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If $A$ and $B$ are two real $2\times 2$ matrices with $\det A = 0 $ and $\det B = 0 $ and $\mathrm{tr}(B)$ is non zero. then what will be limit of $$\lim_{t\to0}\frac{\det(A+tI)}{\det(B+tI)}$$ I used the formula $\lambda^2-\mathrm{tr} A+\det A = 0$. then i think answer is $\dfrac{\mathrm{tr}(A)}{\mathrm{tr}(B)}$. Am I correct?

What would be expansion of $\det(A+tI)$ for a $2\times 2$ matrices?

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1 Answer 1

up vote 6 down vote accepted

Let $A$ be an $n \times n$ matrix. Then, one may check that $$\det (A + t I) = t^n + t^{n-1} \operatorname{tr} A + \cdots + t \, \lambda(A) + \det A$$ where $\lambda (A)$ is a certain constant depending on the eigenvalues of the matrix $A$. In the case $n = 2$, $\lambda (A) = \operatorname{tr} A$. So, if $A$ and $B$ are both $n \times n$ matrices, $$\frac{\det (A + t I)}{\det (B + t I)} = \frac{t^n + t^{n-1} \operatorname{tr} A + \cdots + t \, \lambda(A) + \det A}{t^n + t^{n-1} \operatorname{tr} B + \cdots + t \, \lambda(B) + \det B}$$ Thus, if $\det A = \det B = 0$, and $\lambda (B) \ne 0$, $$\lim_{t \to 0} \frac{\det (A + t I)}{\det (B + t I)} = \frac{\lambda (A)}{\lambda (B)}$$ and in the special case $n = 2$ this reduces to exactly what you guessed.

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Thank you sir.. ... –  srijan May 10 '12 at 11:02
    
What if in place of $I$ we have any matrix $B$? –  srijan May 10 '12 at 11:08
2  
If $B$ is invertible then you can reduce to this case. Otherwise, things are complicated. –  Zhen Lin May 10 '12 at 11:09
    
It means in that case we can work out by taking B =I as a particular case? What would be change in the polynomial expression? –  srijan May 10 '12 at 11:12
2  
Use the multiplicative property of the determinant: $\det (A + t C) = \det (C) \det (C^{-1} A + t I)$. –  Zhen Lin May 10 '12 at 11:17

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