Solving A^2 + 3A + I = 0: Inverse & -A-3I

[EvLer]

I have not come across this kind of problem, hence, not sure:

A is nxn, such that A^2 + 3A + I = 0
show that A is invertible and A^(-1) = -A-3I

So far I have this:

-A^2 - 3A = I

but now I am not sure if I can factor out A:

(-1)A [A + 3] = I

because I do not think that A + 3 is defined for matrices, since 3 is a scalar and not actually a vector.

Any hints are very much appreciated.

 

[Hurkyl]

3 A = 3 A I

 

[EvLer]

Thanks I will try to work it out.
But just to confirm, it's wrong to factor in this case?
So far I have this solution, even though I factored afterall

(-1) A [A + 3] = I

(-1)A A^(-1) [A + 3] = I A^(-1)

(-1) [A + 3] = I A^(-1)

I^(-1) (-1)[A + 3] = I^(-1) I A^(-1)

-AI - 3I = A^(-1)

 

[Hurkyl]

Factoring is fine for matrices, they satisfy the distributive law:

P(Q + R) = PQ + PR
and
(P + Q)R = PR + QR

 

[EvLer]

Oh, so I can represent 3A as 3 A I and then factor out A and end up with 3 I which is theoretically a matrix?

 

[Hurkyl]

It's not just theoretically a matrix.

 

[Hurkyl]

Anyways, an easier way to do the problem is to simply multiply A and the hypothesized A^-1, and check if you get I. (Don't forget to apply the polynomial identity A satisfies)

However, the work you did would be necessary if you weren't given a guess for A^-1.

 

[EvLer]

Shish! I see it now .
Thanks!

 

[HallsofIvy]

However, you appear to be assuming that A has an inverse in your work which is what you want to prove.

What you can do is simply start with A^2 + 3A + I = 0, write that
as -A^2- 3A= I or (-A-3I)A= A(-A-3I)= I. Now what is the DEFINITION of inverse?

 

[EvLer]

so, then from this by DEFINITION of inverse:
AB=I=BA, where B is inverse, and given A(-A-3I)=I i can conclude that B in this case is (-A-3I).
I hope I am not making assumptions again...
Thank you all.

 

[HallsofIvy]

Yes, of course. The inverse of A is just -A-3I.

 

[mathwonk]

so the general principle is that if A satisfies a polynomial with non zero constant coiefficient then A is invertible. (over a field).

think about a diagonal matrix, with diagonal entries c1,...cn.

Then A satisfies (A-c1)(A-c2)...(A-cn), which has non zero constant term if and only if all the diagonal entries ci were non zero.