Is the composition of two linear functions is a linear function?
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Suppose I have $2$ linear functions $f(x)$ and $g(x)$ would the $g(f(x))$ be a linear function as well?
An example I wrote out was: $f(x) = 3x_1+4x_2$ and $g(x) = 7x_1+8x_2$
Is $g(f(x)) = 7(3x_1+4x_2)+8(3x_1+4x_2)= 21x_1+28x_2+24x_1+32x_2= 45x_1+60x_2$? If I'm wrong, can you please explain the reasoning as to why?
linear-algebra
edited Sep 6 at 2:27
Anton Grudkin
2,211719
asked Sep 6 at 2:14
bigpimpin206
61
add a comment |Â
up vote
1
down vote
favorite
Suppose I have $2$ linear functions $f(x)$ and $g(x)$ would the $g(f(x))$ be a linear function as well?
An example I wrote out was: $f(x) = 3x_1+4x_2$ and $g(x) = 7x_1+8x_2$
Is $g(f(x)) = 7(3x_1+4x_2)+8(3x_1+4x_2)= 21x_1+28x_2+24x_1+32x_2= 45x_1+60x_2$? If I'm wrong, can you please explain the reasoning as to why?
linear-algebra
edited Sep 6 at 2:27
Anton Grudkin
2,211719
asked Sep 6 at 2:14
bigpimpin206
61
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose I have $2$ linear functions $f(x)$ and $g(x)$ would the $g(f(x))$ be a linear function as well?
An example I wrote out was: $f(x) = 3x_1+4x_2$ and $g(x) = 7x_1+8x_2$
Is $g(f(x)) = 7(3x_1+4x_2)+8(3x_1+4x_2)= 21x_1+28x_2+24x_1+32x_2= 45x_1+60x_2$? If I'm wrong, can you please explain the reasoning as to why?
linear-algebra
edited Sep 6 at 2:27
Anton Grudkin
2,211719
asked Sep 6 at 2:14
bigpimpin206
61
Suppose I have $2$ linear functions $f(x)$ and $g(x)$ would the $g(f(x))$ be a linear function as well?
An example I wrote out was: $f(x) = 3x_1+4x_2$ and $g(x) = 7x_1+8x_2$
Is $g(f(x)) = 7(3x_1+4x_2)+8(3x_1+4x_2)= 21x_1+28x_2+24x_1+32x_2= 45x_1+60x_2$? If I'm wrong, can you please explain the reasoning as to why?
linear-algebra
linear-algebra
edited Sep 6 at 2:27
Anton Grudkin
2,211719
asked Sep 6 at 2:14
bigpimpin206
61
edited Sep 6 at 2:27
Anton Grudkin
2,211719
asked Sep 6 at 2:14
bigpimpin206
61
edited Sep 6 at 2:27
Anton Grudkin
2,211719
edited Sep 6 at 2:27
Anton Grudkin
2,211719
edited Sep 6 at 2:27
Anton Grudkin
2,211719
2,211719
asked Sep 6 at 2:14
bigpimpin206
61
asked Sep 6 at 2:14
bigpimpin206
61
asked Sep 6 at 2:14
bigpimpin206
61
61
add a comment |Â
add a comment |Â
1 Answer
1
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Suppose $f:U to V$ and $g:W to U$ are two linear functions, where $U,V,W$ are vector spaces over a field $k$. Then
$$fcirc g(w_1 + w_2) = f(g(w_1 + w_2)) = f(g(w_1) + g(w_2)) = f(g(w_1)) + f(g(w_2)) = fcirc g(w_1) + fcirc g(w_2)$$ where $w_i in W$.
Similarly, for $lambda in k$, $$fcirc g(lambda w) = f(g(lambda w)) = f(lambda g(w)) = lambda f(g(w)) = lambda fcirc g(w)$$ for every $w in W$.
answered Sep 6 at 2:20
Benjamin Tighe
328314
Thank you for responding. In an attempt to explain your answer to myself, the way I might've done it is incorrect and that it should actually be 7(3x_1)+8(4x_2) right?
– bigpimpin206
Sep 6 at 2:26
I think the issue with your example is that $f$ and $g$ have the same domain and codomain. For example, if you are working over $mathbb R$, then $f,g:mathbb R^2 to mathbb R$. Thus it doesn't make sense to compose them.
– Benjamin Tighe
Sep 6 at 2:28
My bad, I forgot to include the domain. I know that f: R^n-> R and g: R->R.
– bigpimpin206
Sep 6 at 2:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Suppose $f:U to V$ and $g:W to U$ are two linear functions, where $U,V,W$ are vector spaces over a field $k$. Then
$$fcirc g(w_1 + w_2) = f(g(w_1 + w_2)) = f(g(w_1) + g(w_2)) = f(g(w_1)) + f(g(w_2)) = fcirc g(w_1) + fcirc g(w_2)$$ where $w_i in W$.
Similarly, for $lambda in k$, $$fcirc g(lambda w) = f(g(lambda w)) = f(lambda g(w)) = lambda f(g(w)) = lambda fcirc g(w)$$ for every $w in W$.
answered Sep 6 at 2:20
Benjamin Tighe
328314
Thank you for responding. In an attempt to explain your answer to myself, the way I might've done it is incorrect and that it should actually be 7(3x_1)+8(4x_2) right?
– bigpimpin206
Sep 6 at 2:26
I think the issue with your example is that $f$ and $g$ have the same domain and codomain. For example, if you are working over $mathbb R$, then $f,g:mathbb R^2 to mathbb R$. Thus it doesn't make sense to compose them.
– Benjamin Tighe
Sep 6 at 2:28
My bad, I forgot to include the domain. I know that f: R^n-> R and g: R->R.
– bigpimpin206
Sep 6 at 2:30
add a comment |Â
up vote
1
down vote
Suppose $f:U to V$ and $g:W to U$ are two linear functions, where $U,V,W$ are vector spaces over a field $k$. Then
$$fcirc g(w_1 + w_2) = f(g(w_1 + w_2)) = f(g(w_1) + g(w_2)) = f(g(w_1)) + f(g(w_2)) = fcirc g(w_1) + fcirc g(w_2)$$ where $w_i in W$.
Similarly, for $lambda in k$, $$fcirc g(lambda w) = f(g(lambda w)) = f(lambda g(w)) = lambda f(g(w)) = lambda fcirc g(w)$$ for every $w in W$.
answered Sep 6 at 2:20
Benjamin Tighe
328314
Thank you for responding. In an attempt to explain your answer to myself, the way I might've done it is incorrect and that it should actually be 7(3x_1)+8(4x_2) right?
– bigpimpin206
Sep 6 at 2:26
I think the issue with your example is that $f$ and $g$ have the same domain and codomain. For example, if you are working over $mathbb R$, then $f,g:mathbb R^2 to mathbb R$. Thus it doesn't make sense to compose them.
– Benjamin Tighe
Sep 6 at 2:28
My bad, I forgot to include the domain. I know that f: R^n-> R and g: R->R.
– bigpimpin206
Sep 6 at 2:30
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Suppose $f:U to V$ and $g:W to U$ are two linear functions, where $U,V,W$ are vector spaces over a field $k$. Then
$$fcirc g(w_1 + w_2) = f(g(w_1 + w_2)) = f(g(w_1) + g(w_2)) = f(g(w_1)) + f(g(w_2)) = fcirc g(w_1) + fcirc g(w_2)$$ where $w_i in W$.
Similarly, for $lambda in k$, $$fcirc g(lambda w) = f(g(lambda w)) = f(lambda g(w)) = lambda f(g(w)) = lambda fcirc g(w)$$ for every $w in W$.
answered Sep 6 at 2:20
Benjamin Tighe
328314
Suppose $f:U to V$ and $g:W to U$ are two linear functions, where $U,V,W$ are vector spaces over a field $k$. Then
$$fcirc g(w_1 + w_2) = f(g(w_1 + w_2)) = f(g(w_1) + g(w_2)) = f(g(w_1)) + f(g(w_2)) = fcirc g(w_1) + fcirc g(w_2)$$ where $w_i in W$.
Similarly, for $lambda in k$, $$fcirc g(lambda w) = f(g(lambda w)) = f(lambda g(w)) = lambda f(g(w)) = lambda fcirc g(w)$$ for every $w in W$.
answered Sep 6 at 2:20
Benjamin Tighe
328314
answered Sep 6 at 2:20
Benjamin Tighe
328314
answered Sep 6 at 2:20
Benjamin Tighe
328314
answered Sep 6 at 2:20
Benjamin Tighe
328314
328314
Thank you for responding. In an attempt to explain your answer to myself, the way I might've done it is incorrect and that it should actually be 7(3x_1)+8(4x_2) right?
– bigpimpin206
Sep 6 at 2:26
I think the issue with your example is that $f$ and $g$ have the same domain and codomain. For example, if you are working over $mathbb R$, then $f,g:mathbb R^2 to mathbb R$. Thus it doesn't make sense to compose them.
– Benjamin Tighe
Sep 6 at 2:28
My bad, I forgot to include the domain. I know that f: R^n-> R and g: R->R.
– bigpimpin206
Sep 6 at 2:30
add a comment |Â
Thank you for responding. In an attempt to explain your answer to myself, the way I might've done it is incorrect and that it should actually be 7(3x_1)+8(4x_2) right?
– bigpimpin206
Sep 6 at 2:26
I think the issue with your example is that $f$ and $g$ have the same domain and codomain. For example, if you are working over $mathbb R$, then $f,g:mathbb R^2 to mathbb R$. Thus it doesn't make sense to compose them.
– Benjamin Tighe
Sep 6 at 2:28
My bad, I forgot to include the domain. I know that f: R^n-> R and g: R->R.
– bigpimpin206
Sep 6 at 2:30
Thank you for responding. In an attempt to explain your answer to myself, the way I might've done it is incorrect and that it should actually be 7(3x_1)+8(4x_2) right?
– bigpimpin206
Sep 6 at 2:26
Thank you for responding. In an attempt to explain your answer to myself, the way I might've done it is incorrect and that it should actually be 7(3x_1)+8(4x_2) right?
– bigpimpin206
Sep 6 at 2:26
I think the issue with your example is that $f$ and $g$ have the same domain and codomain. For example, if you are working over $mathbb R$, then $f,g:mathbb R^2 to mathbb R$. Thus it doesn't make sense to compose them.
– Benjamin Tighe
Sep 6 at 2:28
I think the issue with your example is that $f$ and $g$ have the same domain and codomain. For example, if you are working over $mathbb R$, then $f,g:mathbb R^2 to mathbb R$. Thus it doesn't make sense to compose them.
– Benjamin Tighe
Sep 6 at 2:28
My bad, I forgot to include the domain. I know that f: R^n-> R and g: R->R.
– bigpimpin206
Sep 6 at 2:30
My bad, I forgot to include the domain. I know that f: R^n-> R and g: R->R.
– bigpimpin206
Sep 6 at 2:30
add a comment |Â
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