From what I understand, the inverse of a matrix only exists if the matrix is square. I recently learned however that the inverse of a quaternion is the quaternion vector (1xn dimensions) where each element has been divided by the length of the vector squared. In other words: $\displaystyle q^-=(\frac{a}{|a^2|+|v^2|},\frac{v}{|a^2|+|v^2|})$
I am struggling to rectify this difference in what the inverse is and I really can't see an intuitive answer as to what the inverse of the quaternion is. Can someone rectify this confusion?

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    $\begingroup$ Do you understand what the inverse of a complex number is? $\endgroup$ Jul 27, 2015 at 15:50
  • $\begingroup$ No, but after having looked, I see that is the definition of the inverse of complex numbers. $\endgroup$
    – djax
    Jul 27, 2015 at 15:52
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    $\begingroup$ Also, to clarify, it is necessary that a matrix be square for it to have an inverse, but not sufficient. It must be square and have nonzero determinant to have an inverse. $\endgroup$
    – walkar
    Jul 27, 2015 at 16:13
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    $\begingroup$ Mind that the quaternions can be identified to certain matrices in $M_2(\Bbb C)$. $\endgroup$
    – AdLibitum
    Jul 27, 2015 at 16:22
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    $\begingroup$ Quaternion multiplication is not matrix multiplication of the corresponding vectors, as that's undefined. $\endgroup$
    – Mark S.
    Jul 27, 2015 at 16:24

2 Answers 2


In mathematics in general, if we have some object $\alpha$, we usually define its inverse $\alpha^{-1}$ to be the object that "undoes" $\alpha$. That is to say, $\alpha^{-1}$ returns the identity element when it is combined with $\alpha$.

$$\alpha \alpha^{-1} = identity$$

This identity element is sort of a "do nothing" object, and the form it takes depends on the objects we're talking about and on the operations used to combine them.

So in a sense, $\alpha^{-1}$ does the opposite of whatever $\alpha$ does and gets us back to where we started.

That might sound abstract, so here are some concrete examples:

  • the additive identity for the real numbers is $0$ because when we add it to some number $r$, we just get $r$ back again: $r + 0 = r$. The additive inverse for a real number $r$ is the number that produces $0$ after an addition. Obviously this number is $(-r)$ since $r + (-r) = 0$

  • the multiplicative identity for the real numbers is $1$ because $r \cdot 1 = r$. The multiplicative inverse for a real number $r$ is the number that produces $1$ after multiplication, that is $\frac{1}{r}$ because $r \cdot \frac{1}{r} =1$. Note that $0$ has no multiplicative inverse.

  • the multiplicative identity for $n \times n$ matrices is the $n \times n$ identity matrix $I_n$ with $1$'s on the diagonal and $0$ elsewhere. The multiplicative inverse of a matrix $A$ is a matrix $A^{-1}$ such that $AA^{-1} = I_n$

  • the multiplicative identity for a complex number $z = a+ib$ is $1+i0$ since $z \cdot 1 = z$. The multiplicative inverse is some complex number $w$ such that $zw = 1$. You can show that $w = \frac{1}{z} = \frac{z^*}{zz^*} = \frac{z^*}{\|z^{}\|^{2}}$

Finally, quaternions...

  • the multiplicative identity for a quaternion $q = a+ib+jc+kd$ is $1+i0+j0+k0$ since $q1 = q$. The multiplicative inverse is some quarternion $x$ such that $qx = 1$.

Writing a formula for this number $x$ might not be immediately obvious, but you can check that $x = \frac{1}{q} = \frac{q^*}{qq^*} = \frac{q^*}{\|q\|^{2}}$, with $q^* = a-ib-jc-kd$ does satisfy the formula.

  • $\begingroup$ So really we're dealing with a "dot product inverse" and not a multiplicative inverse, right? $\endgroup$
    – djax
    Jul 27, 2015 at 17:49
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    $\begingroup$ @djax The word "multiplicatoin" can mean different things in different cases. "Matrix multiplication" is not the same as "quaternion multiplication". Matrix multiplication is defined by $AB = \sum_k A_{i,k}B_{k,j}$. Quaternion multiplication is defined by using the ordinary FOIL rules with this multiplication table for figuring out the $i,j,k$ bits. If you like thinking of quaternions as 4x1 matrices, then "quaternion multiplication" would be a special non-standard multiplication rule for 4x1 matrices. $\endgroup$
    – eigenchris
    Jul 27, 2015 at 17:56
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    $\begingroup$ This is why I find the representaiton (a,v) misleading because it masks the fact that we are really just using an imaginary polynomial $\endgroup$
    – djax
    Jul 27, 2015 at 18:48
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    $\begingroup$ @djax I would agree. There are times when it is useful to think of a quaternion as having two parts, a "real"/"scalar" part $a$ and an "imaginary"/"vector" part $(b,c,d)$. But when learning about them for the first time and multiplying them out on paper, it's best to just think $q = a +bi + cj + dk$. $\endgroup$
    – eigenchris
    Jul 27, 2015 at 18:52

If $\alpha = a+bi+cj+dk$ is a quaternion, $\alpha^{-1} = \frac{\overline{\alpha}}{\|\alpha\|}$ where $\overline{\alpha} = a-bi-cj-dk$ and $\|\alpha\| = a^2+b^2+c^2+d^2$.

This makes intuitive sense when you understand some basic properties of the norm, as $\alpha \overline{\alpha} = \|\alpha\|$ (this is easily, if tediously, confirmable by calculation) and so $\alpha \cdot \left( \frac{\overline{\alpha}}{\|\alpha\|} \right)= 1$.

  • $\begingroup$ Since conjugation reverses the order of multiplication, you immediately get that $\overline{\alpha\overline\alpha}=\overline{\overline\alpha}\,\overline\alpha=\alpha\overline\alpha$, i.e., $\alpha\overline\alpha$ is real. This reduces the calculation to just checking the real part, which is not tedious at all. $\endgroup$ Nov 3, 2022 at 10:59

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