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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 $x$?

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  • $\begingroup$ 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. $\endgroup$
    – Kevin Arlin
    Oct 3, 2012 at 3:43

1 Answer 1

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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 $det(P)=0$.

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  • $\begingroup$ Thanks. I removed the illogical part. I'd delete this all together, but I don't see how. $\endgroup$
    – James S. Cook
    Oct 3, 2012 at 8:17
  • $\begingroup$ If you want to delete it, and you can't find another way, you can flag it for moderator attention, and ask for it to be deleted. $\endgroup$
    – Gerry Myerson
    Oct 3, 2012 at 13:31

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