# $SU(3)$ acts transitively on $S^5$.

I'm told that $$SU(n)$$ acts transitively on $$S^{2n-1} \subset \mathbb{C}^n$$ by matrix multiplication. Yet I can't find a proof of this anywhere, so I was trying to construct a proof on my own by mimicking a version of the proof that I know for showing $$SO(n)$$ acts transitively on $$S^n$$.

My proof is outlined as follows:

To show transitive, it suffices to show that the point $$x= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \in S^5 \subset \mathbb{C}^3$$ can be taken to any other point $$p=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix} \in S^5 \subset \mathbb{C}^3.$$

A complex matrix lies in $$SU(3)$$ if and only if its columns form an orthonormal basis for $$\mathbb{C}^3$$.

If $$A \in SU(3)$$ has the property that $$Ax=p$$, then $$A$$ must have $$p$$ as its first column. Completing $$p$$ to a basis for $$\mathbb{C}^3$$, then running the Gram-Schmidt algorithm on this basis (starting with $$p$$ in the first step of the algorithm) completes $$p$$ to an orthonormal basis for $$\mathbb{C}^3$$. Then sticking these basis elements in as the columns of $$A$$, with $$p$$ as the first column, yields a matrix in $$U(n)$$ that takes $$x$$ to $$p$$.

This proof in general shows that $$U(n)$$ acts transitively on $$S^{2n-1}$$, but the matrix I have constructed does not necessarily have determinant $$1$$. In the completely analogous proof for showing $$SO(n)$$ acts on $$S^n$$ transitively, the matrix has determinant $$\pm1$$ so if the determinant is $$-1$$, we need only interchange the last two columns to get a matrix with determinant $$1$$.

This proof breaks down here because we only know that the determinant lies on the unit circle in $$\mathbb{C}$$, that is $$|\det(A)|=1$$. Is there some way to produce a matrix with determinant dead equal to $$1$$ with this process? Otherwise, is there another simple proof of transitivity?

You can multiply the last column of your matrix with any complex number of absolut value $$1$$. This does not change the orthonormality of the columns nor the first column (if $$n>1$$), but multiplies the determinant by that number, so we can certainly make the determinant $$=1$$.