In reviewing linear algebra for an exam, I encountered the following problem:

Let $A \in \mathbb{R}^{n\times n}$ be symmetric positive definite. If $x$ is any nonzero vector, show that $$ \det\begin{pmatrix} a_{11} & \cdots & a_{1n} & x_1 \\ \vdots & \ddots & \vdots & \vdots \\ a_{n1} & \cdots & a_{nn} & x_n \\ x_1 & \cdots & x_n & 0\end{pmatrix} < 0$$

I solved the problem using this identity:

$$ \begin{pmatrix}A& B\\ C& D\end{pmatrix} = \begin{pmatrix}A& 0\\ C& I\end{pmatrix} \begin{pmatrix}I& A^{-1} B\\ 0& D - C A^{-1} B\end{pmatrix} $$

Applied to the case above, the original determinant is equivalent to $\det(A)\det(-x^T A^{-1} x)$; since $A$ is symmetric positive definite, $A^{-1}$ is as well, so $x^T A^{-1} x$ is a positive number, and it follows that the expression is negative.

Now to my question: is there another (better) way of solving this problem? My first thought was to approach the problem by induction, but that seemed to be a dead end. I had to look up the above identity which won't be an option for an exam.


1 Answer 1


Call your $(n+1)\times(n+1)$ matrix $B$. If $B$ is singular, then its last column is a linear combination of the first $n$ columns, i.e. $Au=x$ and $x^Tu=0$ for some vector $u$. Thus $x^TA^{-1}x=x^Tu=0$, which is impossible because $A$ is positive definite and $x$ is nonzero.

Therefore $\det(B)\ne0$. If $\det(B)>0$, then all leading principal minors of $B$ would be positive and hence $B$ is positive definite. But this is a contradiction because $B$ has a zero on its diagonal. Therefore we conclude that $\det(B)<0$.

  • $\begingroup$ Very nice. Thank you. $\endgroup$
    – bosmacs
    Apr 24, 2014 at 14:05

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