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Problem. I am given a symmetric matrix \begin{equation} B_t \doteq \begin{bmatrix}2t & -t & -3t & 0 \\ -t & 2t & 2t & 0 \\ -3t & 2t & 10t & 1 \\ 0 & 0 & 1 & 1-t \end{bmatrix} \end{equation} and my task is to discuss its signature $\operatorname{sign}B_t=(s,k)$ as $t$ varies in the real numbers.

One method could be straight-away diagonalization, but ain't nobody got time. Another trick is the method of principal minors, which makes use of this proposition:

Proposition. Let $V$ be a real vector space. Let $q : V \to \mathbb R$ be a quadratic form associated with the symmetric matrix $A$. Call $A_i$ the $i$-th principal minor of $A$ (starting from the upper left or lower right corner ad libitum), and let $\alpha_i = \det A_i$, $\alpha_0 \doteq 1$. Finally, call $k$ the number of sign changes between each $\alpha_i$ and $\alpha_{i+1}$. Then, if $\alpha_i \neq 0$ for all $i$, the signature of $A$ is $$\operatorname{sign}(A) = (n-k,k) $$ where $n$ is the rank of the matrix.

My doubts concern how lengthy and error-prone this method can be in a parameter-dependent situation, and the existence of another criterion (besides the not-always-practical Babylonian method) that would be of use in determining the signature of a matrix when all else fails. I've gone through it anyway:


List the determinants. We compute the determinants of the principal minors of the matrix, starting from the top left, generating the following list of numbers: \begin{equation} \begin{split} \alpha_0(t) &\doteq \color{red}1 \\ \alpha_1(t) &= |2t| = 2t \\ \alpha_2(t) &= \begin{vmatrix} 2t & -t \\ -t & 2t \end{vmatrix} = 4t^2 - t^2 = \color{red}{3t^2} \\ \alpha_3(t) &= (10t)\alpha_2(t) -(2t)\begin{vmatrix} 2t & -t \\ -3t & 2t \end{vmatrix} + (-3t)\begin{vmatrix} -t & -3t \\ 2t & 2t \end{vmatrix} \\ &= (10t)(3t^2) - (2t)(4t^2-3t^2)+(-3t)(-2t^2+6t^2)=30t^3-2t^3-12t^3= \color{red}{16t^3} \\ \alpha_4(t) &=\det B_t = (1-t)\alpha_3(t)-(1)\begin{vmatrix}2t & -t & 0 \\ -t & 2t & 0 \\ -3t & 2t & 1\end{vmatrix} \\ &= (1-t)(16t^3)-(1)\left[(1)\alpha_2(t)-0+0\right]=16t^3-16t^4-3t^2=\color{red}{(-t^2)(16t^2-16t+3)} \end{split} \end{equation}

Discuss the sign of determinants. The method of principal minors consists in counting how many sign changes and how many sign permanences occur as we scroll through the list of determinants. Of course, in our example things are complicated by the fact that the determinants depend on $t$, so we need to discuss their sign beforehand and then we can evaluate the signature on a case-by-case basis.

  • $\alpha_0(t)$ is positive $\forall t$.
  • $\alpha_1(t)$ is negative when $t<0$, vanishes when $t=0$, and is positive when $t>0$.
  • $\alpha_2(t)$ is positive when $t\neq 0$, and vanishes when $t=0$.
  • $\alpha_3(t)$ is negative when $t<0$, vanishes when $t=0$, and is positive when $t>0$.
  • $\alpha_4(t)$ is negative when $t<0$, vanishes when $t=0$, is negative when $0<t<0.25$, vanishes when $t= 0.25$, is positive when $0.25 < t < 0.75$, vanishes when $t = 0.75$, is negative when $t > 0.75$.

Calculate the signature. (almost) Then I get:

  • When $t < 0$, the signature is $(3,1)$.
  • When $0<t<0.25$, the signature is $(1,3)$.
  • When $0.25 < t < 0.75$, the signature is $(4,0)$.
  • When $t > 0.75$, the signature is $(1,3)$.

Singular values of $t$. Here comes the problematic part: we need discuss the signature of the matrix at the values of $t$ that make at least one of the determinants vanish. This means the method of principal minors cannot be applied. In my example, when $t=0$ the matrix simplifies to $$B_0 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$ and I can easily use the Babylonian method: if $q_t$ is the quadratic form associated with $B_t$,

$$ q_{0}(x) = x_4^2 + 2x_3x_4 = (x_4 + x_3)^2 - x_3^2 = - y_3^2 + y_4^2 $$

using the change of variables (change of basis) $x_3 \mapsto y_3$, $(x_3+x_4) \mapsto y_4$. So $B_0$ is congruent to the matrix

$$\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

which means its signature is also $(1,1)$. This is consistent with the fact that the rank and nullity of $B_0$ are both $2$.

However, the matrices $B_{0.25}$ and $B_{0.75}$ are an utter mess. The Babylonian method is impractical, and my notes do not provide any other means of calculating the signature of a matrix. I have thought of applying the following lemma:

Lemma. Let $V$ be an $(n+1)$-dimensional real vector space and $q : V \to \mathbb R$ be the quadratic form associated with the symmetric matrix $A$, having signature $(s,k)$. Let $W$ be an $n$-dimensional subspace of $V$. Denote $q|_W$ the restriction of $q$ to $W$, $B$ the symmetric matrix associated with it, and $(\sigma,\kappa)$ its signature. Then

  • The determinant of $A$ is zero (i.e. $q$ is degenerate) iff $s=\sigma$ and $k = \kappa$;
  • $\det A \det B > 0$ iff $s =\sigma + 1$ and $k = \kappa$;
  • $\det A \det B < 0$ iff $s =\sigma$ and $k = \kappa + 1$.

My intuition is that it could be of use, though I am not exactly sure how to use this information in practice.


The fact that there is no unique, a priori way of determining the signature of a matrix, and the fact that most systems can be very lengthy and error-prone, makes me wonder:

1. Have I been making wrong assumptions in the procedure above? Are there any calculation or conceptual mistakes?

2. Is there a faster and/or safer way to do any one of the steps in the usual methods?

3. Is there a miraculous lemma or theorem concerning matrix signature that could be of help, especially in a parameter-dependent situation?

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    $\begingroup$ Your matrix, written as $B=C+tD$, describes a so-called "pencil" of matrices. The eigenvalues of a pencil of "symmetric" matrices and its inertia have been studied. Connected also is the "generalized eigenvalues problem". Use the keywords "eigenvalues inertia pencil symmetric matrices" $\endgroup$ – Jean Marie Feb 14 '17 at 1:11
  • $\begingroup$ I found a few interesting works, but nothing that could be used e.g. in a test with limited time available. Most of them outlined complicated iterative algorithms that made use of Matlab. Could you redirect me to the paper you had in mind? Thanks $\endgroup$ – giobrach Feb 14 '17 at 10:22
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There is no need to diagonalize the matrix if you only care about the signature. The easiest way is probably to use simultaneous row/column operations (which preserve congruence and thus signature) to transfer your matrix $B_t$ into a congruent matrix $C_t$ which can be analyzed more easily (you can even go all the way and diagonalize it in the sense of diagonalizing quadratic forms). In your case, you have:

$$ B_t = \begin{pmatrix} 2t & -t & -3t & 0 \\ -t & 2t & 2t & 0 \\ -3t & 2t & 10t & 1 \\ 0 & 0 & 1 & 1 - t\end{pmatrix} \xrightarrow{R_2 = R_2 + \frac{1}{2}R_1, C_2 = C_2 + \frac{1}{2}C_1} \begin{pmatrix} 2t & 0 & -3t & 0 \\ 0 & \frac{3}{2}t & \frac{1}{2}t & 0 \\ -3t & \frac{1}{2}t & 10t & 1 \\ 0 & 0 & 1 & 1 - t \end{pmatrix} \xrightarrow{R_3 = R_3 + \frac{3}{2}R_1, C_3 = C_3 + \frac{3}{2}C_1} \\ \begin{pmatrix} 2t & 0 & 0 & 0 \\ 0 & \frac{3}{2}t & \frac{1}{2}t & 0 \\ 0 & \frac{1}{2}t & \frac{11}{2}t & 1 \\ 0 & 0 & 1 & 1 - t \end{pmatrix} \xrightarrow{R_3 = R_3 - \frac{1}{3}R_2, C_3 = C_3 - \frac{1}{3}C_2} \begin{pmatrix} 2t & 0 & 0 & 0 \\ 0 & \frac{3}{2}t & 0 & 0 \\ 0 & 0 & \frac{16}{3}t & 1 \\ 0 & 0 & 1 & 1 - t \end{pmatrix} := C_t. $$

To analyze the signature of $C_t$, it is enough to analyze the signature of the $2 \times 2$ lower matrix

$$ A_t := \begin{pmatrix} \frac{16}{3}t & 1 \\ 1 & 1 -t \end{pmatrix}.$$

We have $$\det(A_t) = \lambda_1(t)\lambda_2(t) = \frac{16}{3}t(1-t) - 1 = -\frac{1}{3}(4t-1)(4t-3), \\ \operatorname{tr}(A_t) = \lambda_1(t) + \lambda_2(t) = \frac{13}{3}t + 1. $$

Hence, if $t < \frac{1}{4}$ or $t > \frac{3}{4}$ the signature of $A_t$ is $(1,1)$ ($\det(A_t) < 0$ so the eigenvalues of $A_t$ have distinct signs). If $t = \frac{1}{4}$ or $t = \frac{3}{4}$ the signature of $A_t$ is $(0,1)$ (one eigenvalue is zero, the other is positive). If $\frac{1}{4} < t < \frac{3}{4}$ then the signature of $A_t$ is $(0,2)$ ($\det(A_t) > 0$ and $t > 0$ so both eigenvalues are positive). Finally, for $B_t$ we get the following cases:

$$ \operatorname{sign}(B_t) = \operatorname{sign}(C_t) = \begin{cases} (3,1) & t < 0, \\ (1,1) & t = 0, \\ (1,3) & 0 < t < \frac{1}{4}, \\ (0,3) & t = \frac{1}{4}, \\ (0,4) & \frac{1}{4} < t < \frac{3}{4}, \\ (0,3) & t = \frac{3}{4}, \\ (1,3) & t > \frac{3}{4}. \end{cases} $$

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  • $\begingroup$ I have assumed that in your notation $(s,k)$ for the signature, $s$ counts the number of negative eigenvalues and $k$ counts the number of positive eigenvalues. $\endgroup$ – levap Feb 14 '17 at 20:47
  • $\begingroup$ Actually, the opposite. But your answer has been very helpful anyway! $\endgroup$ – giobrach Feb 14 '17 at 22:25
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    $\begingroup$ You cannot use only row operations (or only column operations), you must use simultaneous row and column operations. The reason is that if you perform a row operation you move from $B$ to $PB$ but if you perform a simultaneous row and column operation you move from $B$ to $PBP^{T}$ which is congruent to $B$. The signature is not invariant under row operations but it is invariant under simultaneous row and column operations. $\endgroup$ – levap Feb 14 '17 at 23:56
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    $\begingroup$ Of course. And it wouldn't make sense to calculate the signature of a non-symmetric matrix. $\endgroup$ – giobrach Feb 15 '17 at 0:07
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    $\begingroup$ @giobrach: I actually wrote it in a comment. An elementary row operation corresponds to multiplying $B$ by an elementary matrix $P$ from the left resulting in $PB$. If you follow this by the same column operation, you move from $PB$ to $PBP^T$ (elementary column operations corresponding to multiplying by elementary matrices from the right) which by definition is congruent to $B$. $\endgroup$ – levap Feb 22 '17 at 8:50

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