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​)=

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.

its electric no smoke

step-by-step explanation: