# Linear Algebra

I am trying to solve this,,,need help....

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But the S* is linearly Independent, How can we prove that?

HallsofIvy
Homework Helper
Use the definition of "linearly independent". If $a_1\phi_1(f)+ a_2\phi_2(f)+ a_3\phi_3(f)= 0$ for all f, must $a_1, a_2, a_3$ all be 0?

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what is the values of $$\phi0$$ $$\phi1$$ $$\phi-1$$ ???

HallsofIvy
Homework Helper
Your attachment says that phi-1(f)= f(-1), phi0(f)= f(0), and phi1(f)= f(1)!

a phi-1+ b phi0+ c phi1= 0 means that, for any function f, [a phi-1(f)+ b phi0(f)+ c phi1= af(1)+ bf(0)+ cf(1)= 0.

Since P3 is the space of polynomials of degree 2 or less, ax2+ bx+ c. Take f above to be x2, x, and 1. What do a, b, and c have to be?

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What was the values of a b c ?

HallsofIvy
Homework Helper
That is the question that pnaik2008 needs to answer! IF he/she can show that a= b= c= 0, then S* is linearly independent.

How could we show that a=b=c=0?,and if we proof that S is linearly independent, what are the basis of the S??

HallsofIvy
Homework Helper
I've already explained that in my response #6. af(-1)+ bf(0)+ cf(1)= 0 for all polynomials of order 2 or less. Take f to be x2, x, and 1 and you get three equations to solve for a, b, and c.

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I've already explained that in my response #6. af(-1)+ bf(0)+ cf(1)= 0 for all polynomials of order 2 or less. Take f to be x2, x, and 1 and you get three equations to solve for a, b, and c.
Since it has taken already too much time, I will give a shot.

If f = x^2, then
f(0) = 0
f(-1) = 1
f(1) = 1
=> a*0 + b*1 + c*1 = 0

If f = x, then
f(0) = 0
f(-1) = -1
f(1) = 1
=> a*0 + b*(-1) + c*1 = 0

If f = 1, then
f(0) = 1
f(-1) = 1
f(1) = 1
=> a*1 + b*1 + c*1 = 0

So we have three equations
b + c = 0 (1)
-b + c = 0 (2)
a + b + c = 0 (3)

We get from (2)
b = c

Thus, we get from (1)
2b = 0
b = 0 (4)

Putting that to (1)
c = 0 (5)

Putting (4) and (5) to (3), we get
c = 0

Thus,
a + b + c = 0

This means that S* is linearly independent.

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I leave the rest for the questioner.

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HallsofIvy
Homework Helper
Since it has taken already too much time, I will give a shot.

If f = x^2, then
f(0) = 0
f(-1) = 1
f(1) = 1
=> a*0 + b*1 + c*1 = 0
So you are taking af(0)+ bf(-1)+ cf(1)= 0? That's not what I had before but perfectly legal.

If f = x, then
f(0) = 0
f(-1) = -1
f(1) = 1
=> a*0 + b*1 + c*1 = 0
You mean a*0+ b*(-1)+ c*1= -b+ c= 0, right?

If f = 1, then
f(0) = 1
f(-1) = 1
f(1) = 1
=> a*1 + b*1 + c*1 = 0

So we have three equations
b + c = 0
b + c = 0
No, -b+ c= 0.
a + b + c = 0

Simplifying
b + c = 0 (1)
a + b + c = 0 (2)

Substituting the equation (1) to (2)
a = 0

By trial and error, we know that
b = 0
c = 0

Thus,
a + b + c = 0

This means that S* is linearly independent.
The three equations you got, b+ c= 0, b+ c= 0, a+ b+ c= 0, the first two being the same, would give b= -c so that a+ b+ c= a- c+ c= a= 0 but does NOT mean b and c must be 0. for example b= 1, c= -1, a= 0 satisfy b+ c= 1- 1= 0, a+ b+ c= 0+ 1- 1= 0. That would prove S* is NOT linearly independent.

However, the correct equations are -b+ c= 0, b+ c= 0, a+ b+ c= 0. Adding the first two equations, 2c= 0 so c= 0. Putting c= 0 into the first equation -b+ 0= 0 so b= 0. Putting b= c= 0 into the third equation, a+ b+ c= a+ 0+ 0= a= 0. Since a= b= c= 0, S* is linearly independent.

But I would have liked to have seen pnaik2008 at least try that calculation.

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I leave the rest for the questioner.

So you are taking af(0)+ bf(-1)+ cf(1)= 0? That's not what I had before but perfectly legal.

You mean a*0+ b*(-1)+ c*1= -b+ c= 0, right?

I fixed the mistake.

I can't stand in waiting for b).

My attempt is:

The basis of S of (P_3, R) is (x^2, x, 1).

According to the definition of dual basis, for example the dual basis of the vector (1 0) is (1 0)^T.

Thus, the dual basis of (x^2, x, 1) is (x^2, x, 1).

HallsofIvy
Homework Helper
The three given functionals are $\phi_{-1}(f)= f(-1)$, $\phi_0(f)= f(0)$, $\phi)_1(f)= f(1)$. A "dual basis" would be functions

$f-1(x)$ such that
$\phi_{-1}(f_{-1})= f_{-1}(-1)= 1, [math]\phi_{0}(f_{-1})= f_{-1}(0)= 0, [math]\phi_1(f_{-1})= f_{-1}(1)= 0 [math]f0(x)$ such that
$\phi_{0}(f_0)= f_{0}(-1)= 0, [math]\phi_{0}(f_0)= f_0(0)= 1, [math]\phi_1(f_0)= f_0(1)= 0 [math]f1(x)$ such that
[math]\phi_{0}(f_1)= f_{1}(-1)= 0,
[math]\phi_{0}(f_1)= f_1(0)= 0,
[math]\phi_1(f_1)= f_1(1)= 1

Since we are talking about polynomials of degree 2 or less, this reduces to
1) Find $f_{-1}(x)= ax^2+ bx+ c$ satisfying
a) $f_{-1}(-1)= a- b+ c= 1$
b) $f_{-1}(0)= c= 0$
c) $f_{-1}(1)= a+ b+ c= 0$

2) Find $f_{0}(x)= ax^2+ bx+ c$ satisfying
a) $f_{0}(-1)= a- b+ c= 0$
b) $f_{0}(0)= c= 1$
c) $f_{0}(1)= a+ b+ c= 0$

3) Find $f_{1}(x)= ax^2+ bx+ c$ satisfying
a) $f_{1}(-1)= a- b+ c= 0$
b) $f_{1}(0)= c= 0$
c) $f_{1}(1)= a+ b+ c= 1$

Fixing the formatting of HallOfIvy:
$$f_{-1}(x)$$ such that
$$\phi_{-1}(f_{-1})= f_{-1}(-1)= 1,$$
$$\phi_{0}(f_{-1})= f_{-1}(0)= 0,$$
$$\phi_1(f_{-1})= f_{-1}(1)= 0$$

$$f_{0}(x)$$ such that
$$\phi_{-1}(f_0)= f_{0}(-1)= 0,$$ # I fixed -1 here for phi
$$\phi_{0}(f_0)= f_0(0)= 1,$$
$$\phi_1(f_0)= f_0(1)= 0$$

$$f_{1}(x)$$ such that
$$\phi_{-1}(f_1)= f_{1}(-1)= 0,$$ # I fixed -1 here for phi
$$\phi_{0}(f_1)= f_1(0)= 0,$$
$$\phi_1(f_1)= f_1(1)= 1$$

Since we are talking about polynomials of degree 2 or less, this reduces to
1) Find $f_{-1}(x)= ax^2+ bx+ c$ satisfying
a) $f_{-1}(-1)= a- b+ c= 1$ # I am not sure should this be 1
b) $f_{-1}(0)= c= 0$
c) $f_{-1}(1)= a+ b+ c= 0$

2) Find $f_{0}(x)= ax^2+ bx+ c$ satisfying
a) $f_{0}(-1)= a- b+ c= 0$
b) $f_{0}(0)= c= 1$
c) $f_{0}(1)= a+ b+ c= 0$

3) Find $f_{1}(x)= ax^2+ bx+ c$ satisfying
a) $f_{1}(-1)= a- b+ c= 0$
b) $f_{1}(0)= c= 0$
c) $f_{1}(1)= a+ b+ c= 1$
Reformatting what HallsOfIvy says

$$\begin{tabular}{ | f_-1(x) | f_0(x) | f_1(x) | } & f_{-1}(x) & f_0(x) & f_1(x) \\ \hline \phi_{-1} & 1 & 0 & 0 \\ \phi_0 & 0 & 1 & 0 \\ \phi_1 & 0 & 0 & 1 \\ ax^2 + bx + c: x=-1 & a -b + c = 1 & c = 0 & a + b + c = 0 \\ ax^2 + bx + c: x =0 & - - 0 & - - 1 & - - 0 \\ ax^2 + bx + c: x=1 & - - 0 & - - 0 & - - 1 \\ \end{tabular}$$

I did not write variables to the last two rows again.

I am not sure about the #4 line.
a -b + c =1? # should it be 0.