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How can I find the inverse of $A$ using Cayley Hamilton Theorem?

    A=    0 1 0 0
          0 0 1 0
          0 0 0 1
          1 0 0 0

The Characteristic equation of $A$, I get is $A^4=0$, which implies $A=0$ which is clearly not true. Please help.

Ali
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Animesh Tiwari
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  • Two comments: first of all $A$ seems not a square matrix, how do you evaluate the determinant (and the characteristic polynomial)? Also, if $A$ is a matrix, $A^4=0$ does not necessarily imply $A=0$. – GaC Feb 09 '17 at 07:53
  • You have the wrong characteristic equation. Check your algebra. – quasi Feb 09 '17 at 07:57
  • Also, if it was the case that $A^4=0$ (it's not, but suppose it was), then what could you say about $\text{det}(A)$? Based on your answer to that, what could you say about the invertibility of $A$? – quasi Feb 09 '17 at 08:00
  • @quasi thank you. It was a stupid mistake. So I got A^3=A^(-1). So for inverse I would have to find cube of matrix A right? There is no shortcut around that is there? – Animesh Tiwari Feb 09 '17 at 08:01
  • It's not the worst thing in the world. – quasi Feb 09 '17 at 08:02
  • Follow the previous question of @quasi. Does the inverse always exist? – GaC Feb 09 '17 at 08:03
  • @quasi No not by far. Thanks! – Animesh Tiwari Feb 09 '17 at 08:04
  • A characteristic equation A^n=0 will always imply A=0, because multiplication by a non singular matrix does not change rank of the matrix. – Animesh Tiwari Feb 09 '17 at 08:09
  • Computing $A^3$ is easy. Observe that left-multiplying a vector by $A$ rotates that vector’s components one slot “up,” so performing this three times does ...? – amd Feb 09 '17 at 19:16
  • @amd 3 slots up. Yay, thanks! – Animesh Tiwari Feb 10 '17 at 08:31

2 Answers2

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I'll expand my above comment. Suppose we have a $4$-dimensional square matrix $A$ (which is not the zero matrix) with characteristic polynomial $p(\lambda)=\lambda^4$; as I previously said if $B$ is a matrix, $B^n=0$ does not imply in general that $B$ is the zero matrix. Indeed, Cayley-Hamilton itself provides examples of this fact: take our matrix $A$, we have $p(A)=A^4=0$, while $A$ is non-zero (perform a direct calculation if you're sceptic).

GaC
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Expanding $|xI-A|$ along the first row gives $$x(x^3)+1(-1)=x^4-1$$ for the characteristic polynomial. Thus, $$A^4-I=O,$$ $$\text{so that }\ \ \ \ \ A^{-1}=A^3.$$

David
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