What dimension is P3,
the set of polynomials degree three or less.
The dimension of Pn is equal to n plus one,
and therefore the dimension of P3
is equal to three plus one, or four.
This means any basis for P3
must be a set of four polynomials.
So for the next part, which of the following sets
form a basis for P3,
we can eliminate all of the sets
that don't contain four polynomials.
So for example, this first set contains
one, two, three, four, five polynomials,
this set can not be a basis for P3,
we can eliminate this set from consideration.
Also notice how the third set
only contains three polynomials,
and therefore it can not be a basis for P3,
we can eliminate the third set from consideration.
The remaining sets do contain four polynomials,
but if you look at the set containing
two x squared, x, zero, and x cubed,
we should be able to recognize this will not span P3,
if it did span P3, all the polynomials in P3
would be linear combinations of these four polynomials.
But since the constant here is zero,
any linear combination of zero is still going to be zero,
which eliminates this as being a possible basis for P3,
although polynomials that are linear combinations
of these four polynomials will always have a zero
for a sub zero in the form of the polynomial
given here above,
and of course there are polynomials
where the constant is not zero.
So we can also eliminate this set,
but I'll go ahead and leave it in for consideration.
Let's consider the set one, two x negative three x squared,
and two x cubed, as a possible basis for P3.
To do this, because we already know it has four polynomials,
we only have to check to see,
if the set of polynomials is independent.
If it is, and again it contains four polynomials,
it would span all of P3 and be a basis.
To perform the independence test,
we will form the homogeneous vector equation
shown here below,
and if the solution is only the trivial solution,
the set is independent,
if it has more solutions than the trivial solution,
the set is dependent and does not form a basis.
And we are going to convert the polynomials to vectors,
where, if every polynomial, the vector is a sub zero,
a1, a2 and a3, or a0 is the constant
and a1 through a3 are the coefficients.
So, considering the first set,
one is the vector one, zero, zero, zero,
two x is a vector zero, two, zero, zero,,
negative three x squared is the vector
zero, zero, negative three, zero,
and two x cubed is the vector zero, zero, zero, two.
Next, we form the homogeneous vector equation,
which is the linear combinations of the given four vectors
equal to the zero vector.
From here, we can form an augmented matrix
where the first row is going to be
one, zero, zero, zero, zero,
the second row is zero, two, zero, zero, zero,
the third row is zero, zero, negative three, zero, zero,
and the fourth row is zero, zero, zero, two, zero,
which we see here on the left
Next, we write the augmented matrix
and reduced row-echelon form.
which I've already done to save time.
Let's go ahead and label the columns C sub one
through C sub four.
And notice how, there is a pivot
in each of the variable columns,
where row one indicates that C sub one is equal to zero,
row two indicates that C sub two is equal to zero,
row three indicates that C sub three is equal to zero
and row four indicates that C sub four is equal to zero.
Notice, how we only have the trivial solution,
which indicates the set of vectors is independent.
And because we have four vectors in the set,
it also spans our four,
which in our case indicates,
the set of polynomials is independent and spans P3
and therefore forms the basis for P3.
So, going back to our first slide,
we can go ahead and check this set of vectors as a basis
for P3, and now check the set two x squared, x, zero,
and x cubed,
which again, we know is not going to be a basis
because the only constant is zero.
Let's go ahead and check it anyway.
First step is to form the vectors that correspond
to each of the polynomials
in the set, which I've already done here.
Notice how zero is just a zero vector.
Then we form the homogeneous vector equation.
From here, we can write the have augmented matrix,
where the first row is a row of zeros.
The second row is zero, one, zero, zero, zero.
The third row is two, zero, zero, zero, zero
and the fourth row is zero, zero, zero, one, zero.
Writing the matrix in reduced row-echelon form.
Notice we only have one, two, three pivots,
which indicates that C sub one, C sub two
and C sub four are basic variables,
but C sub three is a free variable.
Also notice, how we have a row of zeroes along the bottom,
indicating we have an infinite number of solutions,
which therefore means we have more than the trivial solution
to the homogeneous vector equation.
And therefore the set of vectors is dependent
and cannot form a basis,
which indicates a set of polynomials
can not be a basis for P3.
If you did want to show some extra work here, though,
the first row indicates that C sub one is equal to zero.
The second row indicates that C sub two is equal to zero
third row indicates C sub four is equal to zero,
C sub three is a free variable.
C sub three is equal to C sub three,
which means C sub three can be any real number.
And we still have a solution
to the homogeneous vector equation, which makes sense,
because notice how we have C sub three times zero vector.
So once again, we know the set of polynomials
does not form the basis for P3.
We have one more set to consider the set two x squared,
x, negative five and two plus x cubed.
Two x squared is the vector zero, zero, two, zero.
x is the vector zero, one, zero, zero,
negative five is the vector negative five, zero, zero, zero,
and two plus x cube is the vector two, zero, zero, one.
Next you write the homogeneous vector equation
and then form the augmented matrix
where the first row is zero, zero, negative five, two, zero,
the second row is zero, one, zero, zero, zero,
third row is two, zero, zero, zero, zero,
and the fourth row is zero, zero, zero, one, zero.
Next you write the augmented matrix
and reduced row-echelon form,
which I've already done labeling the columns C sub one
through C sub four and identifying the pivots.
The number of pivots matches the number of variables,
which indicates we have one solution.
And because we have a column of zeros in the fifth column,
we only have the trivial solution where C sub one
through C sub four
are all equal to zero.
So because we only have the trivial solution,
they said the vectors as independent.
And because we have four vectors in the set,
this set spans our four,
which also indicates the set of polynomials is independent
and spans P3, and therefore is a basis for P3,
which means of the five set of vectors,
two of them are basis for P3.
I hope you found this helpful.