Solutions

First, we compute some basics facts about the matrix M.

In[161]:=

NumberOfRows = Length[M] ;

NumberOfColumns = Length[First[M]] ;

R = RowReduce[M] ;

NumberOfPivots = Length[Select[R, Max[#] ≥1&]] ; (* the only way a row of the rref could not have a pivot is if it is all zero *)

(Question A) What is the value of d?

In[165]:=

dimV

Out[165]=

5

(Question B) What is the dimension of the row space of M?

In[166]:=

NumberOfPivots

Out[166]=

4

(Question C) Give a basis for the row space of M.

In[167]:=

Table[MatrixForm[{R[[i]]}], {i, NumberOfPivots}]

Out[167]=

{( {{1, 0., 0., 0., 0.663855}} ), ( {{0, 1, 0., 0., 0.105571}} ), ( {{0, 0, 1, 0., 0.938285}} ), ( {{0, 0, 0, 1, -0.508933}} )}

(Question D) What is the dimension of the column space of M?

In[168]:=

NumberOfPivots

Out[168]=

4

(Question E) Give a basis for the column space of M.

In[169]:=

PivotColumnNumbers = Table[Last[Last[Select[Position[R, 1], First[#] i&, 1]]], {i, NumberOfPivots}] ;

ColumnSpaceSpanners = Transpose[M][[PivotColumnNumbers]] ;

(MatrixForm[#] &) /@ ColumnSpaceSpanners

Out[171]=

{( {{0.517635}, {0.849726}, {0.981629}, {0.674927}} ), ( {{0.4112}, ... 2}, {0.406154}} ), ( {{0.370429}, {0.898778}, {0.111716}, {0.182176}} )}

(Question F) What is the dimension of the null space of M?

In[172]:=

NumberOfColumns - NumberOfPivots

Out[172]=

1

(Question G) Give a basis for the null space of M.

We can get a basis for the Null space of a matrix M in Mathematica by using the command "NullSpace[M]". Here's how it comes out of the reduce row echelon form.

First, identify the free variables (columns of the rref that do not contain pivots).

In[173]:=

Complement[Range[dimV], PivotColumnNumbers]

Out[173]=

{5}

Each nonzero row of the rref corresponds to an equation with exactly one bound variable (bound=not free). Solve each equation for its bound variable.

In[174]:=

Table[Solve[R[[i]] . Table[x_i, {i, NumberOfColumns}] 0], {i, NumberOfPivots}]

Out[174]=

{{{x_10. - 0.663855 x_5}}, {{x_20. - 0.105571 x_5}}, {{x_30. - 0.938285 x_5}}, {{x_40. + 0.508933 x_5}}}

Now write the vector

In[175]:=

MatrixForm[Table[x_i, {i, NumberOfColumns}]]

Out[175]//MatrixForm=

( {{x_1}, {x_2}, {x_3}, {x_4}, {x_5}} )

in the form

free
var
*(
v
e
c
t
o
r
)+
free
var
*(
v
e
c
t
o
r
)+
free
var
*(
v
e
c
t
o
r
)+... , using the equations you solved for above. Note that if x_7 is a free variable, then the seventh component of each of the vectors will be 0, except for the one multiplied by x_7, which will have seventh component equal to 1.

Here's the final answer for the matrix in this quiz:

In[176]:=

freevars = (x_#& /@Complement[Range[NumberOfColumns], PivotColumnNumbers]) ;

ns = MatrixForm[Table[x_i, {i, NumberOfColumns}]]/.Flatten[Table[Solve[R[[i]] . Table[x_i, {i, NumberOfColumns}] 0], {i, NumberOfRows}]] ;

Table[(ns/.freevars[[i]] 1)/.x__0, {i, Length[freevars]}]

Out[178]=

{( {{-0.663855}, {-0.105571}, {-0.938285}, {0.508933}, {1}} )}

(Question H) What is the dimension of the span of the vectors given in Question A?

In[179]:=

NumberOfPivots

Out[179]=

4

(Question I) Give a basis for the span of the vectors given in Question A.

In[180]:=

If[flag, Table[MatrixForm[R[[i]]], {i, NumberOfPivots}], PivotColumnNumbers = Table[Last[Las ... rs = Transpose[M][[PivotColumnNumbers]] ;  (MatrixForm[#] &) /@ ColumnSpaceSpanners]

Out[180]=

{( {{1}, {0.}, {0.}, {0.}, {0.663855}} ), ( {{0}, {1}, {0.}, {0.}, { ... }, {0}, {1}, {0.}, {0.938285}} ), ( {{0}, {0}, {0}, {1}, {-0.508933}} )}

(Question J) Let W be a vector space, and U and V be subspaces. Prove or disprove the following two statements:
        (1)    U∩V, the set of all vectors in both U and V, is a subspace of W.
        (2)    U∪V, the set of all vectors in either U or V, is a subspace of W.

The set U∪V is not a subspace. For example, let U=Span{(1,0)}, and V=Span{(0,1)}. Since U∪V is a subset of the real vector space W, we need only to check if it is closed under addition and multiplication by a scalar. It is not closed under addition: (1,0)+(0,1)=(1,1) is a vector that is not in U∪V, although it is the sum of two vectors in U∪V.

The U∩V is a subspace. Since it is a subset of the real vector space W, we need only to check if it is closed under addition and multiplication by a scalar. Suppose that Overscript[x, ⇀] and Overscript[y, ⇀] are both in U and in V. Then Overscript[x, ⇀]+Overscript[y, ⇀] is in U because U is a subspace, and  Overscript[x, ⇀]+Overscript[y, ⇀] is in V because V is a subspace. Thus,  Overscript[x, ⇀]+Overscript[y, ⇀] is in U∩V because it is in both U and V. Likewise, let c be any real number. The vector c Overscript[x, ⇀] is in U because Overscript[x, ⇀] is in U and U is a subspace, and likewise  c Overscript[x, ⇀] is in V because Overscript[x, ⇀] is in V and V is a subspace. Therefore, c Overscript[x, ⇀] is in both U and V, and so is in U∩V.

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