BRAINSCAPE1
APMA 3080 Final Exam with
complete solutions latest version
How to show a set of vectors are linearly independent? - CORRECT ANSWER-(1) B =
RREF([A | 0])
(2) Check if the only answer for B is x1, x2, x3... = 0
(3) If is, it is linearly independent. Otherwise, no.
difference between span, subspace and basis - CORRECT ANSWER-Subspace and
span are linear combinations of vectors. They contain infinitely many vectors. THEY
ARE THE SAME!
A basis is a only set of vectors.
How to show subset S is a subspace? - CORRECT ANSWER-Check 3 conditions:
(1) if 0 is a part of S.
(2) if closed under addition: if (u in S) and (v in S), then (u + v in S).
(3) if closed under multiplication: if (u in S), then (cu in S).
Shortcuts:
- If S is a null space OR S is a span, it is automatically a subspace.
How to check if a set of vectors A form a basis of a subspace S? - CORRECT
ANSWER-Check 2 conditions:
(1) A spans S. (That is, A can generate every vector in S.)
(2) A is linearly independent
How to find basis for a space? - CORRECT ANSWER-Method 1: Find basis for S =
span(u1, u2, u3)
(1) Use the vectors {u1, u2, u3} to form rows of a matrix A
(2) B = REF(A)
(3) The non-zero rows of B are the basis
Method 1: Find basis for S = span(u1, u2, u3)
(1) Use the vectors {u1, u2, u3} to form columns of a matrix A
BRAINSCAPE1
APMA 3080 Final Exam with
complete solutions latest version
How to show a set of vectors are linearly independent? - CORRECT ANSWER-(1) B =
RREF([A | 0])
(2) Check if the only answer for B is x1, x2, x3... = 0
(3) If is, it is linearly independent. Otherwise, no.
difference between span, subspace and basis - CORRECT ANSWER-Subspace and
span are linear combinations of vectors. They contain infinitely many vectors. THEY
ARE THE SAME!
A basis is a only set of vectors.
How to show subset S is a subspace? - CORRECT ANSWER-Check 3 conditions:
(1) if 0 is a part of S.
(2) if closed under addition: if (u in S) and (v in S), then (u + v in S).
(3) if closed under multiplication: if (u in S), then (cu in S).
Shortcuts:
- If S is a null space OR S is a span, it is automatically a subspace.
How to check if a set of vectors A form a basis of a subspace S? - CORRECT
ANSWER-Check 2 conditions:
(1) A spans S. (That is, A can generate every vector in S.)
(2) A is linearly independent
How to find basis for a space? - CORRECT ANSWER-Method 1: Find basis for S =
span(u1, u2, u3)
(1) Use the vectors {u1, u2, u3} to form rows of a matrix A
(2) B = REF(A)
(3) The non-zero rows of B are the basis
Method 1: Find basis for S = span(u1, u2, u3)
(1) Use the vectors {u1, u2, u3} to form columns of a matrix A
BRAINSCAPE1
