# Show that if a square matrix A satisﬁes the equation …then A must be invertible.

(a) Show that if a square matrix $A$ satisfies the equation $A^2 + 2A + I = 0$, then $A$ must be invertible. What is the inverse?

(b) Show that if $p(x)$ is a polynomial with a nonzero constant term, and if $A$ is a square matrix for which $p(A) = 0$, then $A$ is invertible.

What am i supposed to do here? plug a square matrix with a b c d in the problem?.. but then what? and i dont have a clue how to do the second one either...

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– Git Gud Jun 16 '14 at 1:18
"plug a square matrix with a b c d in the problem" No, obviously not. The matrix $A$ is not assumed to be $2\times 2$. – Andrés Caicedo Jun 16 '14 at 1:32

You could use that approach, but it sounds pretty miserable. Rather, consider the fact that

$$I = -A^2 - 2A = A(-A - 2I)$$

For the second part, something essentially the same will work: Move the constant term to the other side and factor out an $A$.

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i dont know if its cuz im extremely tired or not.. but I'm just not understanding the question to even begin with maybe.. i just dont get it – J L Jun 16 '14 at 1:37
I see that you solved for I right there then factored out the A... but how does that Show that a square matrix A satisfies the equation and A is invertible??? – J L Jun 16 '14 at 1:53
@JL: the question states that IF a square matrix $A$ satisfies the equation, then $A$ is invertible. $A$ is invertible if there is a matrix $B$ such that $AB = I$. Hence, since $A(-A-2I) = I$, $A$ is invertible with inverse... – Alex Wertheim Jun 16 '14 at 1:56
but how did you find out that there is a matrix B such that AB=I??? – J L Jun 16 '14 at 2:12
@JL think of it like this: the inverse of a matrix $A$ is the matrix which, when multiplied by $A$, gives you $I$. According to the equation in this answer, what can you multiply by $A$ to get $I$? – David Z Jun 16 '14 at 2:59

If $A$ is not invertible, then $0$ is an eigenvalue of $A$. Thus, $p(0)$ must be an eigenvalue of $p(A) = 0_{n\times n}$. But all of the eigenvalues of $p(A) = 0_{n\times n}$ are $0$. So, we must have $p(0) = 0$. This is a contradiction since $p$ has a non-zero constant term, and so, $p(0) \neq 0$. Therefore, $A$ is invertible, as desired.

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