LINEAR ALGEBRA EXAM 3 T/F
PRACTICE EXAM QUESTIONS AND
ANSWERS
A set of vectors: {v1, v2, v3, ... vn} is said to be linearly independent when ... - Answer-
the vector equation: x1v1 + x2v2+...xpvp = 0 (where the x's are constants and the v's
are vectors) has only trivial solutions. so matrix equation Ax has only trivial solutions.
non-trivial solutions exists means... - Answer-implies that there are infinitely many
solutions, so when a column has no pivot, the system has one free variable (and is also
dependent system)
trivial solutions exists means... - Answer-the zero vector is a solution
a pivot positions is - Answer-after row reduction, the position in a matrix contains a
leading one
inconsistant matrix - Answer-if it has no solutions
consistant matrix - Answer-a system which has at least one solution is said to be
consistant
A set of vectors is said to be linearly dependent when - Answer-a nontrival solutions
exists
standard basis vectors are linearly independent? - Answer-Yes, because the matrix A=
[e1,e2,e3,...en] is really In where In x = 0 has only trival solution with vector x=0 and
matrix In is the identity matrix.
And the Standard vectors for a basis of R^n and they are linearly independent
Fundamental Theorem of Linear Algebra: Part I. - Answer-for an m×n matrix A:
dim(Col(A))+dim(Nul(A)) = n.
basic theorem: for an m×n matrix A: - Answer-1) the columns of a matrix A
corresponding to the pivot columns of rref[A 0] form a basis for Col(A),
2) the dimension of Nul(A) is the number of free variables in the equation Ax=0,
3) the dimension of Col(A) is the number of basic variables in the equation Ax=0.
Basis - Answer-A basis for a vector space is a sequence of vectors that form a set that
is linearly independent and that spans the space.
, Span - Answer-the span of vectors v1, v2, ... , vn is the set of linear combinations: c1v1
+ c2v2 + ... + cnvn and that this is in vector space.
Some other simple criteria for linear independence we met earlier will be also be useful:
- Answer-(1) a set {v} containing only one non-zero vector is linearly independent
because x1v=0 only if x1=0,
(2) a set {v1,v2} containing two vectors is linearly dependent if and only if v2 is a scalar
multiple of v1 because
x1v1+x2v2=0 → v1= −(x2/x1)v2 or v2= −(x1/x2)v1
unless x1=x2=0,
(3) any set {v1,...,vk−1,0,vk+1,...,vp} containing the zero vector is linearly dependent
because:
c1v1+...+ck−1vk−1+ck0+ck+1vk+1+⋯+cpvp = 0
when cj=0 for j≠k but ck=1.
a set B={v1,v2,...,vp} of vectors is said to be a BASIS for a subspace H of R^n when... -
Answer-(1) H contains {v1,v2,...,vp},
(2) H = Span{v1,v2,...,vp},
(3) {v1,v2,...,vp} is linearly independent.
subspace of R^n is... - Answer-is a vector space that is a subset of some other (higher-
dimension) vector space.
Basis Property: - Answer-when a set B={v1,v2,...,vp} of vectors is a BASIS for a
subspace H of R^n, then each x in H has a UNIQUE representation: x = c1v1+c2v2+...
+cpvp in terms of the basis vectors.
find a basis for R^n - Answer-Find the associated augmented matrix for Ax=0 and then
you use the # of pivot columns of rref(A0) to be the basis of the Col(A) or the # of free
variables to be the basis of Nul(A)
The null space of an m x n matrix A, written as Nul A, is the...? - Answer-set of all
solutions to the
homogeneous equation Ax = 0, using free variable terms s and t if needed to then find a
solutions of the form Null(A)=Span{v1,v2}
Col (A) is ...? - Answer-The column space of an mxn matrix A (Col A) is the set of all
linear combinations of the
columns of A.
If A = {a1 ... an} , then Col(A)= Span {a1, ..., an}
PRACTICE EXAM QUESTIONS AND
ANSWERS
A set of vectors: {v1, v2, v3, ... vn} is said to be linearly independent when ... - Answer-
the vector equation: x1v1 + x2v2+...xpvp = 0 (where the x's are constants and the v's
are vectors) has only trivial solutions. so matrix equation Ax has only trivial solutions.
non-trivial solutions exists means... - Answer-implies that there are infinitely many
solutions, so when a column has no pivot, the system has one free variable (and is also
dependent system)
trivial solutions exists means... - Answer-the zero vector is a solution
a pivot positions is - Answer-after row reduction, the position in a matrix contains a
leading one
inconsistant matrix - Answer-if it has no solutions
consistant matrix - Answer-a system which has at least one solution is said to be
consistant
A set of vectors is said to be linearly dependent when - Answer-a nontrival solutions
exists
standard basis vectors are linearly independent? - Answer-Yes, because the matrix A=
[e1,e2,e3,...en] is really In where In x = 0 has only trival solution with vector x=0 and
matrix In is the identity matrix.
And the Standard vectors for a basis of R^n and they are linearly independent
Fundamental Theorem of Linear Algebra: Part I. - Answer-for an m×n matrix A:
dim(Col(A))+dim(Nul(A)) = n.
basic theorem: for an m×n matrix A: - Answer-1) the columns of a matrix A
corresponding to the pivot columns of rref[A 0] form a basis for Col(A),
2) the dimension of Nul(A) is the number of free variables in the equation Ax=0,
3) the dimension of Col(A) is the number of basic variables in the equation Ax=0.
Basis - Answer-A basis for a vector space is a sequence of vectors that form a set that
is linearly independent and that spans the space.
, Span - Answer-the span of vectors v1, v2, ... , vn is the set of linear combinations: c1v1
+ c2v2 + ... + cnvn and that this is in vector space.
Some other simple criteria for linear independence we met earlier will be also be useful:
- Answer-(1) a set {v} containing only one non-zero vector is linearly independent
because x1v=0 only if x1=0,
(2) a set {v1,v2} containing two vectors is linearly dependent if and only if v2 is a scalar
multiple of v1 because
x1v1+x2v2=0 → v1= −(x2/x1)v2 or v2= −(x1/x2)v1
unless x1=x2=0,
(3) any set {v1,...,vk−1,0,vk+1,...,vp} containing the zero vector is linearly dependent
because:
c1v1+...+ck−1vk−1+ck0+ck+1vk+1+⋯+cpvp = 0
when cj=0 for j≠k but ck=1.
a set B={v1,v2,...,vp} of vectors is said to be a BASIS for a subspace H of R^n when... -
Answer-(1) H contains {v1,v2,...,vp},
(2) H = Span{v1,v2,...,vp},
(3) {v1,v2,...,vp} is linearly independent.
subspace of R^n is... - Answer-is a vector space that is a subset of some other (higher-
dimension) vector space.
Basis Property: - Answer-when a set B={v1,v2,...,vp} of vectors is a BASIS for a
subspace H of R^n, then each x in H has a UNIQUE representation: x = c1v1+c2v2+...
+cpvp in terms of the basis vectors.
find a basis for R^n - Answer-Find the associated augmented matrix for Ax=0 and then
you use the # of pivot columns of rref(A0) to be the basis of the Col(A) or the # of free
variables to be the basis of Nul(A)
The null space of an m x n matrix A, written as Nul A, is the...? - Answer-set of all
solutions to the
homogeneous equation Ax = 0, using free variable terms s and t if needed to then find a
solutions of the form Null(A)=Span{v1,v2}
Col (A) is ...? - Answer-The column space of an mxn matrix A (Col A) is the set of all
linear combinations of the
columns of A.
If A = {a1 ... an} , then Col(A)= Span {a1, ..., an}