Question: Show that if a square matrix $A$ satisfies the equation $A^2 + 2A + I = 0$, then $A$ must be invertible.

My work: Based on the section I read, I will treat I to be an identity matrix, which is a $1 \times 1$ matrix with a $1$ or as an square matrix with main diagonal is all ones and the rest is zero. I will also treat the $O$ as a zero matrix, which is a matrix with all zeros.

So the question wants me to show that the square matrix $A$ will make the following equation true. Okay so I pick $A$ to be $[-1]$, a $1 \times 1$ matrix with a $-1$ inside. This was out of pure luck.

This makes $A^2 = [1]$. This makes $2A = [-2]$. The identity matrix is $[1]$.

$1 + -2 + 1 = 0$. I satisfied the equation with my choice of $A$ which makes my choice of the matrix $A$ an invertible matrix.

I know matrix $A *$ the inverse of $A$ is the identity matrix.

$[-1] * inverse = [1]$. So the inverse has to be $[-1]$.

So the inverse of $A$ is $A$.

It looks right mathematically speaking.

Anyone can tell me how they would pick the square matrix A because I pick my matrix out of pure luck?

  • 1
    $\begingroup$ You don't get to "pick" the square matrix $A$. You have to show that any matrix that satisfies the equation $A^2+2A+I=0$ is invertible. $\endgroup$ – 5xum Mar 2 '14 at 22:49

The inverse of $A$ is almost certainly not $A$. You have $A^2+2A=-I$, or $A(A+2I)=-I$. Multiplying by $-1$ you get $$A(-A-2I)=I$$

Hence the inverse of $A$ is $-A-2I$.

  • $\begingroup$ When you factor out a A, should it not be A(A + 2)? How did the identity matrix get in to the statement? If you multiplied -1 from both side, how come the first A does not become negative? $\endgroup$ – Nicholas Mar 2 '14 at 22:56
  • $\begingroup$ You can't add just 2, because 2 is a scalar and $A$ is a matrix. The answer to your second question is: $-(xy)=(-x)y=x(-y)$. $\endgroup$ – vadim123 Mar 2 '14 at 23:38
  • $\begingroup$ You got a point there. Now I get it. $\endgroup$ – Nicholas Mar 3 '14 at 0:26

$$\det{A} \cdot \det{(A+2I)} =\det{[A(A+2I)]}=\det{(-I)}=\pm 1 \implies \det{A} \neq 0 \iff A \space \text{is invertible}$$

  • $\begingroup$ what is $\left|\cdot\right|$? - edit: ah, it's $\det$ $\endgroup$ – Max Mar 2 '14 at 23:09
  • $\begingroup$ @Max It's a shorthand for the determinant. $\endgroup$ – Emily Mar 2 '14 at 23:09
  • $\begingroup$ @Max It is the Determinant $\endgroup$ – user76568 Mar 2 '14 at 23:09
  • $\begingroup$ How come you need to put the bars over the matrix? It does not seem to be absolute value sign. $\endgroup$ – Nicholas Mar 2 '14 at 23:15
  • $\begingroup$ $|A|$, where $A$ is a square matrix, is the symbol for the determinant of $A$, which is some scalar. $\endgroup$ – user76568 Mar 2 '14 at 23:17

(an alternative to your approach)

if $0=A^{2}+2A+I=\left(A+I\right)\left(A-I\right)$you have that the set $\left\{+1,-1\right\}$ contains all the eigenvalues of $A$. thus $0$ is not an eigenvalue of $A$ and this is equivalent to being invertible.


as LutzL has commented correctly, $A-I$ has to be replaced by $A+I$ and thus $\left\{-1,+1\right\}$ by $\left\{-1\right\}$. The arguments still work.

  • $\begingroup$ This question is before eigenvalues. So I can only use the concepts of square matrix, invertible matrix, identity matrix and zero matrix. $\endgroup$ – Nicholas Mar 2 '14 at 23:00
  • $\begingroup$ ah, what a pitty... but i think vadims approach does it quite nicely. $\endgroup$ – Max Mar 2 '14 at 23:03
  • $\begingroup$ Wrong statement of binomial theorem. $x^2+2x+1=(x+1)^2$, $x^2-1=(x+1)(x-1)$. So the first formula mixes two quite different binomial formulas. $\endgroup$ – LutzL Mar 8 '14 at 7:26
  • $\begingroup$ yes, indeed, thanks. but the argument still works. (edited) $\endgroup$ – Max Mar 8 '14 at 7:28

Here you have to show $A$ is invertible. For this only you have to show $ det (A) \neq 0$.

Now since $A$ satisfies the polynomial equation $ x^{2} + 2x + 1 = 0$ (as $ A^{2} + 2 A + I = 0$ is given), which contains non-zero constant term $1$ so $ det (A) = 1 \neq 0$. Therefore $A$ has an inverse. [Q.E.D]

** Now if you have to find $A^{-1}$, then you can proceed as follows

$A^{2} +2 A + I = 0 $ .......(1)

Since A has an inverse, so multiplying both side of equation (1) by $A^{-1}$ we have

$A^{-1} (A^{2} + 2 A + I) = A^{-1}. 0 \implies A^{-1} . A^{2} +2 A^{-1}. A + A^{-1} .I=0 \implies A + 2I + A^{-1}=0 \implies A^{-1} = -A-2I $


Since $A^2 + 2A + I = 0$, $(A + I)^2 = 0$. Hence $A = -I$, for any size square matrix. Since the $det(-I) = 1$, the inverse exists.

Since $I* I = I$ and $-I * -I = I$

$A = -I * -I = A$

Thus, $A*A = I$, making $A$ its own inverse.

Another solution to this problem is as follows:

$A^2 + 2A = -I$

$A(A+2)= -I$ and thus $A(-A -2I) = I$ (Since the identity is basically 1), and thus $-A -2I$ is the inverse of $A$.

From earlier conclusion $A = -I$, and thus $-(-I) - 2I = -I = A$, so it remains that $A$ is its own inverse.

  • $\begingroup$ $(A+I)^2=0$ does not imply $A=-I$, unless the size of the matrices is $1\times 1$. $\endgroup$ – Arnaud Mortier Feb 9 '18 at 22:27

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