# Determine signature of quadratic form based on the determinants of the leading minors

Let $A$ be an $n$ by $n$ Hermitian matrix and let, for $j = 1,2,\dots,n$, $A_j$ be the $j \times j$ submatrix consisting of the entries of $A$ in the first $j$ rows and columns of $A$. Suppose that det$(A_j) \neq 0$ and det$(A_1) > 0.$

Give and prove a rule in terms of the signs of the det$(A_j)$ to determine the signature of the Hermitian form defined by $A$.

By signature, I mean if we diagonalize $A = S\Lambda S^{-1},$ the signature would be $(n_+,n_-,n_0)$ where $n_+,n_-,n_0$ is the number of positive, negative, and zero entries of the diagonal matrix $\Lambda.$

My guess is that the number of submatrices with positive determinant corresponds to $n_+,$ and the number of submatrices with negative determinant corresponds to $n_-,$ and $n_0$ will be zero under the hypothesis.

This intuition comes from the fact that if all diagonal entries are positive, the quadratic form is positive $(x^TAx \geq 0, \forall x),$ and this is also true of a matrix with exclusively positive leading minors.

Providing a proof of this fact though has me stumped. Any suggestions/solutions? Thanks in advance.

• Do you mean $\det(\mathbf A_j)\ne 0$ for all $j=1,2,\dots,n$? If so, $n_0$ is always $0$. – Ignat Domanov Sep 3 '16 at 18:33
• Yes, I mean $j$ ranges through $n.$ Yes, I have guessed that, but I'm hung up on how to prove this. – Merkh Sep 3 '16 at 18:43

Your guess doesn't hold already in the case when $A$ is diagonal as can be seen by considering

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}.$$

What important is the number of sign changes. If we set $\det(A_0) = 1$ and

$$m_{+}(A) = \left| \left \{ 1 \leq j \leq n \, \big| \, \frac{\det(A_j)}{\det(A_{j-1})} > 0 \right \} \right|,\\ m_{-}(A) = \left| \left \{ 1 \leq j \leq n \, \big| \, \frac{\det(A_j)}{\det(A_{j-1})} < 0 \right \} \right|$$

then $m_{\pm} = n_{\pm}$ and $n_0 = 0$. To prove this, argue by induction. Write

$$A_{n \times n} = \begin{pmatrix} \tilde{A}_{(n-1)\times(n-1)} & B_{(n-1) \times 1} \\ B^T & a_{nn} \end{pmatrix}.$$

By induction hypothesis, the signature of the quadratic form associated to $\tilde{A}$ is $(m_{+}(\tilde{A}), m_{-}(\tilde{A}), 0)$. Thus, we can find an invertible $(n-1)\times(n-1)$ matrix $\tilde{U}$ such that $\tilde{U}^T \tilde{A} \tilde{U} = \tilde{\Lambda}$ is diagonal with $m_{+}(\tilde{A})$ entries equal to $+1$ and $m_{-}(\tilde{A})$ entries equal to $-1$. But then

$$\begin{pmatrix} \tilde{U} & 0_{(n-1) \times 1)} \\ 0_{1 \times (n-1)} & 1 \end{pmatrix}^T \begin{pmatrix} \tilde{A}_{(n-1)\times(n-1)} & B_{(n-1) \times 1} \\ B^T & a_{nn} \end{pmatrix} \begin{pmatrix} \tilde{U} & 0_{(n-1) \times 1)} \\ 0_{1 \times (n-1)} & 1 \end{pmatrix} = \begin{pmatrix} \tilde{\Lambda} & \tilde{U}^T B \\ B^T \tilde{U} & a_{nn} \end{pmatrix}.$$

By performing simultaneous row and column operations, this matrix is congruent to the diagonal matrix

$$\begin{pmatrix} \tilde{\Lambda} & 0 \\ 0 & a \end{pmatrix}.$$

We see that if $a > 0$ ($a < 0$) then the signature of $A$ will be $(m_{+}(\tilde{A}) + 1, m_{-}(\tilde{A}), 0)$ ($(m_{+}(\tilde{A}), m_{-}(\tilde{A}) + 1, 0)$). But $\det(A_n)$ has the same sign as the determinant of the matrix above which is $a (-1)^{m_{-}(\tilde{A})}$ and $\det(A_{n-1})$ has the same sign as the determinant of $\tilde{A}$ which is $(-1)^{m_{-1}(\tilde{A})}$ and so the ratio $\frac{\det(A_n)}{\det(A_{n-1})}$ has the same sign as $a$ proving the required result.

• I assume that $A$ is symmetric in my answer but you can replace $T$ with $*$ and the same argument holds. – levap Sep 3 '16 at 19:14
• Very impressive! Thanks a bunch – Merkh Sep 4 '16 at 13:08