1. ## Working with Inverses

Hi MHF,
I'm having heaps of trouble working out how to solve this question. The text book that I am using is the third addition of Lay's Linear Algebra book and I cant find any examples on this type of question..

Suppose A is a square matrix which satisfies the matrix equation
$\displaystyle A^4+5A^2-5I=0$
Determine whether A is invertible or not. If it is invertible, then provide an expression for the inverse of A.

2. Originally Posted by Robb
Hi MHF,
I'm having heaps of trouble working out how to solve this question. The text book that I am using is the third addition of Lay's Linear Algebra book and I cant find any examples on this type of question..

Suppose A is a square matrix which satisfies the matrix equation
$\displaystyle A^4+5A^2-5I=0$
Determine whether A is invertible or not. If it is invertible, then provide an expression for the inverse of A.
It looks like you have the characteristic equation for $\displaystyle A$. So you can get the eigenvalues of A. Their product is non-zero therefore det(A) is non-zero so the inverse of A exists.

Pre-multiply by $\displaystyle A^{-1}$: $\displaystyle A^4+5A^2-5I=0 \Rightarrow A^3 + 5A - 5A^{-1} = 0 \Rightarrow A^{-1} = \frac{1}{5} A^3 + A$

3. My thought was slightly different:
From $\displaystyle A^4+ 5A^2- 5I= 0$ we have $\displaystyle A^4+ 5A^2= 5I$ so that $\displaystyle \frac{1}{5}A^4+ A^2= I$. Then $\displaystyle A(\frac{1}{5}A^3+ A)= (\frac{1}{5}A^3+ A)A= I$. Since that is the definition of "multiplicative inverse of a matrix", A is invertible and $\displaystyle \frac{1}{5}A^4+ A$ is it's inverse.

That has the slight advantage of not requiring solving for the eigenvalues.

4. Originally Posted by mr fantastic
It looks like you have the characteristic equation for $\displaystyle A$. So you can get the eigenvalues of A. Their product is non-zero therefore det(A) is non-zero so the inverse of A exists.

Pre-multiply by $\displaystyle A^{-1}$: $\displaystyle A^4+5A^2-5I=0 \Rightarrow A^3 + 5A - 5A^{-1} = 0 \Rightarrow A^{-1} = \frac{1}{5} A^3 + A$
Be careful, this is not necessarily the characteristic polynomial. The characteristic polynomial can be writen as $\displaystyle p(x)^k$ (Cayley-Hamilton) where $\displaystyle p(x)$ is minimal polynomial for $\displaystyle A$. Since $\displaystyle x^4 - 5x^2 - 5$ solves for $\displaystyle A$ it means $\displaystyle p(x)$ must divide this polynomial. But this polynomial is irreducible by Eisenstein criterion. Thus, the characteristic polynomial for $\displaystyle A$ must be of form $\displaystyle (x^4 - 5x^2 - 5)^k$. Now we see the constant term is non-zero and so $\displaystyle \det (A)\not = 0$.