Let A, B, C be square matrices. a.) (T/F) A is not invertible iff det(A) = 0. d.
ID: 2967214 • Letter: L
Question
Let A, B, C be square matrices.
a.) (T/F) A is not invertible iff det(A) = 0.
d.) (T/F) Suppose A = PB where P is invertible. Then the columns of A are independent iff the columns of B are independent.
g.) (T/F) The kernel of a linear map is a linear subspace.
h.) (T/F) A linearly independent set of a subspace H is a basis for H.
j.) (T/F) A vector space V with t linearly independent vectors has dimension at least t.
l.) (T/F) For an m n matrix M of rank r, its column span has dimension n - r.
m.) (T/F) For an m n matrix M of rank r, its null space has dimension n - r.
o.) (T/F) A is invertible iff 0 is not an eigenvalue of A.
Let A, B, C be square matrices. a.) (T/F) A is not invertible iff det(A) = 0. b.) (T/F) A linear map from lambdau, and it is a linear subspace. lambda is the set of all vectors u so that Au =Explanation / Answer
1) True, because inv(A)=adj(A)/det(A)
2)False
3)True
4)True
5)False ,its not necessary
6)True
7)True,The kernel of a linear transformation from a vector space V to a vector space W is a subspace of V.
8)True
9) True
10)False
11)Tue
12)True
13)False
14)False
15)True
16)True
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