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Let $A$ be an $n-1 \times n$ matrix. The span of the rows of $A$ define a hyperplane in $\mathbb{R}^n$; let $u$ be the unit normal to this hyperplane.

Now, let $x \in \mathbb{R}^{n-1}$, and replace each element $A_{i1}$ in the first column of $A$ with a variable term $A_{i1}cos(\theta) - x_i sin(\theta)$; let $u(\theta)$ be the resulting unit normal (thus, $u(0) = u$).

Suppose we slide $\theta$ up from $0$ continuously, and stop if/when $u_k(0) - u_k(\theta) = \Delta$ for some index $k \in \{1, \dots, n\}$. In terms of $A$, $x$, $k$, and $\Delta$, how far can we slide $\theta$?

EDIT: I've managed to reduce the problem to a possibly simpler one. If we delete column $k$ from the modified matrix, then can we find an analytic form for its determinant? In other words, what is $\det(A_{*,\sim k}, \theta)$?

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1 Answer 1

Let $A_{\star,\tilde{k}}$ be your matrix of interest (the sub-matrix from removing column $k$). Let it have determinant of $d$. Also let $\mathbf{u}^\top$ be the first row of the inverse for the matrix. Call the first column of the matrix $\mathbf{y}$. Then the determinant is $d\mathbf{u}^\top\mathbf{y}$. If $\mathbf{y}$ is altered while keeping the rest of the matrix, then the determinant still has the same form but with the new column in place of $\mathbf{y}$: $$d_{new}=d\mathbf{u}^\top\left(\mathbf{y}\cos\theta-\mathbf{x}\sin\theta\right) = d\left(\cos\theta-\mathbf{u}^\top\mathbf{x}\sin\theta\right)$$ since from the definition of inverse we have that $\mathbf{u}^\top\mathbf{y} = 1$. Thus your variation on $\theta$ (from $0$ to $\pi$) gives a determinant ranging from $d$ to the quantity of $-d\mathbf{u}^\top\mathbf{x}$ where $\mathbf{x}$ is the vector of elements $x_i$.

My notation of $\mathbf{u}$ here is different than your $u(\theta)$. The point is that the determinant of your altered matrix will take the form of a dot product with $\mathbf{u}$ and the altered column of the matrix. This is simply known to me and may be more difficult to show you in proof form, I hope this helps regardless.

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