Here is something that always bugged me, and I think I have an explanation for it now, but I am wondering if it is correct. Alright, the problem to me was that back when I was in Diff-eq, to use undetermined coefficients with polynomials, we would always group together the terms on one side, and set the undetermined coefficients equal on the other side, as such:

2(x^2)+3x+5=a(x^2)+bx+c

would imply that a=2, b=3 and c=5

However, this method always seemed a little uncertain to me, and when I would ask about why we could set the coefficients equal, I was always given a handwaving argument or "doesn't it seem reasonable?"

Today, I came across an exercise in my linear algebra book that I think may justify this method; the exercise says "prove that if

The proof in the forward direction (all I will prove):

Since a1*

subtract the right side of the equation to end up with

(a1-b1)*

and since these vectors are linearly independent, each ai-bi=0, so ai=bi for i=1,...,n

Now, looking at the space Pn of polynomials of degree n, the set {1,x,(x^2),...,(x^n)} forms a basis, so each of these vectors must be linearly independent. That means that, as above, if we have

2(x^2)+3x+5=a(x^2)+bx+c

Since the set vectors {1,x,(x^2)} is linearly independent, one can set the coefficients equal.

Am I correct about using this theorem to justify this method of setting coefficients equal? I know it seems trivial, but if this is correct, someone who doesn't really understand why this works could try this method on a set of linearly dependent vectors, for which it isn't necessarily true!

2(x^2)+3x+5=a(x^2)+bx+c

would imply that a=2, b=3 and c=5

However, this method always seemed a little uncertain to me, and when I would ask about why we could set the coefficients equal, I was always given a handwaving argument or "doesn't it seem reasonable?"

Today, I came across an exercise in my linear algebra book that I think may justify this method; the exercise says "prove that if

**v1,...,vn**are linearly independent vectors in the vector space U and each ai,bi belongs to the field F then a1***V1**+...an***Vn**=b1***V1**+...bn***Vn**[tex]\leftrightarrow[/tex] a1=b1,...,an=bn withThe proof in the forward direction (all I will prove):

Since a1*

**V1**+...an***Vn**=b1***V1**+...bn***Vn**[tex]subtract the right side of the equation to end up with

(a1-b1)*

**V1**+...(an-bn)***Vn**=0and since these vectors are linearly independent, each ai-bi=0, so ai=bi for i=1,...,n

Now, looking at the space Pn of polynomials of degree n, the set {1,x,(x^2),...,(x^n)} forms a basis, so each of these vectors must be linearly independent. That means that, as above, if we have

2(x^2)+3x+5=a(x^2)+bx+c

Since the set vectors {1,x,(x^2)} is linearly independent, one can set the coefficients equal.

Am I correct about using this theorem to justify this method of setting coefficients equal? I know it seems trivial, but if this is correct, someone who doesn't really understand why this works could try this method on a set of linearly dependent vectors, for which it isn't necessarily true!

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