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Interpret the norms as Euclidean norms and set $B=A^{\mathsf{T}}A$. This is a real symmetric matrix with ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}=\Vert A\mathbf{x}{\Vert }^2\ge 0$, hence a positive semidefinite matrix. Squaring the nonnegative objective does not change its argmin. Let ${\lambda }_{\min }$ be the smallest eigenvalue of $B$ and $E_{\min }$ its eigenspace.

First minimize on the unit sphere. The Lagrange multiplier function $L(\mathbf{x},\lambda )={\mathbf{x}}^{\mathsf{T}}B\mathbf{x}-\lambda ({\mathbf{x}}^{\mathsf{T}}\mathbf{x}-1)$ has stationarity condition $B\mathbf{x}=\lambda \mathbf{x}$. Multiplication by ${\mathbf{x}}^{\mathsf{T}}$ shows that the squared objective at such a point is $\lambda$. The spectral theorem for real symmetric matrices also gives, for every unit vector, ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}\ge {\lambda }_{\min }$, with equality exactly in $E_{\min }$. This proves global minimality, rather than merely finding stationary candidates.

If ${\lambda }_{\min }>0$, every point with $\Vert \mathbf{x}\Vert >1$ has a strictly smaller objective after rescaling to unit length. ThereforeIf ${\lambda }_{\min }=0$, the kernel of a linear map satisfies $\ker B=\ker A$, since $B\mathbf{x}=0$ implies $\Vert A\mathbf{x}{\Vert }^2=0$, and the converse is immediate. The minimum is then zero, attained precisely atIn this case minimizers need not have unit length. This distinction is essential for a rank-deficient $A$, including any real $m\times n$ matrix with $m<n$. The result is the minimum image norm outside a Euclidean unit ball, governed by the least singular value.

Let a real tridiagonal matrix have diagonal entries $a_i$, positive upper entries $b_i$, and positive lower entries $c_i$. Set $d_1=1$, $d_{i+1}=d_i\sqrt{c_i/b_i}$ and $D=\mathrm{diag}(d_i)$. Then $D^{-1}AD$ is a symmetric matrix with diagonal $a_i$ and adjacent entries $\sqrt{b_i{c}_i}$. Indeed its two adjacent entries are $b_i{d}_{i+1}/d_i$ and $c_i{d}_i/d_{i+1}$, which are equal by construction. All other zeros are preserved. This matrix similarity implies real eigenvalues and diagonalizability, by the spectral theorem for real symmetric matrices. No sign assumption on $a_i$ is needed. In the weighted formulation of positive diagonal symmetrization of a matrix, the positive diagonal matrix is $W=D^{-2}$.

The principal moments of inertia about a specified point $O$ are the eigenvalues $I_1,I_2,I_3$ of the inertia tensor ${\mathbf{I}}_O$. Their associated unit vectors that are eigenvectors specify principal axes through $O$. The spectral theorem for real symmetric matrices guarantees an orthonormal basis of these eigenvectors, even for an irregular rigid body. Choose their signs to make a right-handed basis and collect them as the columns of an orthogonal matrix $Q$:Thus diagonalization removes the off-diagonal coupling in the rotational kinetic energy. The eigenvalues belong to the mass distribution and reference point, independently of which orthonormal basis initially describes them. ^[ref]

In this principal body frame, write angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{\mathbf{e}}_i$. ThenUse these as total angular momentum and kinetic energy only under the reference-point conditions stated under inertia tensor. Along a single principal axis with $I_i>0$, angular momentum is parallel to angular velocity. For a mixture of directions with different principal moments of inertia, it generally is not. A principal axis need not be a geometric symmetry axis: it is selected by the complete mass distribution. A principal axis also need not give stable free rotation; for three distinct principal moments of inertia, the intermediate axis theorem supplies the stability distinction.

For any unit vector $\mathbf{n}={\sum }_i{n}_i{\mathbf{e}}_i$,The axial moment of inertia is a weighted average of the principal moments of inertia. At fixed angular speed, the least and greatest rotational kinetic energies occur in the corresponding eigenspaces. At fixed nonzero angular momentum magnitude $L$, instead, $K=\frac{1}{2}{\sum }_i{L}_i^2/I_i$, so the largest principal moment of inertia gives the smallest rotational kinetic energy. These comparisons hold the reference point fixed.

There is an additional physical restriction beyond positive semidefiniteness. In principal axes coordinates,with two analogous cyclic inequalities. Therefore three arbitrary positive numbers do not always represent a possible inertia tensor. For a distribution in the $xy$ plane through $O$, $I_3=I_1+I_2$, as in the planar inertia tensor of paired point masses. A uniform solid ellipsoid of mass $M$ and semiaxes $a,b,c$ supplies a continuous three-dimensional example: using its mass second-moment tensor, its central principal moments of inertia are $M(b^2+c^2)/5$, $M(a^2+c^2)/5$, $M(a^2+b^2)/5$. Setting $a=b=c=R$ recovers the moment of inertia of a uniform solid sphere.

For a two-dimensional symmetric positive-definite matrix $M$ and $R>0$, the region ${\mathbf{x}}^{T}M\mathbf{x}\le R$ is a filled ellipse of area $\pi R/\sqrt{\det M}$. By the spectral theorem for real symmetric matrices, an orthogonal change of coordinates gives positive eigenvalues ${\lambda }_1,{\lambda }_2$ and semiaxes $\sqrt{R/{\lambda }_1}$ and $\sqrt{R/{\lambda }_2}$. Their product and the area of an ellipse prove the formula. Translating the center of an ellipse leaves the area unchanged.

The real case of the finite-dimensional spectral theorem says that a self-adjoint operator on a finite-dimensional real inner product space has a real orthonormal eigenbasis. In orthonormal coordinates this is the spectral theorem for real symmetric matrices, $A=Q\Lambda Q^T$. The worked ellipse under finite-dimensional spectral theorem shows why both orthogonal diagonalization and orthogonality matter: the new coordinates separate the quadratic form into squares without distorting lengths or angles.

A real normal matrix need not have real eigenvalues. For example, the quarter-turn $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ is an orthogonal matrix, but its characteristic polynomial is $z^2+1$ and its eigenvalues are $i,-i$. Over the complex numbers it has an orthonormal eigenbasis; over the real numbers it has no eigenvector. Self-adjointness supplies the reality conclusion in the real spectral theorem.

Let the columns of a centered design matrix $X\in {\mathbb{R}}^{n\times p}$ be variables and write $X^{T}X=V\Lambda V^T$ by the spectral theorem for real symmetric matrices. The columns $u_i$ of $U=XV$ are the sample principal-component score vectors. They satisfy $u_i^T{u}_j={\Lambda }_{ij}$, so distinct score vectors have zero sample covariance and $u_i$ has sample variance ${\Lambda }_{ii}/n$ under the divisor-$n$ convention.

For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.