Hey guys I am struggling in Liner Algebra I have the followi

Hey guys I am struggling in Liner Algebra.

I have the following question

define h: R³ -> R³ by h (a b c ) = (a+c 0 b-a) // the vectors are 3-tall vectors

what is the basis of the range of h and the rank ?

what is the basis of the nullspace of h and the nullity ?

any explnation would be appreciated

Thanks a lot

Solution

If T is a linear transformation from an m-dimension vector space X to an n-dimensional vector space Y, and x₁, x₂, x₃, ..., x_m , a basis for X and y₁, y₂, y₃, ..., y_n be a basis for Y. The image of X, T(X), is called the range of T. The kernel (also known as null space ) of a linear map L : R³ R³, is the set of all elements ( a, b, c) of R³ for which L(v) = 0, where 0 denotes the zero vector in R³ .

The range of h is the set { h(e₁ ), h(e₂ ), h( e₃ )}. Now h(e₁ ) = h ( 1, 0, 0)^T = ( 1, 0, -1)^T , h (e₂ ) = h ( 0, 1, 0)^T = ( 0, 0, 1)^T , h(e₃ ) = h ( 0, 0, 1)^T = ( 1,0 0,)^T . Now, these 3 vectors are not linearly independent as ( 1, 0, -1)^T = ( 1,0 0,)^T - ( 0, 0, 1)^T . Therefore, a basis for the image of h is the set { ( 1,0 0,)^T , ( 0, 0, 1)^T }. The rank oh h is the number of linearly independent columns in the basis of the range, which is 2,

The null space of h is the set of all vectors ( a, b, c)^T such that h (a, b, c)^T = 0.Now, h ( a, b, c)^T = a + c , 0, b –a)^T so that h (a, b, c)^T = 0 means a + c = 0 and b – a = 0 which implies that a = b = - c. If c = t, then ( a , b, c)^T = t ( 1 , 1, -1)^T . Thus, the null space of h = span { ( 1, 1, -1)^T }. The basis of the null space of h is the set { ( 1, 1, -1)^T } . The nullity of h is the dimension of null space of h which is1.