Solutions

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

In[233]:=

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[237]:=

dimV

Out[237]=

7

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

In[238]:=

NumberOfPivots

Out[238]=

4

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

In[239]:=

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

Out[239]=

{( {{1, 0., 0., -6.50674*10^-16, 2.4327, 0.}} ), ( {{0, 1, 0., -0.71 ...  {{0, 0, 1, 1.31169, -1.50345, 0.}} ), ( {{0, 0, 0, 0, 0, 1}} )}

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

In[240]:=

NumberOfPivots

Out[240]=

4

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

In[241]:=

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

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

(MatrixForm[#] &) /@ ColumnSpaceSpanners

Out[243]=

{( {{0.290846}, {0.465627}, {0.161144}, {0.644015}, {0.27398}, {0.0519085}, {0.58772 ... {{0.210909}, {0.32222}, {0.765739}, {0.0119107}, {0.566208}, {0.602279}, {0.969422}} )}

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

In[244]:=

NumberOfColumns - NumberOfPivots

Out[244]=

2

(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[245]:=

Complement[Range[dimV], PivotColumnNumbers]

Out[245]=

{4, 5, 7}

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[246]:=

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

Out[246]=

{{{x_10. + 6.50674*10^-16 x_4 - 2.4327 x_5}}, {{x_20. + 0.716646 x_4 - 0.82141 x_5}}, {{x_30. - 1.31169 x_4 + 1.50345 x_5}}, {{x_60}}}

Now write the vector

In[247]:=

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

Out[247]//MatrixForm=

( {{x_1}, {x_2}, {x_3}, {x_4}, {x_5}, {x_6}} )

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[248]:=

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[250]=

{( {{6.50674*10^-16}, {0.716646}, {-1.31169}, {1}, {0}, {0}} ), ( {{-2.4327}, {-0.82141}, {1.50345}, {0}, {1}, {0}} )}

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

In[251]:=

NumberOfPivots

Out[251]=

4

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

In[252]:=

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

Out[252]=

{( {{0.290846}, {0.465627}, {0.161144}, {0.644015}, {0.27398}, {0.0519085}, {0.58772 ... {{0.210909}, {0.32222}, {0.765739}, {0.0119107}, {0.566208}, {0.602279}, {0.969422}} )}

(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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