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Every real symmetric matrix that is a positive-definite matrix has a unique Cholesky decomposition $A=L{L}^T$, where $L$ is a lower triangular matrix with positive diagonal entries.^[ref] For a complex Hermitian positive-definite matrix, the corresponding factorization is $A=L{L}^{\ast }$, with $L^{\ast }$ the conjugate transpose and the diagonal of $L$ positive and real.^[ref] The factorization turns a linear system into two triangular systems.

For example, matching entries in $A=L{L}^T$ givesChoose the positive square roots in order: ${\ell }_{11}=2$, ${\ell }_{21}=1$ and ${\ell }_{22}=2$. ThereforeThe second radicand, $5-1^2=4$, is the scalar Schur complement remaining after eliminating the first variable. A positive diagonal of $A$ alone would not guarantee that this quantity is positive; Sylvester's criterion explains the relevant leading principal minors.

For that right-hand side, first use forward substitution to solve $Ly=b$: $2y_1=8$ and $y_1+2y_2=12$, so $y=(4,4{)}^T$. Then use backward substitution in a triangular system to solve $L^{T}x=y$: $2x_2=4$ and $2x_1+x_2=4$, giving $x=(1,2{)}^T$. No matrix inverse is required. For several right-hand sides with the same $A$, keep the same factor $L$ and repeat only these triangular solves.

A leading principal minor is a principal minor using the first $k$ rows and columns of a square matrix in a specified ordering. Sylvester's criterion uses these minors to characterize positive-definite matrices.

Work inside $V_{m+1}$, and take the orthogonal complement for a bilinear form of $U=V_m$ relative to this ambient space. If $z\in U\cap U^{\perp }$, then $\phi (z,z)=0$, so positive definiteness on $U$ forces $z=0$.

Part (a)(iv) therefore gives $V_{m+1}=U\oplus U^{\perp }$, with $\dim U^{\perp }=1$. Choose $w\ne 0$ spanning that complement and use the basis $e_1,\dots ,e_m,w$. Symmetry and orthogonality make its Gram matrixBy part (b)(i), $\det B=(\det P{)}^2\det A_{m+1}>0$. Part (ii) applied to the already positive definite restriction gives $\det A_m>0$. ThereforeFor a nonzero vector $u+tw$ with $u\in U$,This proves positive definiteness on $V_{m+1}$.

Finally $\det A_1>0$ says $\phi (e_1,e_1)>0$, so the restriction to $V_1$ is positive definite. Induction using all successive positive leading principal minors reaches $V_n=V$. Combined with part (ii), this proves Sylvester's criterion in both directions.

The restriction of a positive definite form to each $V_m$ is positive definite. By the spectral theorem for real symmetric matrices, choose an orthogonal matrix $P$ such that $P^T{A}_{m}P=\mathrm{diag}(d_1,\dots ,d_m)$.

Each $d_i=\phi (f_i,f_i)>0$, since a basis vector is nonzero. Taking determinants in part (i) givesbecause an orthogonal matrix has determinant $\pm 1$. ConsequentlyThese determinants are the leading principal minors in Sylvester's criterion.

Write $d=\mu -r1$ and let $x$ denote the risky holdings as fractions of total wealth. The remaining fraction $1-1^{T}x$ is in the risk-free asset. Consequently their excess expected return is $d^{T}x$ and their variance is $x^T\Sigma x$. The Cauchy-Schwarz inequality in the inner product defined by $\Sigma$ givesEquality with positive excess return holds precisely for $x=cz$, $c>0$, where $z={\Sigma }^{-1}d$. Thus solving $\Sigma z=d$ determines both the efficient risky direction and the upper capital market line slope. Provided $s=1^{T}z>0$, the normalized market portfolio and its Sharpe ratio areHere the excess means are $(7,14,21{)}^T$, so the required equations areThe leading principal minors of $\Sigma$ are $4$, $35$ and $539$, so Sylvester's criterion verifies positive definiteness and the solution is unique.

Two real symmetric bilinear forms are congruent if an invertible linear map takes one to the other; in matrices, $B=S^{\mathsf{T}}AS$ for some invertible $S$. The spectral theorem for real symmetric matrices gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form toSylvester's law of inertia asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.

The rank of a quadratic form is $p+q$, and its signature of a quadratic form is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\ge 0$ with $p+q\le n$ givesFor $n\ge 1$, exactly two classes have every level set bounded: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.

For the displayed three-dimensional form,Its leading principal minors are $3,14,32$, all positive. Sylvester's criterion makes it positive definite, so the surface is a bounded ellipsoid. Directly, its quadratic expression iswhich also proves positivity and boundedness of the unit level.

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.

Apply symmetric Gaussian elimination without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric Schur complement. If $d_k\le 0$, stop and report that $A$ is not positive definite. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an LDL decompositionwhere $L$ is unit lower triangular and $D=\mathrm{diag}(d_1,\dots ,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square givesChoosing $s=-b^{T}y/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also Sylvester's criterion.

At step $k$, updating the remaining matrix costs $O((n-k{)}^2)$ arithmetic operations. Hence the total iswhich proves the existence of the required algorithm.

A Hermitian form is conjugate-linear in its first argument, linear in its second, and satisfies $\varphi (w,v)=\overline{\varphi (v,w)}$. For $H=\varphi -\psi$, the hypothesis says $H(v,v)=0$. The complex polarization identitytherefore gives $H(v,w)=0$ for all $v,w$.

The displayed matrix is Hermitian. By Sylvester's criterion, its leading principal minors areHence the form is

The displayed simply-laced tree has arms of lengths $2$, $1$, and $3$ from its trivalent vertex, so it is the finite type $E_7$ Coxeter graph. Its Gram matrix is the $E_7$ Cartan matrix, which has positive leading principal minors in a leaf-removal ordering and determinant $2$. By Sylvester's criterion it is positive definite. It is therefore positive semidefinite and nondegenerate as well: the three answers are yes, yes, and yes.

For $\nu =5$, $B$ is real symmetric. Sylvester's criterion says that all eigenvalues are positive exactly when its leading principal matrix minors are positive. These are $1$, $5-4=1$, and $\det B=1-{\mu }^2$. ThereforeAt either endpoint one eigenvalue is zero, so the strict positivity requirement excludes it.

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.