Let T:R2- R2be the transformation T(x1,x2) = (x1,0). The null space (or kernel) N(T)of Tisa)(x1,0) : x1is realb)1c)(0,1)d)(0,x2) : x2is realCorrect answer is option 'D'. Can you explain this answer?

Question

Let T:R2-> R2be the transformation T(x1,x2) = (x1,0). The null space (or kernel) N(T)of Tisa)(x1,0) : x1is realb)1c)(0,1)d)(0,x2) : x2is realCorrect answer is option 'D'. Can you explain this answer?

Answer

R2 be a linear transformation. Let's denote the standard basis vectors in R2 as e1 = (1, 0) and e2 = (0, 1). Then any vector x in R2 can be written as x = x1e1 + x2e2, where x1 and x2 are scalars.

Since T is a linear transformation, we have:

T(x) = T(x1e1 + x2e2) = x1T(e1) + x2T(e2)

Let's denote the images of the standard basis vectors under T as T(e1) = (a, b) and T(e2) = (c, d), where a, b, c, and d are scalars.

Then the transformation T can be represented by the matrix:

[T] = | a c |
| b d |

This matrix is called the standard matrix of the linear transformation T. Each column of the matrix represents the image of the corresponding basis vector under T.

Note that the standard matrix of T depends on the choice of basis vectors in R2. If we choose a different set of basis vectors, the standard matrix representation of T will change accordingly.

Answer

B/z for every  x belong  kernel T iff T (x) =0 now putting  (x1,0) we ge x1=0while x2 can not be defined  so d is correct

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