Find either the nullify or the rank of T and then use the rank thereom to find t
ID: 3116691 • Letter: F
Question
Find either the nullify or the rank of T and then use the rank thereom to find the other. a) T: P2 > R2 defines by T(p(x)) = [p(0) p(9)]b) T: M22 > M22 defined by T(A)= AB, where B = [5 -3 -5 3]
Find either the nullify or the rank of T and then use the rank thereom to find the other. a) T: P2 > R2 defines by T(p(x)) = [p(0) p(9)]
b) T: M22 > M22 defined by T(A)= AB, where B = [5 -3 -5 3]
a) T: P2 > R2 defines by T(p(x)) = [p(0) p(9)]
b) T: M22 > M22 defined by T(A)= AB, where B = [5 -3 -5 3]
Explanation / Answer
(a)T: P2 R2 is defined by T(p(x)) = (p(0), p(9))T. We know that (x2,x,1) is the standard basis of P2. Also, T(x2) = (0,81)T, T(x) = (0,9)T and T(1) = (1,1)T. Hence the standard matrix of T is A =
0
0
1
81
9
1
The RREF of A is
1
1/9
0
0
0
1
Therefore, the rank of A and hence the rank of T is 2. Then, as per the rank-nullity theorem, the nullity of T = the number of columns of A – the rank of A = 3-2 = 1.
(b).
b) T: M22 M22 is defined by T(A)= AB, where
B =
5
-3
-5
3
Now, let A =
a
b
c
d
where a,b,c, d are arbitrary real numbers. Then T(A) = AB =
5a-5b
-3a+3b
5c-5d
-3c+3d
We know that {E11,E12,E21,E22} is the standard basis of M22, where Eij has 1 as the ijth entry and the rest of the entries are 0. Then T(E11) = E11B=
5
-3
0
0
= 5E11-3E12.
Similarly, T(E12) = E12B = -5E11+3E12, T(E21) = E21B = 5E21-3E22 and T(E22) = E22B = -5E21+3E22 .
Now, let M =
5
-5
0
0
-3
3
0
0
0
0
5
-5
0
0
-3
3
It may be observed that the entries in the columns of M are the coefficients of E11,E12,E21,E22 in T(E11) , T(E12) , T(E21) and T(E22) respectively. The RREF of M is
1
-1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
Thus, the rank of M, and therefore, the rank of T is 2. Then, as per the rank-nullity theorem, the nullity of T = the number of columns of M – the rank of M = 4-2 =2.
0
0
1
81
9
1
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