Homework due Class 33

1.Find the rank of the set of vectors in R5 usng only the definition of the rank: (1,2,3,4,5), (6,7,8,9,10), (1,1,1,1,1), (2,2,2,2,2), (5,4,3,2,1), (6,5,4,3,2).

The definition of the rank states:

Let S be a set of vectors in a vector space. Then the dimension of the vector space spanned by S is called the rank of S.

Thus, I need to find the dimension of the vector space spanned by S.

From a theorem about spans in the WebNotes, if S is a subset of a vector space V and a is an element in S which is equal to a linear combination of other elements of S, then S' (the set obtained by removing a from S) is such that span(S)=span(S').

So let S={(1,2,3,4,5), (6,7,8,9,10), (1,1,1,1,1), (2,2,2,2,2), (5, 4,3,2,1), (6,5,4,3,2)}. Then, notice that (2,2,2,2,2) equals a linear combination of other elements of S:

(2,2,2,2,2) = 2*(1,1,1,1,1) + 0*(1,2,3,4,5)


Also notice that (6,7,8,9,10), (5,4,3,2,1), and (6,5,4,3,2) are all linear combinations of the same two elements:
(6,7,8,9,10) = 5*(1,1,1,1,1) + (1,2,3,4,5)
(5,4,3,2,1) = 6*(1,1,1,1,1) + (-1)*(1,2,3,4,5)
(6,5,4,3,2) = 7*(1,1,1,1,1) + (-1)*(1,2,3,4,5)

So let S' be the set obtained by removing (2,2,2,2,2), (6,7,8,9,10), (5,4,3,2,1), and (6,5,4,3,2) from S. Then span(S')=span(S).

Since (1,1,1,1,1) and (1,2,3,4,5) are linearly independent (they are not proportional), they form a basis of the subspace spanned by S', which is identical to the subspace spanned by S. Since S' only contains two elements and is the basis for the subset, this subset has dimension 2.

Also, since the dimension of a vector space spanned by S is called the rank of S, S has a rank of 2.

2.Find the row rank and the column rank of the following matrix using only the definition of the rank:
[ 2 3 4 5 6 ]
[ 1 1 1 1 1 ]
[ 0 0 0 2 3 ]

From the definition of the rank in the web notes, we have:

Let S be a set of vectors in a vector space. Then the dimension of the vector space spanned by S is called the rank of S.

If A is a matrix then the rank of the set of its columns is called the column rank of A and the rank of the set of its rows is called the row rank of A.

Thus, I need to find the dimension of the columns of A and of the rows of A. In other words, we need to find the dimension of the set of the columns of A and the set of the rows of A.

From a theorem about spans in the WebNotes, if S is a subset of a vector space V and a is an element in S which is equal to a linear combination of other elements of S, then S' (the set obtained by removing a from S) is such that span(S)=span(S').

The set of columns C = {(2,1,0), (3,1,0), (4,1,0), (5,1,2), (6,1,3)} Then, notice that (4,1,0) equals a linear combination of other elements of S:

(4,1,0) = (-1)*(2,1,0) + 2*(3,1,0) + 0*(5,1,2)
Also notice that (6,1,3) is also all linear combinations of the same three elements:

(6,1,3) = 0*(2,1,0) + (-1/2)*(3,1,0) + (3/2)*7*(5,1,2)
So let C' be the set obtained by removing (4,1,0) and (6,1,3) from S. Then span(C')=span(C).

Also, from a theorem about linear independence in the webnotes we have that a subset C of a vector space V is linearly independent if and only if there exists exactly one linear combination of elements of C which is equal to 0, the one with all zero coefficients.

Thus, to check linear independence we need to solve the following system:
[ 0 ]
[ 0 ]
[ 0 ]
=
[ 2 ]
[ 1 ]
[ 0 ]
* x1 +
[ 3 ]
[ 1 ]
[ 0 ]
* x2 +
[ 5 ]
[ 1 ]
[ 2 ]
* x3

Which we can do using Maple:

> A:=matrix([[2,3,5,0],[1,1,1,0],[0,0,2,0]]);
A := [ 2 3 5 0 ]
[ 1 1 1 0 ]
[ 0 0 2 0 ]


> gaussjord(A);
[ 1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 1 0 ]

Thus, the only solution is that where x1, x2, and x3=0, so the set is linearly independent. Thus, they form a basis of the subspace spanned by C', which is identical to the subspace spanned by C. Since C' only contains three elements and is the basis for the subset, this subset has dimension 3.

Also, since the dimension of a vector space spanned by C is called the rank of C, C has a rank of 3. Thus, the columnrank of the matrix is three.

The set of rows U = {(2,3,4,5,6), (1,1,1,1,1), (0,0,0,2,3)}

From a theorem about linear independence in the webnotes we have that a subset U of a vector space V is linearly independent if and only if there exists exactly one linear combination of elements of U which is equal to 0, the one with all zero coefficients.

Thus, to check linear independence we need to solve the following system:
[ 0 ]
[ 0 ]
[ 0 ]
[ 0 ]
[ 0 ]
=
[ 2 ]
[ 3 ]
[ 4 ]
[ 5 ]
[ 6 ]
* x1 +
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
[ 1 ]
* x2 +
[ 0 ]
[ 0 ]
[ 0 ]
[ 2 ]
[ 3 ]
* x3

Which we can do using maple:

> A:=matrix([[2,1,0,0],[3,1,0,0],[4,1,0,0],[5,1,2,0],[6,1,3,0]]);
A := [ 2 1 0 0 ]
[ 3 1 0 0 ]
[ 4 1 0 0 ]
[ 5 1 2 0 ]
[ 6 1 3 0 ]

> gaussjord(A);
[ 1 0 0 0 ]
[ 0 1 0 0 ]
[ 0 0 0 0 ]
[ 0 0 0 0 ]
[ 0 0 0 0 ]

Thus, the only solution is that where x1, x2, and x3=0, so the set is linearly independent. Thus, they form a basis of the subspace spanned by U. Since U only contains three elements and is the basis for the subset, this subset has dimension 3.

Also, since the dimension of a vector space spanned by U is called the rank of U, U has a rank of 3. Thus, the rowrank of the matrix is three.

3. Find the rank of the following set of polynomials: x3+x2+x, x3+x+1, x3+x2+x, 2x3+x2+x+1.

The definition of the rank states:

Let S be a set of vectors in a vector space. Then the dimension of the vector space spanned by S is called the rank of S.

Thus, I need to find the dimension of the vector space spanned by S, where S = {x3+x2+x, x3+x+1, x3+x2+x, 2x3+x2+x+1}.

From a theorem about spans in the WebNotes, if S is a subset of a vector space V and a is an element in S which is equal to a linear combination of other elements of S, then S' (the set obtained by removing a from S) is such that span(S)=span(S').

So let S={x3+x2+x, x3+x+1, x3+x2+x, 2x3+x2+x+1}. Then, notice that x3+x2+x equals a linear combination of other elements of S:

x3+x2+x = x3+x2+x + 0*(x3+x+1) + 0*(2x3+x2+x+1)


So let S' be the set obtained by removing one of the x3+x2+x elements from S. Then span(S')=span(S).

From the WebNotes:

Theorem. Let f1, f2,...,fn be functions in C[0,1] each of which has first n-1 derivatives. If the Wronskian of this set of functions is not identically zero then the set of functions is linearly independent.

Thus, we will find the Wronskian of S using Maple:


> f:=x^3+x^2+x: g:=x^3+x+1: h:=2*x^3+x^2+x+1: f1:=3*x^2+2*x+1: g1:=3*x^2+1:

> h1:=6*x^2+2*x+1: f2:=6*x+2: g2:=6*x: h2:=12*x+2:

> A:=matrix([[f,g,h],[f1,g1,h1],[f2,g2,h2]]);
A := [ x3+x2+x x3+x+1 2x3+x2+x+1 ]
[ 3x2+2x+1 3x2+1 6x2+2x+1 ]
[ 6x+2 6x 12x+2 ]

> W:=simplify(det(A));

W := - 2 x3 - 6 x - 2


So W is not identically 0. Thus by the theorem about the Wronskian, the functions f, g, h are linearly independent.

Thus, these functions form a basis of the subspace spanned by S' (from the definition about basis), which is identical to the subspace spanned by S. Since S' only contains three elements and is the basis for the subset, this subset has dimension 3.

Also, since the dimension of a vector space spanned by S is called the rank of S, S has a rank of 3. Thus, the rank of S is three.