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Suppose that $x \in \mathbb R^n \,\, (n \ge 2)$ is fixed and define a matrix $A = (a_{ij})_{i,j=1}^n$ by $$a_{ij} = (1+\lvert x \rvert^2)\delta_{ij} - x_i x_j$$ where $\lvert x \rvert$ denotes the standard Euclidean norm and $\delta_{ij}$ is the Kronecker delta (i.e., $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ if $i \neq j$). Show that $\det(A) = (1+\lvert x \rvert^2)^{n-1}$.

This is not a homework question I found this reading through a book on elliptic PDE (we need to calculate this determinant when proving a version of the maximum principle for the prescribed mean curvature equation) and was simply trying to verify the calculation and I'm finding it very difficult. My first thought was to use induction since it is trivial when $n = 2$, but I could find a clean way to relate the $n$ and $n+1$ cases.

User8128
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1 Answers1

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Let us write $$A= (1+|x|^2) I_n - {\bf x} {\bf x}^t$$ with the column vector ${\bf x}= (x_1, x_2, \dots, x_n)^t$.

We have that $$\det A = \det [(1+|x|^2) I_n - {\bf x} {\bf x}^t] = (1+|x|^2)^n \det (I_n - (1+|x|^2)^{-1} {\bf x} {\bf x}^t] .$$

We have that $$\det A = (1+|x|^2)^n \det [I_1 - (1+|x|^2)^{-1} {\bf x}^t {\bf x} ] = (1+|x|^2)^n [1- |x|^2/(1+|x|^2)] = (1+|x|^2)^{n-1} $$ by Sylvester's determinant identity.

Fabian
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  • Great answer! I totally forgot about Sylvester's identity. – User8128 Nov 02 '16 at 16:19
  • In this case, you don't even need it. As $A$ is a rank 1 correction to the identity matrix (times $1+|x|^2$), you know that the eigenvalues $1+|x|^2$ ($n-1$-times) and $1+|x|^2 - |x|^2$ (1-time). – Fabian Nov 02 '16 at 16:24