= Solution
By the <real spectral theorem>, a real symmetric matrix has an orthogonal diagonalization
$$
A=Q\operatorname{diag}(\lambda_1,\ldots,\lambda_n)Q^T.
$$
Writing $y=Q^Tx$ gives
$$
x^TAx=\sum_i\lambda_i y_i^2.
$$
This is nonnegative for every $x$ exactly when every $\lambda_i\geq0$. Thus the quadratic-form and eigenvalue definitions of a <positive semidefinite matrix> agree.
When $A$ is positive semidefinite, define
$$
\boxed{
\sqrt A
=Q\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})Q^T}.
$$
This matrix is symmetric and positive semidefinite, and its square is $A$, so it is the <principal square root of a positive semidefinite matrix>.
For nonsingular $M$, the matrix $M^TM$ is symmetric and
$$
x^TM^TMx=\|Mx\|^2>0
$$
for every nonzero $x$. Thus $M^TM$ is positive definite, and
$$
P=\sqrt{M^TM}
$$
exists and is nonsingular. Let $R=MP^{-1}$. Since $P$ is symmetric,
$$
R^TR=P^{-1}M^TMP^{-1}
=P^{-1}P^2P^{-1}=I.
$$
Hence $R$ is an <orthogonal matrix> and
$$
\boxed{M=RP},
$$
the <polar decomposition of an invertible real matrix>.
In three dimensions, $P$ stretches or contracts along three mutually perpendicular eigenvector directions by its positive eigenvalues. The orthogonal map $R$ is then applied: it is a rotation when $\det M>0$, and a rotation combined with a reflection when $\det M<0$.
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