- #1
succubus
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The problem:
T(x + yi) = x
C -> C (Complex Numbers)
Show that the above is:
Linear
Isomorphic
This is what i have for showing it's linear:
T(x+yi + a + bi) = x + a + i(y + b) => T(x+a) => T(x) + T(a)
T(k(x+yi)) = k(x) + k(yi) = T(kx) = kT(x)
I assume that i(y+b) = 0. Is this the proper assumption?
Therefore it is not isomorphic because of the fact that i(y+b) = 0?
The answer in the back is states it's not isomorphic because 5i = 0 exists.
So I'm sort of confused as to what they are trying to say.
Aare they saying that by choosing a point on the plane, (x,y) that the component of 5i = 0 (I'm assuming 5 is random) exists for this particular transformation, and since that's impossible the transformation is not isomorphic? Or am I totally on the wrong track.
T(x + yi) = x
C -> C (Complex Numbers)
Show that the above is:
Linear
Isomorphic
This is what i have for showing it's linear:
T(x+yi + a + bi) = x + a + i(y + b) => T(x+a) => T(x) + T(a)
T(k(x+yi)) = k(x) + k(yi) = T(kx) = kT(x)
I assume that i(y+b) = 0. Is this the proper assumption?
Therefore it is not isomorphic because of the fact that i(y+b) = 0?
The answer in the back is states it's not isomorphic because 5i = 0 exists.
So I'm sort of confused as to what they are trying to say.
Aare they saying that by choosing a point on the plane, (x,y) that the component of 5i = 0 (I'm assuming 5 is random) exists for this particular transformation, and since that's impossible the transformation is not isomorphic? Or am I totally on the wrong track.