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.


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 $I+A=A^{-1}$.

  • $\begingroup$ 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. $\endgroup$ – res Nov 18 '12 at 20:36
  • $\begingroup$ @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? $\endgroup$ – Brian M. Scott Nov 18 '12 at 20:44
  • $\begingroup$ @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 $AA^{\prime} = I,\text{ to}\;\; A^{\prime} = A^{-1}$. $\endgroup$ – Namaste Nov 18 '12 at 20:47
  • $\begingroup$ @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.) $\endgroup$ – Brian M. Scott Nov 18 '12 at 20:52
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    $\begingroup$ 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. $\endgroup$ – tst Nov 18 '12 at 21:05

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.

  • $\begingroup$ 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. $\endgroup$ – res Nov 18 '12 at 20:24
  • $\begingroup$ nonpop: I think you just stated what the OP already established. $\endgroup$ – Namaste Nov 18 '12 at 20:26
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    $\begingroup$ @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. $\endgroup$ – nonpop Nov 18 '12 at 20:49

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.

  • $\begingroup$ @mythealias - done, thanks for pointing it out! $\endgroup$ – Namaste Nov 19 '12 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.

  • $\begingroup$ 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. $\endgroup$ – johnny Nov 18 '12 at 21:26
  • $\begingroup$ 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. $\endgroup$ – ncmathsadist Nov 18 '12 at 21:41
  • $\begingroup$ This is a nice proof, wish people would use minimal polynomial more often. $\endgroup$ – Gabriel Romon Apr 11 '16 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


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 $A+A^2 \neq I$


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