Let the set B be an ordered basis for P3.
Find the coordinate vector of F of X equals 7X cubed,
plus 4X squared minus 5X plus 9 relative to B.
Look in the notes below if the set B is an ordered basis
for a vector space V and vector X is in V
then vector X is equal to some linear combination
of the vectors in set B.
Which means in our case,
F of X is equal to some linear combination
of the four polynomials in set B.
And the scalars C sub 1 through C sub n
are the coordinated of vector X relative to the basis B.
Which means in our case, as scalars C sub 1 through C sub 4
will give us the coordinate vector of F of X relative to B.
Looking at set B, you'll notice how
the polynomials by degree are degree 0,
degree one, degree two and degree three.
So let's first change the order of the function
until the terms go from degree zero to a degree three.
Once we do this,
we will set the polynomial equal to a linear combination
of the polynomials in set B.
So on the left we have F of X
ordered from degree zero to degree three as 9 minus 5X
plus 4X squared, plus 7X cubed.
And this is equal to some linear combination
of the polynomials in set B.
So on the right, we have C1 times 1
where 1 is the first polynomial on set B
plus C2 times the quantity X plus 8.
X plus 8 is a second polynomial in set B.
And then we have plus C3
times the square of the quantity X minus 1,
the square of the quantity X minus 1
is the third polynomial in set B
and then following we have C4
times the fourth polynomial in set B which is 3X cubed.
And now we need to expand and simplify the right side
so that we can then equate the coefficients
that will give us a system of equations
that we can then solve to find C sub 1 through C sub 4.
So simplifying on the right, we have C1.
distributing C2, we have C2X plus 8C2.
Next we square the quantity X minus 1,
which gives us C3 times X squared minus 2X plus 1.
And then we have plus 3C4X cubed.
Next we'll distribute the C3 to clear the parentheses.
Distributing C3 we have C3X squared
minus 2C3X plus C3.
Next we'll group the constant terms,
the X terms, X squared terms and X cubed terms on the right,
so that we can then equate the coefficients.
So on the right, there are three constant terms.
We have C1 plus 8C2 plus C3
and there are two X terms.
We have C2X minus 2C3X.
If we factor out the X we have the quantity
C2 minus 2C3 times X.
There's only one X squared term which is C3 times X squared.
And there's also only one X cubed term
which is 3C4 times X cubed.
Now that we have it in this form,
we can equate the coefficients, meaning C1 plus 8C2 plus C3
must equal 9.
C2 minus 2C3, the coefficient of X must equal -5.
C3 must equal the coefficient of X squared which is 4
and 3C4 must equal the coefficient of X cubed, which is 7.
And now that we have a system of equations,
we can solve the system by writing an augmented matrix
and then writing the augmented matrix
and reduce to echelon form.
So because we have four equations and four variables,
we have a four by five augmented matrix
where the first row is 1, 8, 1, 0, 9.
The second row is 0, 1, -2, 0, -5.
The third row is 0, 0, 1, 0, 4.
And the fourth row is 0, 0, 0, 3, 7.
The next step is write the augmented matrix
and reduce to echelon form
which I've already done here to save time.
The first row indicates that C1 equals -19.
The second row indicates C2 equals 3.
The third row indicates C3 equals 4,
and the fourth one indicates C4 equals 7/3.
And these are the values that we need
to form the coordinate vector of F of X relative to B.
The coordinate vector of F of X relative to B
is the vector -19, 3, 4, 7/3.
I hope you found this helpful.