# Prove that matrix A have inverse

Let $A \in M_{n}(\mathbb{R})$ such that

$$A^2+ \alpha A+\beta I_{n}=0$$

where $\space \alpha, \beta \in \mathbb{R} \space$and$\space \beta \neq 0$.

Prove that $A$ have inverse, and find it.

I done this way:

$A^2+\alpha A+\beta I_{n}=0$

$(A+ \alpha)A=-\beta I_{n}$

$-\frac{1}{\beta}(A+\alpha)A=I_{n} \space$ (because $\beta \neq0$)

$(-\frac{1}{\beta}A-\frac{\alpha}{\beta})A=I_{n}$

So I've assumed that $(-\frac{1}{\beta}A-\frac{\alpha}{\beta})$ is the inverse of A, because its product with A is $I_{n}$

This is correct? Thanks

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Yes it is correct. –  awllower Jan 17 '13 at 18:10
Yes, you proved a left inverse. The question as stated does not specify the difference. If you know already that a left inverse is also a right inverse (hence just inverse) then you have done well. –  adam W Jan 17 '13 at 21:34

## 3 Answers

This is not really an answer. It is a discussion of Git Gud's answer.

My immediate reaction to Git Gud's answer was: That's ridiculous! You don't need all that! And then the comments got bogged down in an argument about whether the matrices had to be square and so on.

But that is not the point. The point is that the OP's proof $-$ that $(-\frac{1}{\beta}A-\frac{\alpha}{\beta})A=I_{n} -$ can be trivially modified to show that $A(-\frac{1}{\beta}A-\frac{\alpha}{\beta})=I_{n}$. So Git Gud's answer really was needlessly complicated $-$ determinants are unnecessary here.

Perhaps my comments were off the point. If so, I hope I have made myself clearer.

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Ok, my lemma was unnecessary. A shorter proof exists. But what the OP did by itself doesn't prove that the matrix we expect to be $A^{-1}$is in fact the inverse of $A$, so I made that point clear the way I saw fit. Plus my argument is universal so there's something to be gained from it, I believe. –  Git Gud Jan 17 '13 at 19:50
Please have a look at mine. Is it proper for this question? Thanks –  Babak S. Jan 17 '13 at 19:59
@Babak: You don't need determinants! The result is true for any linear transformation on a vector space, whether it has a well-defined determinant or not. (As long as you interpret $I_n$ as the identity transformation.) –  TonyK Jan 17 '13 at 20:56
@TonyK: Yes. In fact $M_n(\mathbb R)$ is a PI. –  Babak S. Jan 18 '13 at 3:10
Thanks for the help.$AB=I_{n}$ is assume to be true. I guess if the product of two matrices is equal to $I_{n}$, then they are inverse of each other, and they are commutative. –  João Jan 20 '13 at 18:07

I am adding a just a small point for the OP that is: Your approach is a bit similar to the equation we encounter during Cayley-Hamilton Theorem. There, we have: $$A^2-tr(A)A+\det(A)I_2=0$$ and when $A$ is invertible so $\det(A)\neq0$ and then we have: $$A^{-1}=\frac{-1}{\det(A)}(A+tr(A)I_2)$$

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Another theorem to the rescue! +1 –  amWhy Feb 14 '13 at 1:53
@amWhy: Thanks for your kind deeds. I don't know what to say, just saying thanks...... –  Babak S. Feb 14 '13 at 6:00

What you did is correct, but if you wanna be picky about it you should prove the following

$\textbf{Lemma}:$ If $M,N$ are square matrices such that $MN=I$, then both $M$ and $N$ are non sigular and $M^{-1}=N$. First we prove that $M$ and $N$ are non singular: From $MN=I$ it follows that $\det (MN)=1$, therefore $det(M)det(N)=1$ and finally $det(M)\neq 0\neq det(N)$. So $M$ has an inverse $M^{-1}$. To prove that $M^{-1}=N$, just multiply by $M^{-1}$ on the left of both sides of the equality $MN=I$.

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If you wanna be picky, you might want to specify square matrices in this Lemma :). –  Erick Wong Jan 17 '13 at 18:21
@Git, that is totally unnecessary! The OP exhibited a matrix $B$ such that $BA = I$, and that is exactly what it means to say that $B$ is the inverse of $A$. –  TonyK Jan 17 '13 at 18:24
@ErickWong Funny, but you're right :) –  Git Gud Jan 17 '13 at 18:24
@TonyK Usually the definition of invertibility requires that the product is the identity, but also the product in the reverse order. That's why I think this is necessary. –  Git Gud Jan 17 '13 at 18:26
I use the the fact that $M$ and $N$ are square matrices when I apply $\det$ to them. The equality $NM=I$ is a consequence of my argument, since after proving that $M^{-1}=N$ I just need to multiply by $M$ on the right to get $NM=I$. Note that I didn't prove that $N=M^{-1}$ by definition, so there is not circularity here. –  Git Gud Jan 17 '13 at 19:21