbf x) = cT(mathbf x)
$$

But here how can I apply it? I am confused. . .

linear-algebra

share|cite|improve this question

edited Aug 6 at 6:48

Brahadeesh

3,74931449

asked Aug 6 at 4:19

Messi fifa

1718

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 6:48

Brahadeesh

3,74931449

edited Aug 6 at 6:48

Brahadeesh

3,74931449

edited Aug 6 at 6:48

Brahadeesh

3,74931449

3,74931449

asked Aug 6 at 4:19

Messi fifa

1718

asked Aug 6 at 4:19

Messi fifa

1718

asked Aug 6 at 4:19

Messi fifa

1718

1718

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take two arbitrary vectors and try to check the first rule.
  – Anik Bhowmick
  Aug 6 at 4:35
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: consider $mathbfx neq 0$ and $c = 2$.
  – Theo Bendit
  Aug 6 at 4:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take two arbitrary vectors and try to check the first rule.
  – Anik Bhowmick
  Aug 6 at 4:35
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Hint: consider $mathbfx neq 0$ and $c = 2$.
  – Theo Bendit
  Aug 6 at 4:47
  

  
  
  
  

1

1

Take two arbitrary vectors and try to check the first rule.
– Anik Bhowmick
Aug 6 at 4:35

Take two arbitrary vectors and try to check the first rule.
– Anik Bhowmick
Aug 6 at 4:35

2

2

Hint: consider $mathbfx neq 0$ and $c = 2$.
– Theo Bendit
Aug 6 at 4:47

Hint: consider $mathbfx neq 0$ and $c = 2$.
– Theo Bendit
Aug 6 at 4:47

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

Since there is a factor of $| u |$ in the definition of $A(u)$, it means that $A$ is not scaling linearly. The magnitude of the scaling increases as the magnitude of the vector $u$ increases. But this contradicts the second property of a linear transformation, which essentially says that scaling is uniform.

So, we expect that $A$ is not linear. A good choice of $u$ to verify this is to take a nonzero vector $u$ and multiply it by a factor $c$ which is not equal to $1$, and then check. So, let's fix a nonzero $u in mathbbC^n$, say $u = (1,dots,1)$, and $c = 2$. Then, we get
$$
A(2u) = | 2u | cdot 2u = 4 | u | u.
$$
But if $A$ were linear, then we would have had $$A(2u) = 2 A(u) = 2 |u | u.$$ Since this is not the case, we have shown that $A$ is not linear.

---

Also note that the other answer by @GinoCHJ is incomplete. Merely knowing that $$A(u+v) leq A(u) + A(v) + | v| u + |u|v$$ does not imply that $A$ is not a linear transformation. Perhaps by some wicked magic it turns out that $|v| u + |u|v = 0$ for all $u,v in mathbbC^n$? Their answer would be satisfactorily completed if specific vectors $u$ and $v$ are produced such that the inequality is strict.

share|cite|improve this answer

edited Aug 6 at 6:59

answered Aug 6 at 6:45

Brahadeesh

3,74931449

add a comment | 

up vote
-1
down vote

Let $u, v in mathbbC^n$, then
beginalign*
&A(u + v) = lVert u + v rVert(u + v) \
Rightarrow &A(u + v) = lVert u + v rVert u + lVert u + v rVert v leq lVert u rVert u + lVert v rVert u + lVert u rVert v + lVert v rVert v \
Rightarrow &A(u + v) leq A(u) + lVert v rVert u + lVert u rVert v + A(v)
endalign*
Hence, $A$ is not a linear transformation.

share|cite|improve this answer

answered Aug 6 at 6:10

GinoCHJ

794

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Since there is a factor of $| u |$ in the definition of $A(u)$, it means that $A$ is not scaling linearly. The magnitude of the scaling increases as the magnitude of the vector $u$ increases. But this contradicts the second property of a linear transformation, which essentially says that scaling is uniform.

So, we expect that $A$ is not linear. A good choice of $u$ to verify this is to take a nonzero vector $u$ and multiply it by a factor $c$ which is not equal to $1$, and then check. So, let's fix a nonzero $u in mathbbC^n$, say $u = (1,dots,1)$, and $c = 2$. Then, we get
$$
A(2u) = | 2u | cdot 2u = 4 | u | u.
$$
But if $A$ were linear, then we would have had $$A(2u) = 2 A(u) = 2 |u | u.$$ Since this is not the case, we have shown that $A$ is not linear.

---

Also note that the other answer by @GinoCHJ is incomplete. Merely knowing that $$A(u+v) leq A(u) + A(v) + | v| u + |u|v$$ does not imply that $A$ is not a linear transformation. Perhaps by some wicked magic it turns out that $|v| u + |u|v = 0$ for all $u,v in mathbbC^n$? Their answer would be satisfactorily completed if specific vectors $u$ and $v$ are produced such that the inequality is strict.

share|cite|improve this answer

edited Aug 6 at 6:59

answered Aug 6 at 6:45

Brahadeesh

3,74931449

add a comment | 

up vote
3
down vote



accepted

Since there is a factor of $| u |$ in the definition of $A(u)$, it means that $A$ is not scaling linearly. The magnitude of the scaling increases as the magnitude of the vector $u$ increases. But this contradicts the second property of a linear transformation, which essentially says that scaling is uniform.

So, we expect that $A$ is not linear. A good choice of $u$ to verify this is to take a nonzero vector $u$ and multiply it by a factor $c$ which is not equal to $1$, and then check. So, let's fix a nonzero $u in mathbbC^n$, say $u = (1,dots,1)$, and $c = 2$. Then, we get
$$
A(2u) = | 2u | cdot 2u = 4 | u | u.
$$
But if $A$ were linear, then we would have had $$A(2u) = 2 A(u) = 2 |u | u.$$ Since this is not the case, we have shown that $A$ is not linear.

---

Also note that the other answer by @GinoCHJ is incomplete. Merely knowing that $$A(u+v) leq A(u) + A(v) + | v| u + |u|v$$ does not imply that $A$ is not a linear transformation. Perhaps by some wicked magic it turns out that $|v| u + |u|v = 0$ for all $u,v in mathbbC^n$? Their answer would be satisfactorily completed if specific vectors $u$ and $v$ are produced such that the inequality is strict.

share|cite|improve this answer

edited Aug 6 at 6:59

answered Aug 6 at 6:45

Brahadeesh

3,74931449

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Since there is a factor of $| u |$ in the definition of $A(u)$, it means that $A$ is not scaling linearly. The magnitude of the scaling increases as the magnitude of the vector $u$ increases. But this contradicts the second property of a linear transformation, which essentially says that scaling is uniform.

So, we expect that $A$ is not linear. A good choice of $u$ to verify this is to take a nonzero vector $u$ and multiply it by a factor $c$ which is not equal to $1$, and then check. So, let's fix a nonzero $u in mathbbC^n$, say $u = (1,dots,1)$, and $c = 2$. Then, we get
$$
A(2u) = | 2u | cdot 2u = 4 | u | u.
$$
But if $A$ were linear, then we would have had $$A(2u) = 2 A(u) = 2 |u | u.$$ Since this is not the case, we have shown that $A$ is not linear.

---

Also note that the other answer by @GinoCHJ is incomplete. Merely knowing that $$A(u+v) leq A(u) + A(v) + | v| u + |u|v$$ does not imply that $A$ is not a linear transformation. Perhaps by some wicked magic it turns out that $|v| u + |u|v = 0$ for all $u,v in mathbbC^n$? Their answer would be satisfactorily completed if specific vectors $u$ and $v$ are produced such that the inequality is strict.

share|cite|improve this answer

edited Aug 6 at 6:59

answered Aug 6 at 6:45

Brahadeesh

3,74931449

Since there is a factor of $| u |$ in the definition of $A(u)$, it means that $A$ is not scaling linearly. The magnitude of the scaling increases as the magnitude of the vector $u$ increases. But this contradicts the second property of a linear transformation, which essentially says that scaling is uniform.

So, we expect that $A$ is not linear. A good choice of $u$ to verify this is to take a nonzero vector $u$ and multiply it by a factor $c$ which is not equal to $1$, and then check. So, let's fix a nonzero $u in mathbbC^n$, say $u = (1,dots,1)$, and $c = 2$. Then, we get
$$
A(2u) = | 2u | cdot 2u = 4 | u | u.
$$
But if $A$ were linear, then we would have had $$A(2u) = 2 A(u) = 2 |u | u.$$ Since this is not the case, we have shown that $A$ is not linear.

---

Also note that the other answer by @GinoCHJ is incomplete. Merely knowing that $$A(u+v) leq A(u) + A(v) + | v| u + |u|v$$ does not imply that $A$ is not a linear transformation. Perhaps by some wicked magic it turns out that $|v| u + |u|v = 0$ for all $u,v in mathbbC^n$? Their answer would be satisfactorily completed if specific vectors $u$ and $v$ are produced such that the inequality is strict.

share|cite|improve this answer

edited Aug 6 at 6:59

answered Aug 6 at 6:45

Brahadeesh

3,74931449

share|cite|improve this answer

share|cite|improve this answer

edited Aug 6 at 6:59

edited Aug 6 at 6:59

edited Aug 6 at 6:59

answered Aug 6 at 6:45

Brahadeesh

3,74931449

answered Aug 6 at 6:45

Brahadeesh

3,74931449

answered Aug 6 at 6:45

Brahadeesh

3,74931449

3,74931449

add a comment | 

add a comment | 

up vote
-1
down vote

Let $u, v in mathbbC^n$, then
beginalign*
&A(u + v) = lVert u + v rVert(u + v) \
Rightarrow &A(u + v) = lVert u + v rVert u + lVert u + v rVert v leq lVert u rVert u + lVert v rVert u + lVert u rVert v + lVert v rVert v \
Rightarrow &A(u + v) leq A(u) + lVert v rVert u + lVert u rVert v + A(v)
endalign*
Hence, $A$ is not a linear transformation.

share|cite|improve this answer

answered Aug 6 at 6:10

GinoCHJ

794

add a comment | 

up vote
-1
down vote

Let $u, v in mathbbC^n$, then
beginalign*
&A(u + v) = lVert u + v rVert(u + v) \
Rightarrow &A(u + v) = lVert u + v rVert u + lVert u + v rVert v leq lVert u rVert u + lVert v rVert u + lVert u rVert v + lVert v rVert v \
Rightarrow &A(u + v) leq A(u) + lVert v rVert u + lVert u rVert v + A(v)
endalign*
Hence, $A$ is not a linear transformation.

share|cite|improve this answer

answered Aug 6 at 6:10

GinoCHJ

794

add a comment | 

up vote
-1
down vote

up vote
-1
down vote

Let $u, v in mathbbC^n$, then
beginalign*
&A(u + v) = lVert u + v rVert(u + v) \
Rightarrow &A(u + v) = lVert u + v rVert u + lVert u + v rVert v leq lVert u rVert u + lVert v rVert u + lVert u rVert v + lVert v rVert v \
Rightarrow &A(u + v) leq A(u) + lVert v rVert u + lVert u rVert v + A(v)
endalign*
Hence, $A$ is not a linear transformation.

share|cite|improve this answer

answered Aug 6 at 6:10

GinoCHJ

794

Let $u, v in mathbbC^n$, then
beginalign*
&A(u + v) = lVert u + v rVert(u + v) \
Rightarrow &A(u + v) = lVert u + v rVert u + lVert u + v rVert v leq lVert u rVert u + lVert v rVert u + lVert u rVert v + lVert v rVert v \
Rightarrow &A(u + v) leq A(u) + lVert v rVert u + lVert u rVert v + A(v)
endalign*
Hence, $A$ is not a linear transformation.

share|cite|improve this answer

answered Aug 6 at 6:10

GinoCHJ

794

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 6:10

GinoCHJ

794

answered Aug 6 at 6:10

GinoCHJ

794

answered Aug 6 at 6:10

GinoCHJ

794

794

add a comment | 

add a comment | 

 

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