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Show that the transformation T defined by T(x1, x2)= (x1x2, x1, x2) is not linear. If T is a linear transformation, then T(0)= ___
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Show that the transformation T defined by T(x1, x2)= (x1x2, x1, x2) is not linear. If T is a linear transformation, then T(0)= ____________
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2021-08-24T02:50:28+00:00
2021-08-24T02:50:28+00:00 1 Answers
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Answer:
A linear transformation is defined as:
For a transformation T that goes from R^n to R^m, this transformation is linear if, for two vectors A and B, we have that:
T(A + B) = T(A) + T(B)
In this case, we have the transformation:
T(x1, x2)= (x1x2, x1, x2).
If we define two vectors:
A = (a1, a2)
B = (b1, b2)
if the transformation is lineal, we will have that:
T(A + B) = T(A) + T(B)
Or
T(0) = T(A – A) = T(A) + T(-A)
(just two different ways of writing the same thing, I will use the first one, because it is the general way)
We want to see that, for our transformation, this equation is false.
first the left side:
A + B = (a1, a2) + (b1, b2) = (a1 + b1, a2 + b2)
Then the transformation applied to that vector gives:
T(A + B) = T(a1 + b1, a2 + b2) = ( (a1 + b1)*(a2 + b2), (a1 + b1), (a2 + b2))
= (a1*a2 + a1*b2 + b1*a2 + b1*b2, a1 + b1, a2 + b2)
While for the right side, we have:
T(A) + T(B) = T(a1, a2) + T(b1, b2) = (a1*a2, a1, a2) + (b1*b2, b1, b2)
= (a1*a2 + b1*b2, a1 + b1, a2 + b2)
Then we can rewrite:
T(A + B) = T(A) + T(B)
as:
(a1*a2 + a1*b2 + b1*a2 + b1*b2, a1 + b1, a2 + b2) = (a1*a2 + b1*b2, a1 + b1, a2 + b2)
We can see that the first part of these vectors is different, thus, the equality is false.
Then we can conclude that:
T(A + B) ≠ T(A) + T(B)
Then the transformation T is not linear.