This problem is divided into two, of which I am stuck on the second part.

Part (a) wants us to write down a matrix of the linear transformation $T: \mathcal{P}_n(\mathbb{F}) \to \mathbb{F}^{n+1}$ where $p \to (p(a_0), p(a_1), ..., p(a_n))$ where $a_0, a_1, ..., a_n$ are scalars in $\mathbb{F}$. This is all with respect to basis $1, x, ..., x^n$ of $\mathcal{P}_n(\mathbb{F})$ and the standard basis of $\mathbb{F}^{n+1}$. This ends up looking like $\begin{bmatrix} 1 & a_0 & a_0^2 &. & . & . & a_0^n \\ 1 & a_1 & a_1^2 & . & . & . & a_1^n \\ .&.&.&.&.&.&.\\ .&.&.&.&.&.&.\\ 1 & a_n & a_n^2 & .&.&. & a_n^n\\ \end{bmatrix}$

Let's call this $\mathcal{M}(T)$.

From a previous problem, we know that if $a_0, ..., a_n$ are distinct, then the matrix is invertible.

Part (b) lets $a_0, ..., a_n \in \mathbb{R}$ be distinct scalars. Then it gives us functions $$ f_i(t) := e^{a_it}$$ which are elements of $\mathbb{R^{\mathbb{R}}}$. We want to prove that $f_0, ..., f_n$ are linearly independent, with the hint to start with a linear combination of these functions, set them to zero, and differentiate until we have $n+1$ equations and set $t = 0$.

My attempt at a solution (part (b)):

Suppose we have $c_0, ..., c_n \in \mathbb{R}$. Following the hint, we start from $c_0f_0 + c_1f_1 + ... c_nf_n = 0$ and then differentiate to get $n+1$ linear equations that look like $$c_0a_0^if_0 + c_1a_1^if_1 + ... + c_na_n^if_n = 0$$ for $0 \leq i \leq n$. Setting $t = 0$, we get $$c_0a_0^i + c_1a_1^i + ... + c_na_n^i = 0$$ for $0 \leq i \leq n$. If we encode this system in a matrix, we end up with $\begin{bmatrix} 1 & 1& .&.&.& 1 \\ a_0 & a_1 & . & . & . & a_n \\ a_0^2&a_1^2&.&.&.&a_n^2\\ .&.&.&.&.&.\\ .&.&.&.&.&.\\ a_0^n & a_1^n & .&.&. & a_n^n\\ \end{bmatrix}$

with the entries being the coefficients of $c_0, ..., c_n$. This is exactly the transpose of $\mathcal{M}(T)$ (or the matrix of the dual map $T'$).

This is the point where I wonder how to proceed to show that $f_0, ... f_n$ are linearly independent. My original goal is to show that $c_0, ..., c_n$ must all be $0$, but I don't see how the new matrix I created leads to that result. Part of my confusion in this problem is that I don't know how it could relate two matrices that share the same values in each entry but are derived from entirely different vector spaces and bases.

More exactly, my questions:

1) How does this new matrix help in showing $f_0, ..., f_n$ are linearly independent, or how $c_0, ... , c_n$ are all $0$?

2) Are we allowed to say that if the matrix in part (a) ($\mathcal{M}(T)$), for whatever it encodes, is invertible, then its transpose, the matrix in part (b), is also invertible for whatever it encodes? Or do matrix properties like invertibility only hold depending on the bases they are derived from?

3) What exactly does the matrix in part (b) encode? It seems strange to treat the entries of that matrix as coefficients of $c_0, ..., c_n$, which are scalars.


A matrix $A$ of $n+1$ by $n+1$, as you have, can represent a linear map $\mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$ (e.g. both in standard bases, but this does not matter).

$\det(A) \neq 0$ iff $A$ interpreted as a linear map (taking any basis on domain or codimain $\mathbb{R}^{n+1}$ that we like) is invertible as a function.

If $A^t$ is the transposed matrix, then a standard fact is that $\det(A^t)= \det(A)$, so if one is non-zero, so is the other.

In a previous part you've shown (presumably) that $\det(A) \neq 0$ for distinct $a_i$, where $A$ is a so-called Vandermonde matrix

$$\begin{bmatrix} 1 & a_0 & a_0^2 &. & . & . & a_0^n \\ 1 & a_1 & a_1^2 & . & . & . & a_1^n \\ .&.&.&.&.&.&.\\ .&.&.&.&.&.&.\\ 1 & a_n & a_n^2 & .&.&. & a_n^n\\ \end{bmatrix}$$

Now what's happening in part b) is that we're actually working over a function space and the fact that $$c_0f_0+ c_1f_1 + \ldots + c_nf_n = 0\tag{0}$$ where $0$ is the $0$-function, allows us to derive by taking derivatives to derive

$$c_0a_0^i + c_1a_1^i + ... + c_na_n^i = 0$$ for $0 \leq i \leq n$ which is not an equation system in the function space any more but an (easier) equation system in $n+1$ unknowns (the $c_0,c_1,\ldots,c_n$) over $\mathbb{R}^{n+1}$. Your target is still to show all $c_i$ must be $0$ because that is what the proof of linear independence requires.

So you write it as

$$\begin{bmatrix} 1 & 1& .&.&.& 1 \\ a_0 & a_1 & . & . & . & a_n \\ a_0^2&a_1^2&.&.&.&a_n^2\\ .&.&.&.&.&.\\ .&.&.&.&.&.\\ a_0^n & a_1^n & .&.&. & a_n^n\\ \end{bmatrix} \begin{bmatrix}c_0 \\ c_1 \\ c_2 \\ . \\ . \\ c_n \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\ . \\ . \\ 0 \end{bmatrix}\tag{1}$$

Now we have the same matrix $A$ occurring in this auxiliary system, in its transposed form. So its determinant is $\neq 0$ and this means that the only solution to $(1)$ is the zero vector for $(c_0,\ldots, c_n)$ (interpreted as maps: the associated linear map is 1-1, so has trivial null space, e.g.) and that concludes the proof.

As to point 3 I'm suprised you're surprised: matrices always have scalars as entries. But the equation $(0)$ we cannot handle directly, but taking derivatives and $t=0$ we get a system of equations where the scalars are the unknowns, so we reduce to a finite dimensional system, as I explained.

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