Find the rank and nullities of the following matrices
Solution
Solution method: First find the rank of the matrix, then to find the nullity, use the relation
The rank of a matrix can be found using one of the following methods
Hence for using the second method above, we see that the minor found by omitting the
first columns and the second row is
but the full determinant is clearly zero (since
has one columns which is all zeros, since any square matrix which has all zero's in one of its rows or
columns must have zero determinant). Hence the largest size of a minor which is not zero is 2,
hence the rank is 2
Since the rank is 2, then
For ,
hence the rank is 3
Therefor
For , the rank can not be more than 3, and since the last row contains zeros
in the first 3 elements, I will select the minor to test on as the last 3 columns, hence
, hence the rank is 3
question: Find bases of the range spaces and the null spaces of the matrices in problem 3.5
solution method: To find the bases for the range space, all what we need to do is to find
linearly independent vectors where
is the rank of A found above. To find bases for the null space
of A, since we know the rank of the null space, and the dimension of the rank space, we
need to find m linearly independent vectors where
is the rank of the null space of
A.
A1: Since has rank of 2, this means that range(A) has dimension 2. In other words
A maps a 'point' in a 3D volume to a 'point' of in a 2D flat plane. Hence we can use either
or
or
as the basis of the range of A. But since
has zeros in its second row, this means
that there are no points in the range of A which has a component along the
dimension. Hence
the only plane left is the one spanned by
or
. This is illustrated by the following
diagram
since the rank of the null space is 1, then we need to find one vector s.t.
Hence , hence a basic is
where 'any'
could be any number. select 1 to make it normalized, hence a basis is for the null space is
A2: Since has rank of 3, and this is equal to the number of columns in A, then A maps a
point in 3D vector space
to a point in 3D vector space
. Then we can use
as its
bases. i.e.
Since the rank of the null space of A is zero, then the null space of A is , hence no basis
for the null space as it is empty.
A3: Since has rank of 3, then we need to find 3 linearly independent vectors to span the
range of A. Select
as its bases. i.e.
The null space of is 1, then we need to find one vector such
and normalize it as
needed.
Hence and so
Hence if we take , then
and then
Consider the linear algebraic equation , it has 3 equations and 3
unknowns. Does a solution exist in the equation? Is this solution unique, Does a solution exist if
Solution
A solution exist if A spans a space which contains Since
has 2 columns we see that it
takes points from 2D space and send these points to its range space. Since the rank of A is 2 (since
) then the dimension of its range space is 2, i.e. it maps points from 2D space to
2D space. Solve this is by solving for
and to see if we can find a vector
to satisfy this
equation as follows
From second equation we see that , substitute this in either equation 1, we get
that
, hence
Sub this solution in equation 3 we see that is also satisfy
it.
Hence we found a point , which is mapped by A to point
hence a solution
exist.
Since the null space is empty, then this solution is unique.
When , we need to try to see if there is an
such that
From first equation, sub into second equation we get
Hence
Now sub this solution into the third equation we get
which is not valid. Hence no solution exist.
Problem: Consider , where
is an
and has rank
is
a
solution? if not, under what condition will it be a solution? is
a solution?
Solution
First we need to determine if exist.
Since is an
then
is
matrix.
Hence is a square matrix of dimension
. Since we are told the rank is
and the
rank of a matrix is the smaller of its dimensions (the smaller of its rows or columns if they are not
the same), hence there exist only
linearly independent rows, and not
linearly independent
rows. hence
is NOT invertible
is not a solution.
. Since in this case can be inverted.
Second part: is
a solution?
Since is an
then
is
matrix.
Hence, since we are told the rank is then there exist
linearly
independent rows, hence
is invertible. Hence
exist, and so
can be computed. And in addition, if we multiply this by
we get
hence it is a solution.