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Math 4305, Summer 2011,
Midterm II, Practice solutions
Problem 1 Let v = (1, 0 , 1)t. Define the linear transformation T : R^3 → R^3
by T (x) = v × x. Where
a 1 a 2 a 3
×
b 1 b 2 b 3
=
a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1
.
a) Find the standard matrix A of T.
Solution A = [a 1 , a 2 , a 3 ], where ai = T (ei).
T (e 1 ) =
, T^ (e 2 ) =
, T^ (e 3 ) =
.
We thus have
A =
.
b) Find a basis of im(A).
Solution: We do the interchange of r 1 and r 2 , then r 3 + r 2 , we thus reach the REF of A:
.
Therefore, we know that a basis of im(A) is {a 1 , a 2 }.
c) What’s the dimension of ker(A)?
Solution By the Rank Theorem, we know that
dim ker(A) = 3 − dim im(A) = 1.
Problem 2 Consider an m × n matrix A and an n × m matrix B (with n 6 = m) such that AB = Im. Are the columns of B linearly independent? What about columns of A?
Solution: If columns of B are linearly dependent, so are columns of AB. Thus, if AB = Im, then columns of B are linearly independent. Furthermore, we know that n > m (otherwise, columns of B are linearly dependent). Since A is m × n, columns of A are linearly dependent.
Problem 3 Find a basis B of R^2 such that
[ 1 2
]
B
[ 3 5
] ,
[ 3 4
]
B
[ 2 3
] .
Solution: Let the vectors in B be v 1 and v 2 , and the matrix S = [v 1 , v 2 ]. For any x ∈ R^2 , it is clear that S[x]B = x. Therefore, one reads from the problem that
SA = C
where
A =
( 3 2 5 3
) , C =
( 1 3 2 4
) .
Therefore
Proof Since dimP 2 = 3 and B has 3 vectors, it is sufficient to show that B is linearly independet. Fix the standard basis S = { 1 , t, t^2 }, we check the corresponding coordinate vectors are linearly independent. Let p 1 (t) = 1+t^2 , p 2 (t) = t + t^2 , and p 3 (t) = 1 + 2t + t^2 , we have
A = [[p 1 ]S , [p 2 ]S , [p 3 ]S ] =
.
We easily verify that A is invertible and thus proves that B is linearly independent and so is a basis for P 2.
b) Find the matrix of the linear transformation T (f (t)) = f ′^ − 3 f from P 2 to P 2 withe respect to the basis B found in part (a).
Solution: The desired matrix M can be obtained by
AM = [[T (p 1 )]S , [T (p 2 )]S ], [T (p 3 )]S ] = D. Since T (p 1 ) = −3+2t− 3 t^2 , T (p 2 ) = 1−t− 3 t^2 and T (p 3 ) = − 1 − 4 t− 3 t^2 , thus
M = A−^1 D = A−^1
=
− 4 −^12
.
