# Intersection of two subspaces

## Homework Statement

I'm given two subspaces L and K of P2 (R) are given by

L = { f(x) : 19f(0)+f ' (0) = 0 }

K = { f(x) : f(1) = 0 }.

Obtain a non-trivial quadratic n = ax2 + b x +c such that n is element of the intersetion of L and K.

## The Attempt at a Solution

19f(0) =19[a(0)^2+b(0)+c] = 19c
df/dx = 2ax+b...so f'(0) = b. therefore L= {19c+b = 0}.
K = {a+b+c = 0}
let X be an element of both L and K and equate the two equations: So r(19c+b) = s(a+b+c), for r, s reak numbers. 19cr+br =as+bs+cs. c(19r-s)-bs+a(r-s) = 0. I'm stuk from there.. Any help would be appreciated.
Thank you.

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Have you learned Gaussian elimination yet? If so, try using it to solve the system of equations

a + b + c = 0
b+19c=0​

You should have one free variable.

HallsofIvy
Homework Helper

## Homework Statement

I'm given two subspaces L and K of P2 (R) are given by

L = { f(x) : 19f(0)+f ' (0) = 0 }

K = { f(x) : f(1) = 0 }.

Obtain a non-trivial quadratic n = ax2 + b x +c such that n is element of the intersetion of L and K.

## The Attempt at a Solution

19f(0) =19[a(0)^2+b(0)+c] = 19c
df/dx = 2ax+b...so f'(0) = b. therefore L= {19c+b = 0}.
K = {a+b+c = 0}
Yes, those are your two equations.

let X be an element of both L and K and equate the two equations: So r(19c+b) = s(a+b+c), for r, s reak numbers. 19cr+br =as+bs+cs. c(19r-s)-bs+a(r-s) = 0. I'm stuk from there.. Any help would be appreciated.
Thank you.
Where in the world did r and s come from? Since both 19c+ b and a+ b+ c are equal to 0, they are equal to each other. But writing only 19c+ b= a+ b+ c loses the information that they are, individually, equal to 0.

I recommend writing both a and b in terms of c so that you can write $$\displaystyle ax^2+ bx+ c$$ entirely in terms of c. Since you are only asked to find a non-trivial quadratic, choose c to be any non-zero number.

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Check post below...thanks

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Yes, those are your two equations.

Where in the world did r and s come from? Since both 19c+ b and a+ b+ c are equal to 0, they are equal to each other. But writing only 19c+ b= a+ b+ c loses the information that they are, individually, equa; to 0.

I recommend writing both a and b in terms of c so that you can write $$\displaystyle ax^2+ bx+ c$$ entirely in terms of c. Since you are only asked to find a non-trivial quadratic, choose c to be any non-zero number.

HallsofIvy, thanks for pointing that out. I was desperate to use some constants lol. Now, let me see if I can solve it (I'll post the ans shortly). meanwhle, can you please check the solution abve.Cheers

Okay, c=-ax^2-bx.......(I)
But form a+b+c =0 and 19c+b=0, a = 18c and b = -19c. Substituting them into (I) gives:
c =-(18c)x^2-(-19c)x . So 1=-18x^2+19x. Hence g(x) = -18x^2+19x-1. Is that correct? I'm not thinking clearly at the moment....thanks guys.

Okay, c=-ax^2-bx.......(I)
But form a+b+c =0 and 19c+b=0, a = 18c and b = -19c. Substituting them into (I) gives:
c =-(18c)x^2-(-19c)x . So 1=-18x^2+19x. Hence g(x) = -18x^2+19x-1. Is that correct? I'm not thinking clearly at the moment....thanks guys.

I think I still need help..I have a feeling that my soln above was wrong. How about this:
a+b+c =0, so a = -b-c and -19c = b ..
So intersection = [-b-c, -19c, c] = [-b,0,0) +c[-1, -19, 1]. How would I continue from there? Cheers.
Oh wait, b =-19c, so c[19,0,0]+c[1,-19,1] = intersection. So when c = 1, the equation of the intersection would be 18x^2-19x+1...signs were incorrect..
Thanks guys.

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Why do you think your original answer is wrong? Doesn't it satisfy both of the equations?

HallsofIvy
You originally got [/itex]g(x) = -18x^2+19x-1[/itex] but changed to $18x^2-19x+1[/quote] which is just the negative of the other. For any vector space, if v is in the space then so is -v. Why do you think your original answer is wrong? Doesn't it satisfy both of the equations? I was just intamidated by the difference in sign. Yes it satifies both equation..I have just checked. You originally got$g(x) = -18x^2+19x-1[/itex] but changed to [itex]18x^2-19x+1