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Could anyone guide along with this question?

I was trying $(A−I)(A−I)^{−1} = I$ and was figuring if there's a way out to expand $(A−I)^{−1}$. I also tried $(A−I)x=0$ but to no avail.

matthew.j
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  • I was trying $(A-I)(A-I)^{-1}$ = I and was figuring if there's a way out to expand $(A-I)^{-1}$. I also tried $(A-I)x = 0$ but to no avail. – matthew.j Feb 10 '16 at 14:02
  • Please edit your question to include those efforts. See this guide for how to format properly: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Bobson Dugnutt Feb 10 '16 at 14:03
  • We are told that $A^3 = 2I$. See if you can wrestle out $(A-I)\cdot B = I$ from that somehow, for some $B$. – Arthur Feb 10 '16 at 14:03
  • So is it to get something like: $A^{3} = (A-I)(...)(...)$? – matthew.j Feb 10 '16 at 14:10

3 Answers3

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For excercises like this it is always a good idea to start with the classical expansions. In this case: $$ A^3-I=(A-I)(A^2+A+I) $$ We can then write: $$ (A-I)(A^2+A+I)=A^3-I=2I-I=I $$ By comparison with $(A-I)(A-I)^{-1}=I$ we can find that $$ (A-I)^{-1}=(A^2+A+I) $$

N74
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See from the relation you have given. $f (x)=x^3-2$ is a polynomial that annihilates the operator.hence the minimal polynomial must divide this polynomial .and hence as you can see that the eigenvalues are.cube root of 2....i hope this hepls...proceed on these lines

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Of course, in this case, we can just immediately see the inverse after using the factorization of $x^3-1$.

But I think it is good to know the general method:

Let $A$ be a matrix with $f(A)=0$ for some polynomial $f$ and $g$ some other polynomial, which is co-prime to $f$. The euclidean algorithm (so you actually have a general method at hand to compute it!) yields polynomials $a,b$ with

$$af+bg=1,$$ hence we have $$\mathbb I=a(A)f(A)+b(A)g(A)=b(A)g(A).$$

We have shown: $g(A)$ is invertible and the inverse is $b(A)$.

MooS
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  • would you mind elaborating a bit? because this is my first course in linear algebra. if it's too much trouble, it's ok. i'll look up google on some of those terms. – matthew.j Feb 10 '16 at 14:25
  • Have you ever seen polynomials evaluated at matrices? Or even the polynomial ring? If not, it is probably not the time to understand this method now. You will encounter this soon, at least definitely in your second course in linear algebra. And it is really no big deal. – MooS Feb 10 '16 at 14:28
  • ok thanks for the heads up! – matthew.j Feb 10 '16 at 14:34