Define the linear transformation Tby T(x) Ax. Find ker(n, nullityIT rangem, and
ID: 3037314 • Letter: D
Question
Define the linear transformation Tby T(x) Ax. Find ker(n, nullityIT rangem, and rank(T. (a) ker(T) STEP 1: The kernel of T is given by the solution to the equation Tx) o, Let x (r1, x2, x3) and find x such that T(x) 0. (If there are an infinite number of solutions use t and s as your parameters.) STEP 2 use your result from Step 1 to find the kernel of T ar there are an infinite number of solutions use t and e as your parameters (b) nullity(T) STEP 3: Use the fact that nullityIT) dim kertT to compute nullExplanation / Answer
(a) Since T(x) = Ax, hence Ker(T) = null(A), Nullity(T) = nullity(A) and rank(T) = rank(A). Also, range(T) = Col(A).
Since Null(A) is the set of solutions to the equation AX = 0, in order to solve this equation, we will reduce A to its RREF as under:
Multiply the 1st row by ¾
Add 4/3 times the 1st row to the 2nd row
Add -2/3 times the 1st row to the 3rd row
Then the RREF of A is
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If X = (x,y,z)T, then the equation AX = 0 is equivalent to x-y +z/2 = 0 so that z= -2x+2y. Then X = (x,y,-2x+2y)T = x(1,0,-2)T+y(0,1,2)T. Then Ker (T) = Null(A) = span{ (1,0,-2)T, (0,1,2)T }.
(b) Nullity (T) = Dimension of Ker(T) = 2.
(c) The number of columns in A is 3. Then, by the Dimension theorem (Rank-Nullty theorem), rank(T) +nullity(T) = 3. Since nullity(T) =2, hence rank(T) = 3-2=1.
(d) Range (T) = Col(A). Further, in view of the RREF of A, Col(A) = span{ (4/3,-4/3,2/3)T}. Hence, Range (T) = span{ (4/3,-4/3,2/3)T}.
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