Prove that the limit of this matrix expression is 0

[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.

 

[anuttarasammyak]

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[5]
$${\displaystyle p_{A}(t)=\det \left(tI-A\right)}$$
where I denotes the n×n identity matrix.
----
Wiki says parameter of characteristic polynomial t a number though you say it B, a matrix. I am confused.

 

[lriuui0x0]

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[5]
$${\displaystyle p_{A}(t)=\det \left(tI-A\right)}$$
where I denotes the n×n identity matrix.
----
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.

 

[fresh_42]

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

 

[pasmith]

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.

 

[lriuui0x0]

Thanks for the help!