# Prove that the limit of this matrix expression is 0

• I
• lriuui0x0

#### lriuui0x0

Given a singular matrix ##A##, let ##B = A - tI## for small positive ##t## such that ##B## is non-singular. Prove that:

$$\lim_{t\to 0} (\chi_A(B) + \det(B)I)B^{-1} = 0$$

where ##\chi_A## is the characteristic polynomial of ##A##. Note that ##\lim_{t\to 0} \chi_A(B) = \chi_A(A) = 0## by Cayley-Hamilton theorem.

This limit involves the product of a convergent to zero function and a divergent function. I'm not sure how to transform the limit in order to prove this.

In wikipedia characteristic polynomials
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We consider an n×n matrix A. The characteristic polynomial of A, denoted by pA(t), is the polynomial defined by
$$p_{A}(t)=\det \left(tI-A\right)$$
where I denotes the n×n identity matrix.
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Wiki says parameter of characteristic polynomial t a number though you say it B, a matrix. I am confused.

In wikipedia characteristic polynomials
----
We consider an n×n matrix A. The characteristic polynomial of A, denoted by pA(t), is the polynomial defined by
$$p_{A}(t)=\det \left(tI-A\right)$$
where I denotes the n×n identity matrix.
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Wiki says parameter of characteristic polynomial t a number though you say it B, a matrix. I am confused.
Yes, it's passing in the matrix into the polynomial. Please check Cayley-Hamilton theorem.

• anuttarasammyak
I don't know how to tackle the problem, but here are some ideas:
• proof by induction
• replacing ##B^{-1}=\dfrac{1}{A-tI}## by its series
• using the Jordan normal form
• the limit has to be taken for every matrix entry, so maybe choosing a single one could simplify the problem

Let $$\chi_A(x) = \sum_{n=0}^N a_nx^n$$. Then $$(\chi_A(B) + \det(B)I)B^{-1} = (a_0 + \det B)B^{-1} + \sum_{n=0}^{N-1} a_{n+1}B^{n}.$$ Recall that $$B^{-1} = \frac{\operatorname{adj}(B)}{\det(B)}$$ where $\operatorname{adj}(B)$ is the adjugate matrix of $B$, and that $a_0$ vanishes as $A$ is singular. This shows that the limit is finite; there is more to do to show that it vanishes.

EDIT: We also have the identity (stated here) $$\operatorname{adj}(-A) = \sum_{n=0}^{N-1} a_{n+1}A^n$$ and thus $$\lim_{t \to 0} (\chi_A(B) + \det(B)I)B^{-1} = \operatorname{adj}(A) + \operatorname{adj}(-A) = (1 + (-1)^{N-1})\operatorname{adj}(A)$$ which is either $2\operatorname{adj}(A)$ or $0$ depending on whether $N$ is odd or even.

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Thanks for the help!