# Linear Systems - Matrix Powers - Determinants

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 $x$?

• 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. Commented Oct 3, 2012 at 3:43

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 $det(P)=0$.

• Thanks. I removed the illogical part. I'd delete this all together, but I don't see how. Commented Oct 3, 2012 at 8:17
• 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. Commented Oct 3, 2012 at 13:31