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Is there a simple way of determining the determinant of a matrix of the following form?

$$ P=\left[x \mid Ax \mid A^2x \mid \cdots \mid A^{(n-1)}x \right] $$

Here $A$ is an $n\times n$ matrix and $x$ is a $n\times 1$ vector.

Can we represent $\det(P)$ as a function of $A$ and $x$?

Michael Hardy
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ozlsn
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  • In the 2-dimensional case $x=(x_0\ x_1)$ and $A=\left(\begin{matrix} a&b\c&d \end{matrix}\right)$ your determinant is given by $cx_0^2+(d-a)x_0x_1-bx_1^2$ or by $\langle x, (\langle(c,d),x\rangle,-\langle(a,b),x\rangle)$ or as the matrix of a quadratic form by $\left(\begin{matrix} c&\frac{d-a}{2}\\frac{d-a}{2}&-b\end{matrix}\right)$. All nice enough, but already in such a low dimension I don't see how to write it without using the entries of $A$. Is that what you meant by "as a function of $A$?" It's a rather strong request for a determinant. – Kevin Carlson Oct 03 '12 at 03:43

1 Answers1

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If $x$ is an eigenvector of $A$ then $Ax = \lambda x$ hence $P = [x|\lambda x | \lambda^2 x| \cdots | \lambda^{n-1}x]$ and clearly $det(P)=0$ since the columns of $P$ are mere scalar multiples of one another. More generally, if any two columns of $P$ are linearly dependent then $det(P)=0$.

James S. Cook
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