(a) Is it true that $A$ must commute with its transpose?

(b) Suppose that the columns of $A$ (considered as vectors) form an orthonormal set.
Is it true that the rows of $A$ must also form an orthonormal set?

(University of California, Berkeley, Linear algebra qualifying exam)

(a) Is it true that $A$ must commute with its transpose?

The answer is no.

We give a counterexample. Let
\[A=\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}.\]
Then the transpose of $A$ is
\[A^{\trans}=\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}.\]
We compute
\[AA^{\trans}=\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}
=
\begin{bmatrix}
2 & -2\\
-2& 4
\end{bmatrix},\]
and
\[A^{\trans}A=
\begin{bmatrix}
1 & 0\\
-1& 2
\end{bmatrix}
\begin{bmatrix}
1 & -1\\
0& 2
\end{bmatrix}
=
\begin{bmatrix}
1 & -1\\
-1& 5
\end{bmatrix}.
\]
Therefore, we see that
\[AA^{\trans}\neq A^{\trans} A,\]
that is, $A$ does not commute with its transpose $A^{\trans}$.

(b) Is it true that the rows of $A$ must also form an orthonormal set?

The answer is yes.

Note that in general the column vectors of a matrix $M$ form an orthonormal set if and only if $M^{\trans}M=I$, where $I$ is the identity matrix. (Such a matrix is called orthogonal matrix.)

Thus, by assumption we have $A^{\trans} A=I$. Let $B=A^{\trans}$.
Then the column vectors of $B$ is the row vectors of $A$. Hence it suffices to show that $B^{\trans}B=I$.

Since $A^{\trans} A=I$, we know that $A$ is invertible and the inverse $A^{-1}=A^{\trans}$.
In particular, we have $A^{\trans} A=A A^{\trans}=I$.

We have
\begin{align*}
B^{\trans}B=(A^{\trans})^{\trans}A^{\trans}=(AA^{\trans})^{\trans}=I^{\trans}=I.
\end{align*}
Thus, we obtain $B^{\trans}B=I$ and by the general fact stated above, the column vectors of $B$ form an orthonormal set.
Hence the row column vectors of $A$ form an orthonormal set.

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