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For any real vector $u$, a Cholesky decomposition gives the quadratic form identitySince $L$ is an invertible matrix, this squared Euclidean norm is positive whenever $u\ne 0$. The factorization thus certifies positive definiteness and gives coordinates in which the quadratic form is a sum of ordinary squares. For the matrix above,The level set $u^{T}Au=1$ is an ellipse; the linear map $z=L^{T}u$ sends it to the unit circle. The points $u=(1/2,0{)}^T$ and $u=(-1/4,1/2{)}^T$ go to $(1,0{)}^T$ and $(0,1{)}^T$, respectively.

The same factor works in the reverse direction for Gaussian random vectors. If $Z$ has independent unit-variance, zero-mean Gaussian random variables as coordinates, $X=LZ$ has covariance matrix $L{L}^T=A$. In this example $X_1=2Z_1$ and $X_2=Z_1+2Z_2$: their variances are $4$ and $5$, while their covariance is $2$. This is a concrete way to generate specified correlations from independent draws.

A triangular factor is not generally a square root of a matrix in the sense $B^2=A$. Here $L^2=\left(\begin{array}{cc}4& 0\\ 4& 4\end{array}\right)\ne A$. The Cholesky decomposition uses the product with the transpose, which is also what makes the quadratic form a squared Euclidean norm.

For a two-dimensional positive-definite covariance matrix $\Sigma$, the centered covariance ellipse is the image of the unit circle under its positive definite square root of a matrix. Equivalently, it satisfies $x^T{\Sigma }^{-1}x=1$. Its semi-major and semi-minor axes have lengths given by the square roots of the larger and smaller eigenvalues, along the corresponding eigenvectors. The ellipse encodes a second-moment geometry; a probability-coverage interpretation needs additional information about the probability distribution and the chosen scale.

A Hermitian matrix satisfies $A=A^{†}$, where $†$ means conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda \Vert v{\Vert }^2$ is real, since it equals its complex conjugate $v^{†}{A}^{†}v$. Thus every eigenvalue is real. To prove the finite-dimensional spectral theorem, choose a unit eigenvector $v$; one exists because the complex characteristic polynomial has a root. Its orthogonal complement is invariant, since for $w\perp v$,The restriction to that complement is again Hermitian. Induction on dimension supplies an orthonormal basis of eigenvectors, so $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)U^{†}$ for a unitary matrix $U$.

In this basis, $x^{†}Ax={\sum }_i{\lambda }_i|(U^{†}x{)}_i{|}^2$. Therefore $\boxed{A>0⟺{\lambda }_i>0\text{ for every }i}$. In that case the square root of a matrixis Hermitian, positive definite and satisfies $C^2=A$. For uniqueness, any positive-definite Hermitian $D$ satisfying $D^2=A$ commutes with $A$ and hence preserves each eigenspace of $A$. Its restriction to the $\lambda$-eigenspace is Hermitian, and its positive eigenvalues all solve ${\mu }^2=\lambda$. Thus that restriction is $\sqrt{\lambda }I$, forcing $D=C$. This is the principal square root of a positive semidefinite matrix, here strictly positive definite.

Now let $D=\sqrt{B}$ and $X=C-D$. Expanding without assuming that $C$ and $D$ commute givesThe Hermitian matrix $X$ has $X^2\ge 0$ because $v^{†}{X}^{2}v=\Vert Xv{\Vert }^2$, whereas $A-B>0$. Hence $CX+XC>0$. If $Xv=\mu v$, $v\ne 0$, thenSince $C>0$, every real eigenvalue $\mu$ of $X$ is positive. The positive-eigenvalue criterion now gives the strict finite-dimensional operator monotonicity of the square root: