Some Combination of $A$ and $B$ is Positive Definite

This is the general case for the definite matrix pair, and now $B$ may be singular. To be able to handle infinite eigenvalues, it is standard practice [425] to introduce a homogeneous representation of an eigenvalue $\lambda$ by a nonzero pair of numbers $(\alpha,\beta)$:

$$\lambda\equiv\alpha/\beta, \quad\quad \vert\alpha\vert^2+\vert\beta\vert^2>0.$$

When $\beta = 0$, such pairs represent eigenvalue $\infty$, and this occurs when $B$ is singular. Such representations are clearly not unique since $(\xi\alpha,\xi\beta)$ represents the same ratio for any $\xi\ne 0$, and consequently the same eigenvalue. So really a pair $(\alpha,\beta)$ is a representative from a class of pairs that give the same ratio. The difference of two eigenvalues is measured by the chordal metric: for $\lambda = \alpha/\beta$ and $\wtd\lambda=\wtd\alpha/\wtd\beta$,

$$\chi(\lambda,\wtd\lambda)\equiv\chi((\alpha, \beta),(\wtd\al... ...{1 + \vert\lambda\vert^2} \sqrt{1 + \vert\wtd\lambda\vert^2}}.$$ (99)

An equivalent definition for a Hermitian matrix pair $\{A,B\}$ being a definite pair is that the Crawford number

$$\gamma(A,B)\equiv\min_{\Vert x\Vert _2=1}\sqrt{ (x^*Ax)^2+(x^*Bx)^2 } > 0.$$

It can be proved [425] that if $\{A,B\}$ is a definite pair, then

• there is a $\phi\in[0,2\pi)$ such that $B_{\phi}$ is positive definite and $\gamma(A,B)=\lambda_{\min}(B_{\phi})$, the smallest eigenvalue of $B_{\phi}$, where
  

  $$A_{\phi}= A\cos\phi-B\sin\phi, \quad B_{\phi}= A\sin\phi+B\cos\phi.$$ (100)

  
  
  The reader is referred to [86,87] for an algorithm that realizes such a $B_{\phi}$, which in turn yields the lower bound on $\gamma(A,B)$ needed for estimating error bounds below.
• there is an $n \times n$ nonsingular matrix $X$ such that
  

  $$X^*AX =\Lambda_A,\quad X^*BX =\Lambda_B,$$ (101)

  
  
  where
  

  $$\Lambda_A =\diag(\alpha_1,\alpha_2,\ldots,\alpha_n), \quad \Lambda_B =\diag(\beta_1,\beta_2,\ldots,\beta_n).$$ (102)

  
  

The decompositions (5.37) and (5.38) give a complete picture of the underlying eigenvalue problems. In fact, all eigenvalues are given by pairs $(\alpha_i,\beta_i)$ with corresponding eigenvectors $Xe_i$. If, in addition, in (5.37) and (5.38) $\vert\alpha_i\vert^2 + \vert\beta_i\vert^2=1$ for all $i$, then [423]

$$\Vert X\Vert _2\le\frac 1{\sqrt{\gamma(A,B)}},\quad \Vert X^{-1}\Vert _2\le\frac {\Vert(A,B)\Vert _2}{\sqrt{\gamma(A,B)}}.$$ (103)