OpenStudy (anonymous):

Let T : R^5> R^4 be the linear transformation given by the rule T ((v; w; x; y; z)) = (v- w; y -z; x - 3y; -x + 3z): (i) Find the matrix A that represents T relative to the ordered bases Ba = ((1; 1; 1; 1; 1); (1; 1; 1; 1; 0); (1; 1; 1; 0; 0); (1; 1; 0; 0; 0); (1; 0; 0; 0; 0)) and Bb= ((1; 0; 0; 0); (0; 1; 0; 0); (0; 0; 1; 1); (0; 0; 1; 2)) of R^5>R^4 respectively. (ii) Determine the rank of T: (iii) Determine the kernel, ker(T ), and the image, im(T ). (iv) Find a basis of ker(T ) and determine the nullity of T . (v) Find a basis of the image, im(T )

7 years ago
myininaya (myininaya):

go joe!

7 years ago
OpenStudy (anonymous):

Thats a lot of work <.< for i), plug each basis vector into the transformation. Then write your answer as a linear combination of the basis vectors. T(a1) will be the first column, T(a2) will be the second column, etc. You would need to do this for the a basis and the b basis.

7 years ago
OpenStudy (anonymous):

for ii) pick one of the matrices you got for i), and row reduce to reduce echelon form. count pivots. Thats the rank.

7 years ago
OpenStudy (anonymous):

The Ker(T) is the set of all vectors such that T(x) = 0. looking at the row reduced matrix from ii), solve for all the pivot variables in terms of the free variables, and find the basis for the Ker(T) (or Null Space). The Im(T) (or Column Space) is the span of the columns of the matrix. Grab the columns from the original matrix that correspond to the pivot columns in the row reduced echelon matrix.

7 years ago
OpenStudy (anonymous):

wow okay i'll try this and post what i get incase you're interested! check tomorrow morning! thank u

7 years ago
OpenStudy (anonymous):

for i) i wasnt sure how to find Bb?

7 years ago
OpenStudy (anonymous):

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7 years ago