(a) Prove that the polynomials pi(x)=∑j=0^n ai j x^j for i=1, …, k are linearly independent if a

(a) Prove that the polynomials pi(x)=∑j=0^n ai j x^j for...

(a) Prove that the polynomials $p_i(x)=\sum_{j=0}^n a_{i j} x^j$ for $i=1, \ldots, k$ are linearly independent if and only if the $k \times(n+1)$ matrix $A$ whose entries are their coefficients $a_{i j}, 1 \leq i \leq k, 0 \leq j \leq n$, has rank $k$. (b) Formulate a similar matrix condition for testing whether another polynomial $q(x)$ lies in their span. (c) Use (a) to determine whether $p_1(x)=x^3-1, p_2(x)=x^3-2 x+4, p_3(x)=x^4-4 x, p_4(x)=x^2+1$, $p_5(x)=-x^4+4 x^3+2 x+1$ are linearly independent or linearly dependent. (d) Does the polynomial $q(x)=x^3$ lie in their span? If so find a linear combination that adds up to $q(x)$.

(a) Prove that the polynomials $p_i(x)=\sum_{j=0}^n a_{i j} x^j$ for $i=1, \ldots, k$ are linearly independent if and only if the $k \times(n+1)$ matrix $A$ whose entries are their coefficients $a_{i j}, 1 \leq i \leq k, 0 \leq j \leq n$, has rank $k$. (b) Formulate a similar matrix condition for testing whether another polynomial $q(x)$ lies in their span.
(c) Use (a) to determine whether $p_1(x)=x^3-1, p_2(x)=x^3-2 x+4, p_3(x)=x^4-4 x, p_4(x)=x^2+1$, $p_5(x)=-x^4+4 x^3+2 x+1$ are linearly independent or linearly dependent. (d) Does the polynomial $q(x)=x^3$ lie in their span? If so find a linear combination that adds up to $q(x)$.

Key Concepts

Span and Linear Combinations

The span of a set of vectors is the set of all possible linear combinations of those vectors. A polynomial lies in the span of a given set if it can be expressed as a linear combination of the polynomials in that set. This concept is often investigated by setting up and solving a system of linear equations whose coefficients are taken from the given polynomial vectors.

Matrix Rank

The rank of a matrix is the number of linearly independent rows or columns it contains. In the context of polynomial vectors, if the matrix of coefficients has rank equal to the number of polynomials, it indicates that the polynomials are linearly independent. Conversely, a rank less than the number of polynomials implies a linear dependency among them.

Matrix Representation of Polynomials

Polynomials can be represented by their coefficient vectors, which allows a collection of polynomials to be organized into a matrix where each row (or column) corresponds to a polynomial’s coefficients. This representation is particularly useful for studying the set's properties using the framework of linear algebra.

Linear Independence

A set of vectors (or polynomials, in this context) is said to be linearly independent if none of them can be expressed as a linear combination of the others. Equivalently, the only solution to the equation c1*p1(x) + c2*p2(x) + ... + ck*pk(x) = 0 is the trivial solution where all the coefficients c1, c2, …, ck are zero.

Polynomial Vector Space

A polynomial vector space consists of all polynomials up to a given degree, and its elements can be treated as vectors whose components are the coefficients corresponding to the polynomial's basis elements (typically powers of x). This perspective allows one to apply linear algebra techniques to study properties such as linear independence and span.

Transcript

00:01 The question states, prove that if the vector space is polynomials of nth degree with real coefficients, and a subspace is polynomials 1, 2, up to k, that is a set of vectors, each of different degree.

00:28 So these are different degrees.

00:32 P1, p2, dot, dot, dot, pk are different degrees.

00:51 Maybe are of different degrees, polynomials of different degrees.

00:55 Then s is linearly independent.

00:59 So let's do what the hint says.

01:01 Assume without loss of generality that the polynomials are orders.

01:06 In descending degree, such that the degree of polynomial 1 is greater than the degree of polynomial 2, which is greater than we're just going to put them in descending order.

01:32 Now, if they are linearly independent, then some constant times polytend, 1 plus some constant times polynomial 2 plus dot dot dot plus some constant times polynomial k must equal 0.

02:15 So looks like we're proving this by contradiction.

02:20 Let p1, p2, p2, p k, p k, b k, b k be d.

02:41 Dependent.

02:46 And if they are dependent, then a constant times each of them, the sum of that is going to be zero.

02:54 But if that were true, then since c1 is the only polynomial, not c1, p1 is the only polynomial that has the highest degree term in it, well then c1, c1, must be zero.

03:23 And now, if p2 has the second highest term, but p1 has already been multiplied by zero, so if it had the same degree term as p2, it was already multiplied by zero, so that's gone...

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