Show a linear combination of linear transformations is still a linear transformation
$begingroup$
I've been posed the following question:
I am unsure where to begin showing that $2F - G$ is still a linear transformation. Should I show that $2F-G$ conforms to the 3 properties of linear transformations?
Any suggestions/hints would be appreciated
linear-algebra
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
$endgroup$
add a comment |
$begingroup$
I've been posed the following question:
I am unsure where to begin showing that $2F - G$ is still a linear transformation. Should I show that $2F-G$ conforms to the 3 properties of linear transformations?
Any suggestions/hints would be appreciated
linear-algebra
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
$endgroup$
2
$begingroup$
Yeah. Just verify that $2F-G$ still satisfies the properties of linear transformations, using the fact that $F$ and $G$ individually satisfy the properties.
$endgroup$
– twnly
Dec 22 '18 at 2:02
add a comment |
$begingroup$
I've been posed the following question:
I am unsure where to begin showing that $2F - G$ is still a linear transformation. Should I show that $2F-G$ conforms to the 3 properties of linear transformations?
Any suggestions/hints would be appreciated
linear-algebra
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
$endgroup$
I've been posed the following question:
I am unsure where to begin showing that $2F - G$ is still a linear transformation. Should I show that $2F-G$ conforms to the 3 properties of linear transformations?
Any suggestions/hints would be appreciated
linear-algebra
linear-algebra
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
asked Dec 22 '18 at 2:01
lohboyslohboys
9818
9818
2
$begingroup$
Yeah. Just verify that $2F-G$ still satisfies the properties of linear transformations, using the fact that $F$ and $G$ individually satisfy the properties.
$endgroup$
– twnly
Dec 22 '18 at 2:02
add a comment |
2
$begingroup$
Yeah. Just verify that $2F-G$ still satisfies the properties of linear transformations, using the fact that $F$ and $G$ individually satisfy the properties.
$endgroup$
– twnly
Dec 22 '18 at 2:02
2
2
$begingroup$
Yeah. Just verify that $2F-G$ still satisfies the properties of linear transformations, using the fact that $F$ and $G$ individually satisfy the properties.
$endgroup$
– twnly
Dec 22 '18 at 2:02
$begingroup$
Yeah. Just verify that $2F-G$ still satisfies the properties of linear transformations, using the fact that $F$ and $G$ individually satisfy the properties.
$endgroup$
– twnly
Dec 22 '18 at 2:02
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You basically answered your own question; if you want to show that something is a linear transformation, just use the definition!
What will save the day (of course) is the hypothesis that both $F$ and $G$ are already linear transformations. You said three properties, but technically speaking you only need two, as respecting scalar multiplication will give you that $0$ is mapped to $0$.
So now just write it all out:
$$
H(u+v)=2F(u+v)-G(u+v) = 2F(u)+2F(v)-G(u)-G(v) = cdots
$$
$$
H(cv)=2F(cv)-G(cv)=2cF(v)-cG(v) = cdots
$$
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
$endgroup$
$begingroup$
thanks for the reply!
$endgroup$
– lohboys
Jan 4 at 0:41
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
You basically answered your own question; if you want to show that something is a linear transformation, just use the definition!
What will save the day (of course) is the hypothesis that both $F$ and $G$ are already linear transformations. You said three properties, but technically speaking you only need two, as respecting scalar multiplication will give you that $0$ is mapped to $0$.
So now just write it all out:
$$
H(u+v)=2F(u+v)-G(u+v) = 2F(u)+2F(v)-G(u)-G(v) = cdots
$$
$$
H(cv)=2F(cv)-G(cv)=2cF(v)-cG(v) = cdots
$$
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
$endgroup$
$begingroup$
thanks for the reply!
$endgroup$
– lohboys
Jan 4 at 0:41
add a comment |
$begingroup$
You basically answered your own question; if you want to show that something is a linear transformation, just use the definition!
What will save the day (of course) is the hypothesis that both $F$ and $G$ are already linear transformations. You said three properties, but technically speaking you only need two, as respecting scalar multiplication will give you that $0$ is mapped to $0$.
So now just write it all out:
$$
H(u+v)=2F(u+v)-G(u+v) = 2F(u)+2F(v)-G(u)-G(v) = cdots
$$
$$
H(cv)=2F(cv)-G(cv)=2cF(v)-cG(v) = cdots
$$
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
$endgroup$
$begingroup$
thanks for the reply!
$endgroup$
– lohboys
Jan 4 at 0:41
add a comment |
$begingroup$
You basically answered your own question; if you want to show that something is a linear transformation, just use the definition!
What will save the day (of course) is the hypothesis that both $F$ and $G$ are already linear transformations. You said three properties, but technically speaking you only need two, as respecting scalar multiplication will give you that $0$ is mapped to $0$.
So now just write it all out:
$$
H(u+v)=2F(u+v)-G(u+v) = 2F(u)+2F(v)-G(u)-G(v) = cdots
$$
$$
H(cv)=2F(cv)-G(cv)=2cF(v)-cG(v) = cdots
$$
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
$endgroup$
You basically answered your own question; if you want to show that something is a linear transformation, just use the definition!
What will save the day (of course) is the hypothesis that both $F$ and $G$ are already linear transformations. You said three properties, but technically speaking you only need two, as respecting scalar multiplication will give you that $0$ is mapped to $0$.
So now just write it all out:
$$
H(u+v)=2F(u+v)-G(u+v) = 2F(u)+2F(v)-G(u)-G(v) = cdots
$$
$$
H(cv)=2F(cv)-G(cv)=2cF(v)-cG(v) = cdots
$$
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
answered Dec 22 '18 at 2:06
DanielOfJackDanielOfJack
1763
1763
$begingroup$
thanks for the reply!
$endgroup$
– lohboys
Jan 4 at 0:41
add a comment |
$begingroup$
thanks for the reply!
$endgroup$
– lohboys
Jan 4 at 0:41
$begingroup$
thanks for the reply!
$endgroup$
– lohboys
Jan 4 at 0:41
$begingroup$
thanks for the reply!
$endgroup$
– lohboys
Jan 4 at 0:41
add a comment |
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$begingroup$
Yeah. Just verify that $2F-G$ still satisfies the properties of linear transformations, using the fact that $F$ and $G$ individually satisfy the properties.
$endgroup$
– twnly
Dec 22 '18 at 2:02