# Given a Characteristic Polynomial of a Matrix…

This question contains three parts. I have already answered the first two. The last part is confusing me.

Suppose $$A$$ is a $$4 \times 4$$ matrix whose characteristic polynomial is $$p(x) = (x - 1)(x + 2)^2(x - 3)$$.

Part (a): Show that $$A$$ is invertible. Find the characteristic polynomial of $$A^{-1}$$.

We have that the roots of a characteristic polynomial are the eigenvalues of $$A$$. That is, $$\lambda = -2, -2, 1, 3$$ are our eigenvalues. The determinant of an $$n \times n$$ matrix is the product of its eigenvalues. Hence, det$$A = 12$$. An $$n \times n$$ matrix is invertible if and only if its determinant is nonzero. Therefore, $$A$$ is invertible.

Since none of the eigenvalues are zero, we have that $$\lambda$$ is an eigenvalue of $$A$$ if and only if $$\frac{1}{\lambda}$$ is an eigenvalue of $$A^{-1}$$. Then, the characteristic polynomial for $$A^{-1}$$ is $$q(x) = (x - 1)(x + 1/2)^2(x - 1/3)$$.

Part (b): Find the determinant and trace of $$A$$ and $$A^{-1}$$.

This is easy since the determinant of an $$n \times n$$ matrix is the product of its eigenvalues and the trace of an $$n \times n$$ matrix is the sum of its eigenvalues.

Part (c): Express $$A^{-1}$$ as a polynomial in $$A$$. Explain your answer.

Not really sure what part (c) is getting at.

• For the part $a$, you did not address the issue of multiplicities... I suggest instead to note that $\det(xI-A)=\det(A)\det(xA^{-1}-I)=\det(A)*(-1/x)^4\det(x^{-1}I-A)$, which does give your result. – Mindlack Jan 6 at 0:49

## 2 Answers

By the Cayley-Hamilton theorem, we have $$(A-1)(A+2)^2(A-3)=0$$, that is, $$A^4-9A^2-4A+12I=0$$. Multiply both sides by $$A^{-1}$$, and be amazed!

• suggest you type in the $I$ for $12 I$ I see, you also wrote the factored version without any $I$ – Will Jagy Jan 6 at 0:49
• You are right, of course. Thanks. – A. Pongrácz Jan 6 at 1:00
• Much appreciated. Is there an approach that does not use Cayley-Hamilton? Or is this the "obvious" way to go about this problem? – Taylor McMillan Jan 6 at 1:07
• Not sure about obvious, but certainly the simplest. – A. Pongrácz Jan 6 at 1:08

Hint for part (c): Cayley-Hamilton. Then multiply by $$A^{-1}$$ and solve for the inverse.

• What tells you to use Cayley-Hamilton? Just the fact that it gives us a polynomial for $A$, and since $A$ is invertible, we should then be able to find a polynomial $A^{-1}$ in terms of $A$? – Taylor McMillan Jan 6 at 1:06