In Morton L. Curtis's book Matrix Groups, it is stated that the orthonormality of the rows of a quaternionic matrix implies the orthonormality of the columns (Chapter 2 Proposition 4).

However, as the quaternion multiplication is non-commutative, I cannot see why this is true. All I could find is that the orthonormality of the rows of quaternionic matrix $A$ implies that the orthonormality of the columns of $\bar{A}$ which can be observed quite easily. Thus the question reduces to this:

If $x,y$ are orthonormal vectors with quaternionic coefficients, can we say that $\bar{x},\bar{y}$ are orthonormal?

  • $\begingroup$ Hmm. We define inner products to be linear in one argument and conjugate-linear in the other. (My preference: matrices go on the left of vectors, we use right quaternionic vector spaces with right scalar multiplication, and inner products are conjugate-linear in the first argument.) If $A^{\dagger}$ denotes conjugate-transpose and $\overline{A}$ just conjugate, then my inner product for column vectors is $x^{\dagger} y$. For row vectors, $xy^{\dagger}$ is not an inner product (it's conjugate-linear in the second argument, not the first), so would we use $\overline{xy^{\dagger}}$ instead? $\endgroup$ – arctic tern Feb 27 at 5:50
  • $\begingroup$ Thanks for commenting, first of all. Our professor and accordingly the book proceed with row vectors always, so we carry multiplication with matrices on the right of these row vectors, preserving the linearity in non-commutative fields. Consequently, our definition of inner product of (row vectors) $x,y$ is different than what you have written, $\langle x | y \rangle = \sum_i x_i\bar{y_i}$ As you said, if we use a second product for column vectors, as you did with row vectors, the result is clear, yet these would be two different inner products due to non-commutativity of quaternions. $\endgroup$ – sbakirtas Feb 27 at 12:39

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