Which sets below form a basis for a P two?
There may be more than one answer.
Notice, each of those sets contains three polynomials
and this is important to recognize because
the dimension of P two is three,
which indicates any basis for a P two must contain
a set of three polynomials.
So we can approach this problem two ways,
because we have a set of three polynomials and P two.
Number one we can check to see if the set of polynomials
is independent.
If it is then the set must also span P two
and forms a basis for P two.
Number two, we can also check to see
if the set up polynomials spans P two.
In this video though, we will take approach one,
but we're also going to convert the polynomials to vectors
and see if the set of three vectors, is independent.
If the set is independent,
it also spans R three and therefore the set of polynomials
must form a basis for P two.
Each vector will be in the form of a sub zero,
a sub one, a sub two.
A sub zero is the constant.
A sub one is a coefficient, in our case of X
and a sub two in our case is the coefficient of X squared.
Let's begin by considering the first set of polynomials.
Let's first form the three vectors
for the three polynomials
where seven plus two X plus four X squared
is the vector seven, two, four.
Two plus X is the vector two, one, zero.
Notice it's zero because the X squared term
would be zero X squared
and 20 plus four X plus 16 X squared
is the vector 20, 4, 16.
To test for independence,
we now form the homogeneous vector equation,
where we have the linear combinations of the three vectors
equal to the zero vector.
From here, we can write the augmented matrix
where the first row is 7, 2, 20, 0.
The second row is 2, 1, 4, 0
and the third row is 4, 0, 16, 0.
Next, we write the augmented matrix
in reduced row echelon form,
which I've already done here on the right.
Notice because we have three variables and three rows
where the last row is a row of zeros,
we know we have a dependent system
or an infinite number of solutions,
which indicates we have more than the trivial solution
to the homogeneous vector equation,
which indicates the vectors are dependent
and therefore they don't form a basis for R three
and therefore the set of polynomials
does not form a basis for P two.
Let's show a little work here.
Let's label the first three columns C sub one
through C sub three.
The first row indicates that C sub one
plus four C three is equal to zero.
The second row indicates that C sub two
minus four C sub three equals zero.
And if we identify the pivots,
which are in row one column one and row two column two
we know C sub one and C two of the basic variables.
And C sub three is a free variable.
Which begin state as C sub three equals C sub three.
Let's go ahead and solve the first two equations
for C sub one and C sub two.
C sub one is equal to negative four C sub three.
C sub two is equal to positive four C sub three.
And again, C sub three is equal to C sub three.
This indicates begin that C sub three be any real number
and then find C sub one and C sub two.
Again, indicating we have an infinite number of solutions
and therefore the set of vectors is dependent
and so is the set polynomials,
which means they don't form a basis for P two.
I do want to show how you would do this,
if we did not convert to vectors.
We would form linear combinations of the three polynomials
instead of equal to the zero polynomial
written as zero plus zero X plus zero X squared.
Next we would distribute C sub one, C sub two
and C sub three.
Then we would group the terms,
so that we can equate the coefficients,
meaning we group the terms that have zero factors of X,
group the terms that have one factor of X
and group the terms that have two factors of X
or a factor of X squared.
Then we factor out X and X squared.
And then finally equate the coefficients,
which gives us the equations seven C sub one,
plus two C sub two, plus 20 C sub three equals zero.
Two C sub one plus C sub two
plus four C sub three equals zero.
And finally four C sub one plus 16 C sub three equals zero,
which I've already written as equations here on the left.
From here, we can form the augmented matrix
and then write the augmented matrix
in reduced row echelon form.
So notice how this was quite a bit more work
than converting to vectors.
But if we compare these two matrices
to the previous method,
notice how they are exactly the same.
But I think you can see why I decided to convert to vectors
rather than work with the polynomials.
And now check the second set to see
if it forms a basis for P two.
Negative X is the vector 0, negative 1, 0.
One plus two X is the vector 1, 2, 0.
X plus X squared as the vector 0, 1, 1.
And again, we form the homogeneous vector equation
where we have the linear combinations of
the given three vectors equal to the zero vector.
Then we form the augmented matrix
and write the augmented matrix in reduced row echelon form.
Let's again, label the columns C sub one
through C sub three.
Notice, now we have a pivot in each of the variable columns.
Well the first row indicates that C sub one
is equal to zero.
The second row indicates that C sub two equals zero
and third row indicates that C sub three equals zero.
Notice here we only have the trivial solution
to the homogeneous vector equation,
which indicates a set of vectors is independent.
And because we have three vectors in R three,
they also span R three,
which tells us the set of polynomials is independent
and spans P two and therefore forms a basis for P two.
So we do select the second set of polynomials
and now we check the last set.
We first formed the corresponding vectors.
Negative X squared is the vector 0, 0, negative 1.
X plus two X squared is the vector 0, 1, 2.
And one plus X squared is the vector 1, 0, 1.
Next we form the homogeneous vector equation,
set up the augmented matrix
and write the augmented matrix in reduced row echelon form.
And we can quickly see again we have three pivots.
The number of pivots matches the number of variables
where the first show indicates that C sub one equals zero.
The second row indicates that C sub two equals zero,
and the third row indicates that C sub three equals zero.
Once again, we only have the trivial solution
to the homogeneous vector equation.
And therefore the set of vectors is independent
and spans R three,
which indicates a set of polynomials is independent,
spans P two and is a basis for P two.
The third set of polynomials is also a basis for P two.
I hope you found this helpful.