Linear Algebra Linear Algebra Math Forum

 March 6th, 2015, 07:11 AM #1 Newbie   Joined: Mar 2015 From: Lubbock, Texas Posts: 5 Thanks: 0 3 parts (A) Is {x(x-1), (2x - 1)^2, 1} a spanning set for P3? (B) Are the matrices 1 1 1 1 0 -1 1 0 2 0 0 2 0 1 -1 -1 linearly independent? (C) It can be shown that the solutions to the differential equation y'' = y are exactly all functions of the form y = ae^x + be^-x, for constants a and b. These solutions form a vector space. Show that cosh(x) = (e^x + e^-x)/2 and sinh(x) = (e^x - e^-x)/2 form a basis for this vector space. March 6th, 2015, 07:55 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra $\mathbb P _3$ is the set of all polynomials $$p(x) = \sum_{k = 0}^3 a_k x^k$$ Since your proposed basis has no terms in $x^3$ it is clearly not a basis of $\mathbb P_3$. It might be a basis of $\mathbb P _2$, the quadratic polynomials. If it is, then any quadratic $p(x) = a_0 + a_1 x + a_2 x^2$ where the $a_k$ are not all zero can be written as a non-zero linear combination of the proposed basis vectors. Writing $b_0(x) = 1$, $b_1(x) = (2x-1)^2 = 4x^2 -4x +1$ and $b_2(x) = x(x-1) = x^2 - x$ we see that any vector $$p b_0(x) + q b_1(x) + r b_2(x) = \big(p\big) + \big(4qx^2 - 4qx + q\big) + \big(rx^2 - rx\big) = (p + 4q) - (4q + r)x + (4q + r)x^2$$ In particular, any linear combination of the $b_k(x)$ is going to have the coefficient of $x$ being equal to minus the coefficient of $x^2$. That is $p(x) = a - bx + bx^2$. So, for example, we can't make $q(x) = 1 + x + x^2$ or $r(x) = 1 + 2x + 3x^2$. March 6th, 2015, 09:07 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,105 Thanks: 2324 For (C), use e^x = cosh(x) + sinh(x) and e^(-x) = cosh(x) - sinh(x). March 14th, 2015, 11:06 AM   #4
Math Team

Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Quote:
 Originally Posted by v8archie $\mathbb P _3$ is the set of all polynomials $$p(x) = \sum_{k = 0}^3 a_k x^k$$
I agree but- some people use "P3" to mean the space of quadratic polynomials. The "3" because it is a vector space of dimension 3.

Quote:
 Since your proposed basis has no terms in $x^3$ it is clearly not a basis of $\mathbb P_3$. It might be a basis of $\mathbb P _2$, the quadratic polynomials. If it is, then any quadratic $p(x) = a_0 + a_1 x + a_2 x^2$ where the $a_k$ are not all zero can be written as a non-zero linear combination of the proposed basis vectors. Writing $b_0(x) = 1$, $b_1(x) = (2x-1)^2 = 4x^2 -4x +1$ and $b_2(x) = x(x-1) = x^2 - x$ we see that any vector $$p b_0(x) + q b_1(x) + r b_2(x) = \big(p\big) + \big(4qx^2 - 4qx + q\big) + \big(rx^2 - rx\big) = (p + 4q) - (4q + r)x + (4q + r)x^2$$ In particular, any linear combination of the $b_k(x)$ is going to have the coefficient of $x$ being equal to minus the coefficient of $x^2$. That is $p(x) = a - bx + bx^2$. So, for example, we can't make $q(x) = 1 + x + x^2$ or $r(x) = 1 + 2x + 3x^2$. Tags parts Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post beesee Probability and Statistics 3 January 20th, 2015 12:12 PM rain Calculus 3 November 8th, 2013 04:08 PM math221 Calculus 5 February 21st, 2013 02:51 AM aaron-math Calculus 1 September 27th, 2012 05:44 PM zell^ Algebra 1 April 28th, 2012 01:29 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      