I know in general that if a matrix $A$ is positive definite, then there exists a (unique?) square root matrix $B$, which is also positive definite, such that $BB=A$.

Therefore, suppose $A$ is positive definite. It is invertible, and its inverse is also positive definite. Therefore I know there exists $C$ (possibly unique?) so that $CC=A^{-1}$. For simplicity, I will call $C=A^{-1/2}$, so that in this notation $A^{-1/2}A^{-1/2}=A^{-1}$.

My question is, is it also true that $A^{-1/2}AA^{-1/2}=I$? Or is this not necessarily true?

I have posted an attempted solution below -- please let me know what you think.


Edit It occurs to me that at the essence of this question is whether or not $B=C^{-1}$; is this necessarily true? That is, is square root of the inverse of a matrix equal to the inverse of the square root of the matrix? (This also partly depends on whether or not $B$ and $C$ are themselves invertible; would this be true?)

Edit 2 There's now a related topic which asks what kind of matrix decomposition we're using here (and the differences between these different approaches to matrix decomposition. That question can be found here.

  • $\begingroup$ There is a risk of misunderstanding in referring to "the" square root of a matrix, as matrix square roots are typically even less unique than square roots of scalars (where we have a choice of sign in picking a square root). $\endgroup$
    – hardmath
    Oct 23, 2014 at 3:32
  • $\begingroup$ Hi Hardmath! Thanks for your comment. I got the uniqueness fact from here, see #7 under properties: en.wikipedia.org/wiki/Positive-definite_matrix $\endgroup$ Oct 23, 2014 at 4:39
  • 2
    $\begingroup$ Note that in that article "positive definite" assumes (at least for real matrices) that the matrix is also symmetric. Thus a real symmetric positive definite matrix has a unique real symmetric positive definite matrix square root (which is the analog of taking the positive square root of a positive real number). $\endgroup$
    – hardmath
    Oct 23, 2014 at 11:25

2 Answers 2


It is generally true that if $A$ is an $n\times n$ invertible and if $A^{-1}$ has a "square root" $C$, also $n\times n$, such that:

$$ A^{-1} = C^2 $$

then $C^{-1} A^{-1} C^{-1} = I$ holds.

The first fact we need is that since $A$ is invertible, $A^{-1}$ is invertible, and this implies $C$ is invertible. For if not, then there would exist a nonzero vector $x$ in the nullspace of $C$, and $Cx=0$ would imply $A^{-1}x=0$, contradicting the invertibility of $A^{-1}$. Thus $A = (C^2)^{-1} = (C^{-1})^2$.

The second fact we need is that a one-sided inverse of a matrix is a two-sided inverse, so that:

$$ A C^2 = I \; \implies \; C A C = I $$

That is, using associativity of matrix multiplication, the left hand side tells us $(AC)C = I$, so that $C$ is a (right) inverse of $AC$. Thus it must also be a (left) inverse of $AC$, which is what the right hand equation states.

Finally by the same reasoning:

$$ C A C = I \; \implies \; C^{-1} A^{-1} C^{-1} = I $$

In this discussion/proof we have not invoked the symmetry of $A$ nor the uniqueness of a symmetric positive definite square root $C$ for $A^{-1}$, which also would be symmetric positive definite. The reasoning above is correct even if $A$ is not symmetric, and even if $C$ is not positive definite, and relies only on $A^{-1} = C^2$.

  • $\begingroup$ Thank you for the thoughtful answer! What are the conditions under which $A^{-1}$ has a square root $C$ so that $A^{-1}=C^2$? $\endgroup$ Oct 24, 2014 at 0:09
  • $\begingroup$ For real symmetric matrices $A$ you have articulated the necessary and sufficient condition: $A$ must be positive definite. For nonsymmetric matrices we can analyze the conditions in terms of the Jordan normal form and the spectrum (eigenvalues) of $A$. It's worth a new Question if you'd like to pursue that topic. $\endgroup$
    – hardmath
    Oct 24, 2014 at 0:59
  • $\begingroup$ @hardmath towards your "necessary and sufficient condition:" what if $$ C = \pmatrix{0&-1\\1&0}? $$ $\endgroup$ Oct 24, 2014 at 4:15
  • $\begingroup$ @Mathemanic: You have defined $C=A^{-1/2}$, not $C^{-1} = A^{-1/2}$. It is easy to get lost in the exponents, but I believe $CAC = I = C^{-1} A^{-1} C^{-1}$ is correct. $\endgroup$
    – hardmath
    Oct 24, 2014 at 12:18
  • $\begingroup$ @Omnomnomnom: Excellent! If your $C^2$ is computed, then $A = C^{-2} = -I$, which is symmetric but negative definite. I stand corrected. $\endgroup$
    – hardmath
    Oct 24, 2014 at 12:23

We know \begin{equation} A^{-1}=CC \end{equation} Inverting both sides (Is this allowed? Depends on whether $C$ is invertible, which I do not know...), we would have \begin{equation} A=C^{-1}C^{-1} \end{equation} Multiply both sides on the left by $C$, and then on the right by $C$, so we get \begin{equation} CAC=I \end{equation} or, in the "$1/2$" notation, \begin{equation} A^{-1/2}AA^{-1/2}=I \end{equation}


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