# Proving $A$ is invertible if $A + A^2 = I$.

I'm trying to prove $A$ is invertible by proving there is an $A'$, for $AA' = I$.

So I got to this stage $A(I + A) = I$, now I determine that $A' = I + A$, and from that I get $AA' = I$.

I wanted to know if this is valid, casue it doesn't seem to make sense,and I can't find a 'real' solution for such equation.

Note: $A$ is $n \times n$ and $I$ represents the identity matrix and is also of the same dimensions.

From $A+A^2=I$ you do indeed get that $AA'=I$ for some $A'$, namely $A'=I+A$. And since the distribution laws hold "on both sides", you also get $A'A=I$ for the same $A'$, which proves that $A$ is invertible.

• I understand your answer, and I understood it's possible too, I just wonder how will it imply using 'real numbers', and not just theoretically. Nov 18, 2012 at 20:24
• nonpop: I think you just stated what the OP already established. Nov 18, 2012 at 20:26
• @res I don't really understand what you mean by "how will it imply using 'real numbers'". Taking a general, say $2\times 2$ matrix and just calculating is probably very messy. That's the power of abstraction, that we can prove this very cleanly using previous results. For a simple concrete example you can take $\phi=\frac{\sqrt 5-1}2$ and the matrix $A=\left(\matrix{\phi&0\\0&\phi}\right)$ and see that it works. Nov 18, 2012 at 20:49

With $A(I+A)=I$ you’re almost there: this shows that $A$ has a right inverse, $I+A$. But a square matrix has a right inverse if and only if it’s actually invertible, so $A$ is invertible.

If you don’t already know that theorem, you can prove it in a variety of ways. For instance, $$I=I^T=\big(A(I+A)\big)^T=(I+A)^TA^T\;,$$ so $A^T$ has a left inverse. Suppose that $A^Tx=0$. Then $$x=Ix=(I+A)^TA^Tx=(I+A)^T0=0\;,$$ so the null space of $A^T$ is trivial, containing only the zero vector. Therefore $A^T$ is invertible, and its inverse must be $(I+A)^T$. Thus, $I=A^T(I+A)^T=\big((I+A)A\big)^T$, and therefore $$(I+A)A=I^T=I\;,$$ meaning that $I+A$ is also a left inverse of $A$. Since $I+A$ is both a left and a right inverse of $A$, it’s actually $A^{-1}$, and $A$ is invertible.

Added: As tst points out in the comments, I was working much too hard here. We can factor the $A$ out of $A+A^2$ on the right as well as on the left, so not only do we have $A(I+A)=I$, we also have $(I+A)A=I$, and it’s immediate that $I+A=A^{-1}$.

• I did understand that case, sorry for not clarifying that in my original post, But I still don't understand how will this imply for a matrix, lets say of order 2, which is invertible. Nov 18, 2012 at 20:36
• @res: I don’t understand what you mean. If you have a $2\times 2$ matrix $A$ such that $A+A^2=\pmatrix{1&0\\0&1}$ then $A$ is invertible, and $A^{-1}=I+A$; that’s all it says. What more do you think is needed? Nov 18, 2012 at 20:44
• @Brian, I am stating that, having shown that A has a right inverse, since A is square, that right inverse is the inverse of A: i.e., it is also the left inverse (vice versa). I may not have been clear in my comment, so I will delete it. (I shouldn't have referred to the left inverse in my earlier comment as $A^{-1}$, so sorry about that.) It's just that the OP may already be able to move from having found $AA^{\prime} = I,\text{ to}\;\; A^{\prime} = A^{-1}$. Nov 18, 2012 at 20:47
• @amWhy: Yes, and that’s precisely the result that I thought might possibly not yet have been proved in the OP’s course. I thought that the phrasing ‘In case you don’t already know this theorem’ pretty clearly recognized the possibility that the result had been covered. (And yes, I have seen people get that close to the end and not realize that they were there, so that might have been the sticking point even if the result was known.) Nov 18, 2012 at 20:52
• You have $A(I+A)=(I+A)A=I$ so by definition $A^{-1}=I+A$. I think the answer is more complicated than what's needed.
– tst
Nov 18, 2012 at 21:05

We have $1=\det{(I)}=\det{(A+A^2)}=\det{(A(I+A))}=\det({A)}\det{(I+A)}$. Thus $\det({A)}\neq 0$ and $A$ is invertible.

Given: $A + A^2 = I$.

You set out to prove $A$ is invertible by proving there is an $A'$, such that $AA' = I$.

You recognized that $A + A^2 = AI + A^2 = A(I +A) = I.$

So you got to this stage: $A(I + A) = I$. Now you let $A^{\prime} = I + A$, and from that you got $AA' = I$, where $A^{\prime}$ is the right inverse of $A$.

Note that, because addition of matrices is commutative, $$A+A^2 = A^2 + A = (A+I)A = = (I + A)A=I.$$ So by the same strategy we used to show that $(I + A)$ is a right inverse of $A$, it follows that $(I+A)$ is also a left inverse of $A$, and hence is THE unique inverse of $A.$

In general, and as you may already know:

For any square matrix $M$, if $M$ has a right inverse, then the right-inverse is the unique inverse of $M$, and so it is also a left inverse of $M$ (and vice versa).

So you did indeed find the inverse of $A$, in having found a right inverse of $A$, namely $A^{\prime} = I + A = A^{-1}$.

Therefore, since $A$ has an inverse, $A$ is invertible.

An alternate strategy is to take the determinant of each side of the equation:

$A + A^2 = I.$

Note that $$1 = \det{(I)} = \det{(A + A^2)} = \det{(A(I + A))} = \det{(A)}\det{(I + A)} > 0.$$ So $\det{(A)}$ cannot be zero. Thus, $A$ is invertible.

What might be confusing you is that you are proving IF $A + A^2 = I$, THEN $A$ is invertible, but that is NOT to say that if $A$ is invertible, then it is always the case $A + A^2 = I$.

So you shouldn't expect to find that $A+A^2 = I$ for all invertible matrices $A$, and don't worry if an example of an invertible matrix $A$ for which $A+A^2 = I$ doesn't immediately come to mind.

• @mythealias - done, thanks for pointing it out! Nov 19, 2012 at 13:35

Put $f(x) = x^2 + x - 1$. We have $f(A) = 0$. Since $x = 0$ is not a root of $f$, zero is not an eigenvalue of $A$. Thus, $A$ is invertible.

• Could you elaborate what result the last statement is a consequence of? From what I recall, minimal polynomial of a matrix divides its characteristic polynomial but this wouldn't necessarily imply absence of zero eigenvalues. Nov 18, 2012 at 21:26
• Since this polynomial sends $A$ to zero, the minimal polynomial of $A$, $m_A$ must satisfy $m_A|f$. Every eigenvalue must be a root of the minimal polynomial. Since $f$ does not have 0 as a root, and $m_A|f$, 0 is not a root of $f$ either. Nov 18, 2012 at 21:41
• This is a nice proof, wish people would use minimal polynomial more often. Apr 11, 2016 at 20:31

Hint $\$ Over any ring: polynomial $\rm\,f(a) = 0\:$ and $\rm\:f(0)\:$ invertible $\rm\:\Rightarrow\: a\:$ invertible (with two-sided inverse, if $\rm\:a\:$ commutes with all coefficients $\rm\,c_i\,$ of $\rm\,f),\,$ since

$$\rm\:c_n a^n + \cdots + c_1 a + c_0 = 0\,\ \Rightarrow\,\ (c_n a^{n-1}+\cdots + c_1) a\, =\, -c_0$$

thus left-multiplying by $\rm\:-c_0^{-1}$ yields a left-inverse for $\rm\:a.\:$ When $\rm\:a\:$ commutes with the coefficients, we can commute $\rm\:a\:$ to the left, showing that the above inverse is a two-sided inverse (in particular such commutativity is true if the coefficients are integers, so universally commutative).

Prove if $A+A^2 = I$ then $A$ is invertible

Contrapositive

If $A$ is singular then $A+A^2 \neq I$

we have $detA = 0$ and $det|A+A^2| = det|A|det|A+I| = 0$

And $detI = 1$

So $A+A^2 \neq I$