Mixing
Show that f is non linear and g is linear
$begingroup$
I have been given the Basis B of V and C of W in Q with
$$B = {v_1,v_2,v_3}$$
$$C = {w_1,w_2}$$
And f and g with:
$$f:mathbb{V} to mathbb{W}, f(k_1v_1, k_2v_2, k_3v_3)=(3k_1+k_2)w_1+k_3^7w_2$$
$$g:mathbb{V} to mathbb{W}, g(k_1v_1, k_2v_2, k_3v_3)=5k_1w_1+(2k_2+7k_3)w_2$$
I have to show that f is non linear and g is linear.
So I need to check for homogenity and additivity. So
$$f(kv) = kf(v)$$ and $$f(v+w) = f(v) + f(w)$$ has to be right for linearity.
I see that f can't be linear because of the k^7.
But I have no clue how to put this Basis into the definition of linearity for this example.
I hope someone could show me.
Greetings
KLLK
linear-algebra vector-spaces
edited Jan 31 at 9:03
Arthur
122k7122211
asked Jan 31 at 8:48
KLLKKLLK
83
$endgroup$
add a comment |
$begingroup$
I have been given the Basis B of V and C of W in Q with
$$B = {v_1,v_2,v_3}$$
$$C = {w_1,w_2}$$
And f and g with:
$$f:mathbb{V} to mathbb{W}, f(k_1v_1, k_2v_2, k_3v_3)=(3k_1+k_2)w_1+k_3^7w_2$$
$$g:mathbb{V} to mathbb{W}, g(k_1v_1, k_2v_2, k_3v_3)=5k_1w_1+(2k_2+7k_3)w_2$$
I have to show that f is non linear and g is linear.
So I need to check for homogenity and additivity. So
$$f(kv) = kf(v)$$ and $$f(v+w) = f(v) + f(w)$$ has to be right for linearity.
I see that f can't be linear because of the k^7.
But I have no clue how to put this Basis into the definition of linearity for this example.
I hope someone could show me.
Greetings
KLLK
linear-algebra vector-spaces
edited Jan 31 at 9:03
Arthur
122k7122211
asked Jan 31 at 8:48
KLLKKLLK
83
$endgroup$
$begingroup$
In my opinion it ought to be either $f(k_1, k_2, k_3)$, or $f(k_1w_1 + k_2w_2 + k_3w_3)$. The mixture you have there looks strange.
$endgroup$
– Arthur
Jan 31 at 8:58
add a comment |
$begingroup$
I have been given the Basis B of V and C of W in Q with
$$B = {v_1,v_2,v_3}$$
$$C = {w_1,w_2}$$
And f and g with:
$$f:mathbb{V} to mathbb{W}, f(k_1v_1, k_2v_2, k_3v_3)=(3k_1+k_2)w_1+k_3^7w_2$$
$$g:mathbb{V} to mathbb{W}, g(k_1v_1, k_2v_2, k_3v_3)=5k_1w_1+(2k_2+7k_3)w_2$$
I have to show that f is non linear and g is linear.
So I need to check for homogenity and additivity. So
$$f(kv) = kf(v)$$ and $$f(v+w) = f(v) + f(w)$$ has to be right for linearity.
I see that f can't be linear because of the k^7.
But I have no clue how to put this Basis into the definition of linearity for this example.
I hope someone could show me.
Greetings
KLLK
linear-algebra vector-spaces
edited Jan 31 at 9:03
Arthur
122k7122211
asked Jan 31 at 8:48
KLLKKLLK
83
$endgroup$
I have been given the Basis B of V and C of W in Q with
$$B = {v_1,v_2,v_3}$$
$$C = {w_1,w_2}$$
And f and g with:
$$f:mathbb{V} to mathbb{W}, f(k_1v_1, k_2v_2, k_3v_3)=(3k_1+k_2)w_1+k_3^7w_2$$
$$g:mathbb{V} to mathbb{W}, g(k_1v_1, k_2v_2, k_3v_3)=5k_1w_1+(2k_2+7k_3)w_2$$
I have to show that f is non linear and g is linear.
So I need to check for homogenity and additivity. So
$$f(kv) = kf(v)$$ and $$f(v+w) = f(v) + f(w)$$ has to be right for linearity.
I see that f can't be linear because of the k^7.
But I have no clue how to put this Basis into the definition of linearity for this example.
I hope someone could show me.
Greetings
KLLK
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 31 at 9:03
Arthur
122k7122211
asked Jan 31 at 8:48
KLLKKLLK
83
edited Jan 31 at 9:03
Arthur
122k7122211
asked Jan 31 at 8:48
KLLKKLLK
83
edited Jan 31 at 9:03
Arthur
122k7122211
edited Jan 31 at 9:03
Arthur
122k7122211
edited Jan 31 at 9:03
Arthur
122k7122211
122k7122211
asked Jan 31 at 8:48
KLLKKLLK
83
asked Jan 31 at 8:48
KLLKKLLK
83
asked Jan 31 at 8:48
KLLKKLLK
83
83
$begingroup$
In my opinion it ought to be either $f(k_1, k_2, k_3)$, or $f(k_1w_1 + k_2w_2 + k_3w_3)$. The mixture you have there looks strange.
$endgroup$
– Arthur
Jan 31 at 8:58
add a comment |
$begingroup$
In my opinion it ought to be either $f(k_1, k_2, k_3)$, or $f(k_1w_1 + k_2w_2 + k_3w_3)$. The mixture you have there looks strange.
$endgroup$
– Arthur
Jan 31 at 8:58
$begingroup$
In my opinion it ought to be either $f(k_1, k_2, k_3)$, or $f(k_1w_1 + k_2w_2 + k_3w_3)$. The mixture you have there looks strange.
$endgroup$
– Arthur
Jan 31 at 8:58
$begingroup$
In my opinion it ought to be either $f(k_1, k_2, k_3)$, or $f(k_1w_1 + k_2w_2 + k_3w_3)$. The mixture you have there looks strange.
$endgroup$
– Arthur
Jan 31 at 8:58
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
What is $f(v_3) + f(v_3)$, and what is $f(2v_3)$? If $f$ were linear, should they be equal? Can they be equal?
answered Jan 31 at 8:57
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
this just means that you have (for B) three linearly independent vectors that span your vector space V. It is a more general way of writing vecor spaces, you don't always have a standard basis (i.e. $mathbb{R}^{2}=span{((1,0), (0,1))}$).
For your example you can simply handle this the exact way you would if you had been given a standard basis, i.e.:
$f$ is not linear:
Let $vin V, kin K$, then
$$f(k_{1}v_{1}+l_{1}v_1,k_{2}v_2+l_2v_2, k_{3}v_3+l_3v_3) = 3((k_1+l_1+k_2+l_2) w_1 + (k_3+l_3)^7w_2
$$
And continue on from there.
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
$endgroup$
add a comment |
$begingroup$
You're right about the $k_3^7$. What's at work there is the freshman's binomial. That is to say, were $(x+y)^7=x^7+y^7$ true, it would be linear. But as we know, it isn't true.
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
What is $f(v_3) + f(v_3)$, and what is $f(2v_3)$? If $f$ were linear, should they be equal? Can they be equal?
answered Jan 31 at 8:57
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
What is $f(v_3) + f(v_3)$, and what is $f(2v_3)$? If $f$ were linear, should they be equal? Can they be equal?
answered Jan 31 at 8:57
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
What is $f(v_3) + f(v_3)$, and what is $f(2v_3)$? If $f$ were linear, should they be equal? Can they be equal?
answered Jan 31 at 8:57
ArthurArthur
122k7122211
$endgroup$
What is $f(v_3) + f(v_3)$, and what is $f(2v_3)$? If $f$ were linear, should they be equal? Can they be equal?
answered Jan 31 at 8:57
ArthurArthur
122k7122211
answered Jan 31 at 8:57
ArthurArthur
122k7122211
answered Jan 31 at 8:57
ArthurArthur
122k7122211
answered Jan 31 at 8:57
ArthurArthur
122k7122211
122k7122211
add a comment |
add a comment |
$begingroup$
this just means that you have (for B) three linearly independent vectors that span your vector space V. It is a more general way of writing vecor spaces, you don't always have a standard basis (i.e. $mathbb{R}^{2}=span{((1,0), (0,1))}$).
For your example you can simply handle this the exact way you would if you had been given a standard basis, i.e.:
$f$ is not linear:
Let $vin V, kin K$, then
$$f(k_{1}v_{1}+l_{1}v_1,k_{2}v_2+l_2v_2, k_{3}v_3+l_3v_3) = 3((k_1+l_1+k_2+l_2) w_1 + (k_3+l_3)^7w_2
$$
And continue on from there.
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
$endgroup$
add a comment |
$begingroup$
this just means that you have (for B) three linearly independent vectors that span your vector space V. It is a more general way of writing vecor spaces, you don't always have a standard basis (i.e. $mathbb{R}^{2}=span{((1,0), (0,1))}$).
For your example you can simply handle this the exact way you would if you had been given a standard basis, i.e.:
$f$ is not linear:
Let $vin V, kin K$, then
$$f(k_{1}v_{1}+l_{1}v_1,k_{2}v_2+l_2v_2, k_{3}v_3+l_3v_3) = 3((k_1+l_1+k_2+l_2) w_1 + (k_3+l_3)^7w_2
$$
And continue on from there.
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
$endgroup$
add a comment |
$begingroup$
this just means that you have (for B) three linearly independent vectors that span your vector space V. It is a more general way of writing vecor spaces, you don't always have a standard basis (i.e. $mathbb{R}^{2}=span{((1,0), (0,1))}$).
For your example you can simply handle this the exact way you would if you had been given a standard basis, i.e.:
$f$ is not linear:
Let $vin V, kin K$, then
$$f(k_{1}v_{1}+l_{1}v_1,k_{2}v_2+l_2v_2, k_{3}v_3+l_3v_3) = 3((k_1+l_1+k_2+l_2) w_1 + (k_3+l_3)^7w_2
$$
And continue on from there.
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
$endgroup$
this just means that you have (for B) three linearly independent vectors that span your vector space V. It is a more general way of writing vecor spaces, you don't always have a standard basis (i.e. $mathbb{R}^{2}=span{((1,0), (0,1))}$).
For your example you can simply handle this the exact way you would if you had been given a standard basis, i.e.:
$f$ is not linear:
Let $vin V, kin K$, then
$$f(k_{1}v_{1}+l_{1}v_1,k_{2}v_2+l_2v_2, k_{3}v_3+l_3v_3) = 3((k_1+l_1+k_2+l_2) w_1 + (k_3+l_3)^7w_2
$$
And continue on from there.
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
answered Jan 31 at 9:09
IamalgebraicIamalgebraic
364
364
add a comment |
add a comment |
$begingroup$
You're right about the $k_3^7$. What's at work there is the freshman's binomial. That is to say, were $(x+y)^7=x^7+y^7$ true, it would be linear. But as we know, it isn't true.
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
You're right about the $k_3^7$. What's at work there is the freshman's binomial. That is to say, were $(x+y)^7=x^7+y^7$ true, it would be linear. But as we know, it isn't true.
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
You're right about the $k_3^7$. What's at work there is the freshman's binomial. That is to say, were $(x+y)^7=x^7+y^7$ true, it would be linear. But as we know, it isn't true.
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
$endgroup$
You're right about the $k_3^7$. What's at work there is the freshman's binomial. That is to say, were $(x+y)^7=x^7+y^7$ true, it would be linear. But as we know, it isn't true.
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 9:20
Chris CusterChris Custer
14.3k3827
14.3k3827
add a comment |
add a comment |
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$begingroup$
In my opinion it ought to be either $f(k_1, k_2, k_3)$, or $f(k_1w_1 + k_2w_2 + k_3w_3)$. The mixture you have there looks strange.
$endgroup$
– Arthur
Jan 31 at 8:58