Let matrix $A$ be normal. If $A{A^T}$ has $n$ distinct eigenvalues, why is $A$ symmetric? Let matrix $A \in {M_n}(\Bbb R)$ be normal. If $A{A^T}$ has $n$ distinct eigenvalues, why is $A$ symmetric?
 A: Here's a late answer.
First, a bit of notation. Since we will work over $\mathbb C,$ we will need the notion of "conjugate-transpose" of a complex matrix. Namely, if $M\in \mathbb C^{n\times n},$ then its "conjugate-transpose" is defined as
$$
M^H = (\overline M)^T,
$$
where $\overline \cdot$ denotes complex conjugation of the matrix entries and $\cdot^T$ denotes matrix transposition. This is a well known operation. Observe that $\overline \cdot$ and $\cdot^T$ commute. And, of course, for $N\in\mathbb R^{n\times n},$ we have $N^H = N^T.$ In the following, we will use only the operation $\cdot^H.$
On to the actual question. So we are given a matrix $A\in\mathbb R^{n\times n}$ which is normal, i.e. $A^HA = AA^H$ and such that $A^HA$ has $n$ distinct eigenvalues. We want to show that $A$ is symmetric, i.e. $A^H = A.$
Observe that, since $A$ is real, its characteristic polynomial has real coefficients. This means that the eigenvalues of $A,$ i.e. the roots of its characteristic polynomial, are either real or come in complex conjugate pairs. In other words, we can enumerate the eigenvalues of $A$ as a sequence
$$
\lambda_1,\ldots,\lambda_r,\mu_1,\overline{\mu_1},\ldots,\mu_s,\overline{\mu_s},
$$
with $r,s\in \mathbb N_0,$ $\lambda_j\in\mathbb R,$ and $\mu_k\in \mathbb C$ with $\Im\mu_k > 0.$ Repetitions are allowed, i.e. we can have $\lambda_j = \lambda_k$ or $\mu_j = \mu_k$ for $j\neq k.$
Now, since $A$ is normal, it can be diagonalized using a unitary transformation. This is a well known fact. In more detail, we have the following. Put
$$
D = {\rm diag}(\lambda_1,\ldots,\lambda_r,\mu_1,\overline{\mu_1},\ldots,\mu_s,\overline{\mu_s})\in \mathbb C^{n\times n},
$$
i.e. $D$ is the diagonal matrix with the eigenvalues of $A$ on the diagonal. Then, there is a unitary matrix $U\in \mathbb C^{n\times n}$ such that
$$
\tag{1}
A = U^HDU.
$$
Recall that "unitary" means
$$
\tag{2}
U^HU = UU^H = I,
$$
where $I \in \mathbb C^{n\times n}$ is the identity matrix. From $(1),$ we get
$$
\tag{3}
A^H = (U^HDU)^H = U^HD^HU.
$$
From $(1),$ $(2),$ and $(3),$ we conclude
$$
A^HA = \left(U^HD^HU\right)\left(U^HDU\right) = U^HD^HDU.
$$
This shows that the eigenvalues of $A^HA$ are the diagonal elements of the dialgonal matrix
$$
\begin{align}
D^HD & = {\rm diag}(\lambda_1^2,\ldots,\lambda_r^2,\overline{\mu_1}\mu_1,\mu_1\overline{\mu_1},\ldots,\overline{\mu_s}\mu_s,\mu_s\overline{\mu_s}) \\
& = {\rm diag}(\lambda_1^2,\ldots,\lambda_r^2,|\mu_1|^2,|\mu_1|^2,\ldots,|\mu_s|^2,|\mu_s|^2).
\end{align}
$$
Observe that every pair $\mu_j,\overline{\mu_j}$ of complex conjugate eigenvalues of $A$ produces twice the same eigenvalue for $A^HA.$ By assumption, this can't happen. So we must have that $s = 0,$ $A$ has only real eigenvalues, and in particular $D\in \mathbb R^{n\times n},$ i.e. $D$ is real. But then
$$
\tag{4}
D^H = D.
$$
Now we get from $(1),$ $(3),$ and $(4)$
$$
A^H = U^HD^HU = U^HDU = A,
$$
as desired.
A: Since $AA^T$ is symmetric ($(AA^T)^T=AA^T$), $AA^T$ has an orthonormal eigenvector base. Since $AA^T$ has $n$ distinict eigenvalues, each eigenvalue only has one eigenvector.
Suppose $P=[u_1,\cdots,u_n] \hspace{2 mm} (u_i$  is the eigenvector for eigenvalue $\lambda_i$ of $AA^T$). Then
$$
P^{-1}(AA^T)P=P^{T}(AA^T)P=\operatorname{diag}[\lambda_1,\cdots,\lambda_n]
$$
Since $A$ is normal
$$
A \hspace{2 mm}\text{and}\hspace{2 mm} AA^T \hspace{2 mm} \text{are commute, for}\hspace{2 mm} A(AA^T)=A(A^TA)=(AA^T)A
$$
Since
$$
A(AA^T)u_i=(AA^T)(Au_i)=A(\lambda_iu_i)=\lambda_i(Au_i)
$$
If $Au_i\neq0$, $Au_i$ is an eigenvector for eigenvalue $\lambda_i$ of $AA^T$. Since there is only one such eigenvector for eigenvalue $\lambda_i$, $Au_i=\mu_i u_i,\space \mu_i$ is an eigenvalue of $A$. Thus $P$ is also an orthonormal eigenvector base for $A$. So there is
$$P^{T}AP=\operatorname{diag}[\mu_1,\cdots,\mu_n] \hspace{2 mm} \text{and}\hspace{2 mm} A=P\operatorname{diag}[\mu_1,\cdots,\mu_n]P^T
$$
$A$ is symmetric for
$$A^T=(P\operatorname{diag}[\mu_1,\cdots,\mu_n]P^T)^T=P\operatorname{diag}[\mu_1,\cdots,\mu_n]P^T=A
$$
A: The matrix $X\in\mathbb R^{n\times n}$ being normal is equivalent to the existence of a unitary matrix $U\in \mathrm U(n)\subset \mathbb C^{n\times n}$ such that $U^* X U=D\in\mathbb C^{n\times n}$ is a complex diagonal matrix. It is symmetric if and only if all eigenvalues of $X$ (the diagonal entries of $D$) are real numbers.
Hence,
$$
X^T X = UD^* U^* UDU^* = U D^* D U^*
$$
and the matrix $X^T X$ is similar to $D^* D$.
Assume that $X$ is not symmetric for sake of contradiction, then $X$ has at least one pair of complex eigenvalues $\lambda=a\pm b\mathrm i$ with $b\neq 0$. But then $D^* D$ has the repeated diagonal entry
$$
(a-b\mathrm i)(a+b\mathrm i) = a^2 + b^2 = (a+b\mathrm i)(a-b\mathrm i),
$$
which can't be the case since $X^T X$ has $n$ distinct eigenvalues by assumption.
A: Since $AA^T$ is a symmetric matrix, it has an orthogonal eigenbasis.
Be $v$ an eigenvector of $AA^T$ with eigenvalue $\lambda$. Then we have
$$(AA^T)(Av) = A(A^TA)v = A(AA^T)v = A\lambda v = \lambda Av$$
where in the second step I used the normality of $A$.
But since $AA^T$ has $n$ different eigenvalues, each of its eigenspaces has dimension $1$, therefore the above equation implies that $v$ and $Av$ have to be linear dependent, and since $v$ is non-zero, this means there exists a $\mu$ such that $Av=\mu v$. But that equation means that $v$ is an eigenvector of $A$.
Thus $A$ has an orthogonal basis of eigenvectors (namely the same one as $AA^T$). But a matrix that has an orthogonal basis of eigenvectors is symmetric.
A: Let $A=USV^T$ be a singular value decomposition. Then $US^2U^T=AA^T=A^TA=VS^2V^T$. Hence $S^2$ commutes with $V^TU$. However, as all eigenvalues of $A^TA$ are distinct, $S^2$ has distinct diagonal entries. Therefore $V^TU$ is a diagonal matrix. Let $V^TU=D$. Then $A=USV^T=V(DS)V^T$ is symmetric.
A: $A$ is normal and $AA^T$ has $n$ distinct eigenvalues, so $AA^T=X \Lambda^2X^{-1}=A^TA $, where $\Lambda$ is diagonal and has $\lambda$'s on the diagonal.
Now $AA^T=(X \Lambda X^{-1})(X \Lambda X^{-1})=A^TA $ so that $A=A^T$.
