Homogeneity does not suffice for a map between vector spaces to be linear
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The following problem is taken from Sheldon Axler's book Linear Algebra Done Right, more precisely Exercise 1. from Chapter 3:
Problem: Give an example of a function $f : mathbb{R}^2 to mathbb{R}$ such that $f(av) = af(v)$ for all $a in mathbb{R}$ and all $v in mathbb{R}^2$ but $f$ is not linear.
I tried to include things like absolute values and square roots in order to handle the homogeneity, but I did not had any success in constructing such an example yet by doing so.
Do you know such an example? Thank you in advance!
linear-algebra linear-transformations examples-counterexamples
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
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add a comment |
$begingroup$
The following problem is taken from Sheldon Axler's book Linear Algebra Done Right, more precisely Exercise 1. from Chapter 3:
Problem: Give an example of a function $f : mathbb{R}^2 to mathbb{R}$ such that $f(av) = af(v)$ for all $a in mathbb{R}$ and all $v in mathbb{R}^2$ but $f$ is not linear.
I tried to include things like absolute values and square roots in order to handle the homogeneity, but I did not had any success in constructing such an example yet by doing so.
Do you know such an example? Thank you in advance!
linear-algebra linear-transformations examples-counterexamples
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
$endgroup$
$begingroup$
Have you drawn pictures of what homogeneity does and does not require of the function? Not every function has to be given by a formula.
$endgroup$
– Mark S.
Dec 10 '18 at 12:54
add a comment |
$begingroup$
The following problem is taken from Sheldon Axler's book Linear Algebra Done Right, more precisely Exercise 1. from Chapter 3:
Problem: Give an example of a function $f : mathbb{R}^2 to mathbb{R}$ such that $f(av) = af(v)$ for all $a in mathbb{R}$ and all $v in mathbb{R}^2$ but $f$ is not linear.
I tried to include things like absolute values and square roots in order to handle the homogeneity, but I did not had any success in constructing such an example yet by doing so.
Do you know such an example? Thank you in advance!
linear-algebra linear-transformations examples-counterexamples
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
$endgroup$
The following problem is taken from Sheldon Axler's book Linear Algebra Done Right, more precisely Exercise 1. from Chapter 3:
Problem: Give an example of a function $f : mathbb{R}^2 to mathbb{R}$ such that $f(av) = af(v)$ for all $a in mathbb{R}$ and all $v in mathbb{R}^2$ but $f$ is not linear.
I tried to include things like absolute values and square roots in order to handle the homogeneity, but I did not had any success in constructing such an example yet by doing so.
Do you know such an example? Thank you in advance!
linear-algebra linear-transformations examples-counterexamples
linear-algebra linear-transformations examples-counterexamples
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
asked Dec 10 '18 at 2:49
DiglettDiglett
9381521
9381521
$begingroup$
Have you drawn pictures of what homogeneity does and does not require of the function? Not every function has to be given by a formula.
$endgroup$
– Mark S.
Dec 10 '18 at 12:54
add a comment |
$begingroup$
Have you drawn pictures of what homogeneity does and does not require of the function? Not every function has to be given by a formula.
$endgroup$
– Mark S.
Dec 10 '18 at 12:54
$begingroup$
Have you drawn pictures of what homogeneity does and does not require of the function? Not every function has to be given by a formula.
$endgroup$
– Mark S.
Dec 10 '18 at 12:54
$begingroup$
Have you drawn pictures of what homogeneity does and does not require of the function? Not every function has to be given by a formula.
$endgroup$
– Mark S.
Dec 10 '18 at 12:54
add a comment |
1 Answer
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consider the map: $$f:mathbb{R}^2 to mathbb{R}\ (x,y) mapsto begin{cases}
x text{ if } y=0 \
0 text{ else }end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=begin{cases}ax text{ if } y=0 \ a0text{ else }end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 neq f(2,0) =f((1,1)+(1,-1)).$$
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
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1
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You say "this is clearly homogeneous", but it can't be (in the sense given in the problem) unless your norm is constantly zero. Otherwise, there's no way for it to send $-v$ to $-f(v)$ as norms never output negative numbers.
$endgroup$
– Mark S.
Dec 10 '18 at 12:56
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sorry, you are completely right, I think i remmedied that, I was just very tempted by the simplicity of the norm.
$endgroup$
– Enkidu
Dec 10 '18 at 13:05
add a comment |
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1 Answer
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$begingroup$
consider the map: $$f:mathbb{R}^2 to mathbb{R}\ (x,y) mapsto begin{cases}
x text{ if } y=0 \
0 text{ else }end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=begin{cases}ax text{ if } y=0 \ a0text{ else }end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 neq f(2,0) =f((1,1)+(1,-1)).$$
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
$endgroup$
1
$begingroup$
You say "this is clearly homogeneous", but it can't be (in the sense given in the problem) unless your norm is constantly zero. Otherwise, there's no way for it to send $-v$ to $-f(v)$ as norms never output negative numbers.
$endgroup$
– Mark S.
Dec 10 '18 at 12:56
$begingroup$
sorry, you are completely right, I think i remmedied that, I was just very tempted by the simplicity of the norm.
$endgroup$
– Enkidu
Dec 10 '18 at 13:05
add a comment |
$begingroup$
consider the map: $$f:mathbb{R}^2 to mathbb{R}\ (x,y) mapsto begin{cases}
x text{ if } y=0 \
0 text{ else }end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=begin{cases}ax text{ if } y=0 \ a0text{ else }end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 neq f(2,0) =f((1,1)+(1,-1)).$$
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
$endgroup$
1
$begingroup$
You say "this is clearly homogeneous", but it can't be (in the sense given in the problem) unless your norm is constantly zero. Otherwise, there's no way for it to send $-v$ to $-f(v)$ as norms never output negative numbers.
$endgroup$
– Mark S.
Dec 10 '18 at 12:56
$begingroup$
sorry, you are completely right, I think i remmedied that, I was just very tempted by the simplicity of the norm.
$endgroup$
– Enkidu
Dec 10 '18 at 13:05
add a comment |
$begingroup$
consider the map: $$f:mathbb{R}^2 to mathbb{R}\ (x,y) mapsto begin{cases}
x text{ if } y=0 \
0 text{ else }end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=begin{cases}ax text{ if } y=0 \ a0text{ else }end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 neq f(2,0) =f((1,1)+(1,-1)).$$
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
$endgroup$
consider the map: $$f:mathbb{R}^2 to mathbb{R}\ (x,y) mapsto begin{cases}
x text{ if } y=0 \
0 text{ else }end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=begin{cases}ax text{ if } y=0 \ a0text{ else }end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 neq f(2,0) =f((1,1)+(1,-1)).$$
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
edited Dec 10 '18 at 13:01
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
answered Dec 10 '18 at 12:38
EnkiduEnkidu
1,32619
1,32619
1
$begingroup$
You say "this is clearly homogeneous", but it can't be (in the sense given in the problem) unless your norm is constantly zero. Otherwise, there's no way for it to send $-v$ to $-f(v)$ as norms never output negative numbers.
$endgroup$
– Mark S.
Dec 10 '18 at 12:56
$begingroup$
sorry, you are completely right, I think i remmedied that, I was just very tempted by the simplicity of the norm.
$endgroup$
– Enkidu
Dec 10 '18 at 13:05
add a comment |
1
$begingroup$
You say "this is clearly homogeneous", but it can't be (in the sense given in the problem) unless your norm is constantly zero. Otherwise, there's no way for it to send $-v$ to $-f(v)$ as norms never output negative numbers.
$endgroup$
– Mark S.
Dec 10 '18 at 12:56
$begingroup$
sorry, you are completely right, I think i remmedied that, I was just very tempted by the simplicity of the norm.
$endgroup$
– Enkidu
Dec 10 '18 at 13:05
1
1
$begingroup$
You say "this is clearly homogeneous", but it can't be (in the sense given in the problem) unless your norm is constantly zero. Otherwise, there's no way for it to send $-v$ to $-f(v)$ as norms never output negative numbers.
$endgroup$
– Mark S.
Dec 10 '18 at 12:56
$begingroup$
You say "this is clearly homogeneous", but it can't be (in the sense given in the problem) unless your norm is constantly zero. Otherwise, there's no way for it to send $-v$ to $-f(v)$ as norms never output negative numbers.
$endgroup$
– Mark S.
Dec 10 '18 at 12:56
$begingroup$
sorry, you are completely right, I think i remmedied that, I was just very tempted by the simplicity of the norm.
$endgroup$
– Enkidu
Dec 10 '18 at 13:05
$begingroup$
sorry, you are completely right, I think i remmedied that, I was just very tempted by the simplicity of the norm.
$endgroup$
– Enkidu
Dec 10 '18 at 13:05
add a comment |
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$begingroup$
Have you drawn pictures of what homogeneity does and does not require of the function? Not every function has to be given by a formula.
$endgroup$
– Mark S.
Dec 10 '18 at 12:54