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For any general $n$, is there any easy way of finding out the inverse of a square matrix of the form:

$$A=van(a_1,a_2,\cdots , a_n)=\pmatrix{ 1 \quad a_1 \quad a_1^2 \ \dots \ a_1^{n-1} \\ 1 \quad a_2 \quad a_2^2 \ \dots \ a_2^{n-1} \\\vdots \qquad \vdots \qquad \vdots \\ 1 \quad a_n \quad a_n^2 \ \dots \ a_n^{n-1}}.$$

Amin235
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Television
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1 Answers1

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When I look at your matrix, I see a matrix that transforms a polynomial in the standard basis to $p(a_0), p(a_1)\cdots p(a_n)$

$\pmatrix{1 \quad a_0 \quad a_0^2 \ \dots \ a_0^n \\ 1 \quad a_1 \quad a_1^2 \ \dots \ a_1^n \\ 1 \quad a_2 \quad a_2^2 \ \dots \ a_2^n \\ \vdots \\ 1 \quad a_n \quad a_n^2 \ \dots \ a_n^n}\pmatrix{1\\x\\x^2\\\vdots\\x^n} = \pmatrix{p(a_0)\\p(a_1)\\p(a_2)\\\vdots\\p(a_n)}$

So, what is the inverse transformation?

The first column would be the coefficients of the polynomial that goes through $(a_0,1), (a_1,0), a_2,0),\cdots,(a_n,0)$

$p(x) = \frac {(x-a_1)(x-a_2)\cdots(x-a_n)}{(a_0-a_1)(a_0-a_2)\cdots(a_0-a_n)}$

And from this you can build your inverse matrix.

In a more simplified example

$\pmatrix {1&0&0\\1&2&4\\1&3&9}$ takes the ploynomial $p(x) = c+bx + ax^2$ to $(p(0),p(2),p(3)$

i.e. if $p(x) = 3 + 2x + x^2$ $p(0) = 3, p(2) = 11, p(3) = 18$

$\pmatrix {1&0&0\\1&2&4\\1&3&9}\pmatrix{3\\2\\1} = \pmatrix{3\\11\\18}$

checks out.

The inverse. we need a polynomial that goes through $(0,1),(2,0),(3,0)$ $p(x) = \frac 16 (x-2)(x-3) = 1 -\frac 56 x + \frac 16 x^2$

$\pmatrix{1\\-\frac56\\\frac 16}$ is the first column of the inverse matrix.

we need a polynomial that goes through $(0,0),(2,1),(3,0)$ $p(x) = -\frac 12 (x)(x-3) = 0 +\frac 32 x - \frac 12 x^2$

and we need a polynomial that goes through $(0,0),(2,0),(3,1)$

$p(x) = \frac 13 (x)(x-2) = 0 -\frac 23 x + \frac 13 x^2$

$\pmatrix {1&0&0\\1&2&4\\1&3&9}\pmatrix{1&0&0\\-\frac 56&\frac 32&-\frac 23\\\frac 16&-\frac 12&\frac 13} = \pmatrix{1&0&0\\0&1&0\\0&0&1}$

Doug M
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    Thanks for the answer. Shouldn't the column vector of $x$'s be the coefficients of $p_i(x)$ (corresponding to the $i^{th}$ column of $\mathcal{A}^{-1}$? I am not sure if this is what you meant by $1, x, x^2 ...$ – Television Jun 22 '17 at 06:09
  • I have provided an example. – Doug M Jun 22 '17 at 16:41
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    Yes, I understand what you mean. In your example, you multiply the Vandermonde matrix by a column vector holding the coefficients of $1,x,x^2...$. Not by $1,x,x^2...$ itself. But you have multiplied the Vandermonde matrix by the column vector $1,x,x^2..$ instead of p(x) coefficients in the first equation. – Television Jun 23 '17 at 04:27