Solutions

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

In[53]:=

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[57]:=

dimV

Out[57]=

6

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

In[58]:=

NumberOfPivots

Out[58]=

1

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

In[59]:=

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

Out[59]=

{( {{1, 0.543013, 2.11222, 3.19631, 3.89561}} )}

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

In[60]:=

NumberOfPivots

Out[60]=

1

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

In[61]:=

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

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

(MatrixForm[#] &) /@ ColumnSpaceSpanners

Out[63]=

{( {{0.234564}, {0.246998}, {0.151143}, {0.246653}, {0.059681}, {0.108331}} )}

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

In[64]:=

NumberOfColumns - NumberOfPivots

Out[64]=

4

(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[65]:=

Complement[Range[dimV], PivotColumnNumbers]

Out[65]=

{2, 3, 4, 5, 6}

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[66]:=

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

Out[66]=

{{{x_10. - 0.543013 x_2 - 2.11222 x_3 - 3.19631 x_4 - 3.89561 x_5}}}

Now write the vector

In[67]:=

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

Out[67]//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[68]:=

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[70]=

{( {{-0.543013}, {1}, {0}, {0}, {0}} ), ( {{-2.11222}, {0}, {1}, {0} ... {-3.19631}, {0}, {0}, {1}, {0}} ), ( {{-3.89561}, {0}, {0}, {0}, {1}} )}

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

In[71]:=

NumberOfPivots

Out[71]=

1

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

In[72]:=

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

Out[72]=

{( {{0.234564}, {0.246998}, {0.151143}, {0.246653}, {0.059681}, {0.108331}} )}

(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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