Linear transformation over C and R respectively
$begingroup$
I am new to linear algebra, and was asked to determine if the following is a linear transformation.
$T(z_1,z_2)=z_1+overline z_2, T_3> C^2 to C$
a) when we look at $C, C^2$ as linear spaces over C.
b) when we look at $C, C^2$ as linear spaces over R.
I have an answer, but I fear it is lacking in many ways:
a)
I tried to give a counterexample:
$z_1=1+1i$
$z_2=2+2i$
$λ$ is a scalar 3+3i $in C$
Therefore, $λz_1=6i$, $λz_2=12i$
So:
$T(λz_1,λz_2)=(6i,12i)=6i-12i=-6i$
$λT(1+1i,2+2i)=λ(1+1i+2-2i)=(3+3i)(3-i)=12+6i$
$-6i neq 12+6i$
So the answer is no...
b)
using scalars α,β $in R$
$T(α(z_1+z'_1),β(z_2+z'_2))=α(z_1+z'_1)+ β(overline z_2+overline z'_2)$
which is not equal to $αT(z_1+z'_1)+βT(z_2+z'_2)$, so the answer is no.
Could anyone help out?
Thank you!
linear-algebra linear-transformations
asked Jan 1 at 20:21
daltadalta
1508
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, and was asked to determine if the following is a linear transformation.
$T(z_1,z_2)=z_1+overline z_2, T_3> C^2 to C$
a) when we look at $C, C^2$ as linear spaces over C.
b) when we look at $C, C^2$ as linear spaces over R.
I have an answer, but I fear it is lacking in many ways:
a)
I tried to give a counterexample:
$z_1=1+1i$
$z_2=2+2i$
$λ$ is a scalar 3+3i $in C$
Therefore, $λz_1=6i$, $λz_2=12i$
So:
$T(λz_1,λz_2)=(6i,12i)=6i-12i=-6i$
$λT(1+1i,2+2i)=λ(1+1i+2-2i)=(3+3i)(3-i)=12+6i$
$-6i neq 12+6i$
So the answer is no...
b)
using scalars α,β $in R$
$T(α(z_1+z'_1),β(z_2+z'_2))=α(z_1+z'_1)+ β(overline z_2+overline z'_2)$
which is not equal to $αT(z_1+z'_1)+βT(z_2+z'_2)$, so the answer is no.
Could anyone help out?
Thank you!
linear-algebra linear-transformations
asked Jan 1 at 20:21
daltadalta
1508
$endgroup$
1
$begingroup$
I notice that you have not marked any of your previous questions as answered. Here is some information from MSE on what to do when someone answers your questions.
$endgroup$
– Nyfiken
Jan 3 at 12:16
1
$begingroup$
@Nyfiken You are right, I had no idea I was supposed to, as I am new to this sight… I will do it now...
$endgroup$
– dalta
Jan 3 at 12:19
add a comment |
$begingroup$
I am new to linear algebra, and was asked to determine if the following is a linear transformation.
$T(z_1,z_2)=z_1+overline z_2, T_3> C^2 to C$
a) when we look at $C, C^2$ as linear spaces over C.
b) when we look at $C, C^2$ as linear spaces over R.
I have an answer, but I fear it is lacking in many ways:
a)
I tried to give a counterexample:
$z_1=1+1i$
$z_2=2+2i$
$λ$ is a scalar 3+3i $in C$
Therefore, $λz_1=6i$, $λz_2=12i$
So:
$T(λz_1,λz_2)=(6i,12i)=6i-12i=-6i$
$λT(1+1i,2+2i)=λ(1+1i+2-2i)=(3+3i)(3-i)=12+6i$
$-6i neq 12+6i$
So the answer is no...
b)
using scalars α,β $in R$
$T(α(z_1+z'_1),β(z_2+z'_2))=α(z_1+z'_1)+ β(overline z_2+overline z'_2)$
which is not equal to $αT(z_1+z'_1)+βT(z_2+z'_2)$, so the answer is no.
Could anyone help out?
Thank you!
linear-algebra linear-transformations
asked Jan 1 at 20:21
daltadalta
1508
$endgroup$
I am new to linear algebra, and was asked to determine if the following is a linear transformation.
$T(z_1,z_2)=z_1+overline z_2, T_3> C^2 to C$
a) when we look at $C, C^2$ as linear spaces over C.
b) when we look at $C, C^2$ as linear spaces over R.
I have an answer, but I fear it is lacking in many ways:
a)
I tried to give a counterexample:
$z_1=1+1i$
$z_2=2+2i$
$λ$ is a scalar 3+3i $in C$
Therefore, $λz_1=6i$, $λz_2=12i$
So:
$T(λz_1,λz_2)=(6i,12i)=6i-12i=-6i$
$λT(1+1i,2+2i)=λ(1+1i+2-2i)=(3+3i)(3-i)=12+6i$
$-6i neq 12+6i$
So the answer is no...
b)
using scalars α,β $in R$
$T(α(z_1+z'_1),β(z_2+z'_2))=α(z_1+z'_1)+ β(overline z_2+overline z'_2)$
which is not equal to $αT(z_1+z'_1)+βT(z_2+z'_2)$, so the answer is no.
Could anyone help out?
Thank you!
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Jan 1 at 20:21
daltadalta
1508
asked Jan 1 at 20:21
daltadalta
1508
asked Jan 1 at 20:21
daltadalta
1508
asked Jan 1 at 20:21
daltadalta
1508
asked Jan 1 at 20:21
daltadalta
1508
1508
1
$begingroup$
I notice that you have not marked any of your previous questions as answered. Here is some information from MSE on what to do when someone answers your questions.
$endgroup$
– Nyfiken
Jan 3 at 12:16
1
$begingroup$
@Nyfiken You are right, I had no idea I was supposed to, as I am new to this sight… I will do it now...
$endgroup$
– dalta
Jan 3 at 12:19
add a comment |
1
$begingroup$
I notice that you have not marked any of your previous questions as answered. Here is some information from MSE on what to do when someone answers your questions.
$endgroup$
– Nyfiken
Jan 3 at 12:16
1
$begingroup$
@Nyfiken You are right, I had no idea I was supposed to, as I am new to this sight… I will do it now...
$endgroup$
– dalta
Jan 3 at 12:19
1
1
$begingroup$
I notice that you have not marked any of your previous questions as answered. Here is some information from MSE on what to do when someone answers your questions.
$endgroup$
– Nyfiken
Jan 3 at 12:16
$begingroup$
I notice that you have not marked any of your previous questions as answered. Here is some information from MSE on what to do when someone answers your questions.
$endgroup$
– Nyfiken
Jan 3 at 12:16
1
1
$begingroup$
@Nyfiken You are right, I had no idea I was supposed to, as I am new to this sight… I will do it now...
$endgroup$
– dalta
Jan 3 at 12:19
$begingroup$
@Nyfiken You are right, I had no idea I was supposed to, as I am new to this sight… I will do it now...
$endgroup$
– dalta
Jan 3 at 12:19
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
(a) Looks good.
(b) This is incorrect. $T$ is linear in this case. Remember, the criteria for linearity is that $aT(z_1,z_2)=T(a(z_1,z_2))$ and $T((z_1,z_2)+(z'_1,z'_2))=T(z_1,z_2)+T(z'_1,z'_2)$.
answered Jan 1 at 20:33
Ben WBen W
2,321615
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1 Answer
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1 Answer
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$begingroup$
(a) Looks good.
(b) This is incorrect. $T$ is linear in this case. Remember, the criteria for linearity is that $aT(z_1,z_2)=T(a(z_1,z_2))$ and $T((z_1,z_2)+(z'_1,z'_2))=T(z_1,z_2)+T(z'_1,z'_2)$.
answered Jan 1 at 20:33
Ben WBen W
2,321615
$endgroup$
add a comment |
$begingroup$
(a) Looks good.
(b) This is incorrect. $T$ is linear in this case. Remember, the criteria for linearity is that $aT(z_1,z_2)=T(a(z_1,z_2))$ and $T((z_1,z_2)+(z'_1,z'_2))=T(z_1,z_2)+T(z'_1,z'_2)$.
answered Jan 1 at 20:33
Ben WBen W
2,321615
$endgroup$
add a comment |
$begingroup$
(a) Looks good.
(b) This is incorrect. $T$ is linear in this case. Remember, the criteria for linearity is that $aT(z_1,z_2)=T(a(z_1,z_2))$ and $T((z_1,z_2)+(z'_1,z'_2))=T(z_1,z_2)+T(z'_1,z'_2)$.
answered Jan 1 at 20:33
Ben WBen W
2,321615
$endgroup$
(a) Looks good.
(b) This is incorrect. $T$ is linear in this case. Remember, the criteria for linearity is that $aT(z_1,z_2)=T(a(z_1,z_2))$ and $T((z_1,z_2)+(z'_1,z'_2))=T(z_1,z_2)+T(z'_1,z'_2)$.
answered Jan 1 at 20:33
Ben WBen W
2,321615
answered Jan 1 at 20:33
Ben WBen W
2,321615
answered Jan 1 at 20:33
Ben WBen W
2,321615
answered Jan 1 at 20:33
Ben WBen W
2,321615
2,321615
add a comment |
add a comment |
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$begingroup$
I notice that you have not marked any of your previous questions as answered. Here is some information from MSE on what to do when someone answers your questions.
$endgroup$
– Nyfiken
Jan 3 at 12:16
1
$begingroup$
@Nyfiken You are right, I had no idea I was supposed to, as I am new to this sight… I will do it now...
$endgroup$
– dalta
Jan 3 at 12:19