Principal Submatrices And Definiteness

Restricting a Quadratic Form to Selected Coordinates

A real symmetric matrix determines a quadratic form: to a column vector \(\mathbf{x}\), it assigns the scalar \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\). Positive definiteness means this scalar is strictly positive for every nonzero vector. Selecting only some coordinates restricts the quadratic form to vectors whose other coordinates are zero. The corresponding matrix is a principal submatrix, and its determinant is a principal minor. These objects reveal what positive definiteness forces on coordinate restrictions, although checking only proper submatrices cannot establish that the full matrix is positive definite.

Throughout, matrices and vectors are real, and positive definite means symmetric positive definite. An index set \(I\) will be a nonempty subset of \(\{1,\dots,n\}\), written in increasing order. For example, if \(I=\{1,3\}\), we select rows 1 and 3 and columns 1 and 3, keeping those rows and columns in that order.

A submatrix formed using different row and column index sets is not principal. For instance, choosing rows 1 and 2 but columns 2 and 3 does not give a principal submatrix, even though it is a square submatrix. If \(\mathbf{A}\) is symmetric, then \(\mathbf{A}[I,I]\) is symmetric too: its \((r,s)\) entry is \(a_{i_r i_s}=a_{i_s i_r}\), the \((s,r)\) entry.

The Restriction Identity

To relate a principal submatrix to the full quadratic form, insert zeros in the coordinates not in \(I\). More precisely, for \(\mathbf{x}\in\mathbb{R}^k\), let \(\mathbf{y}\in\mathbb{R}^n\) be the vector whose \(i_r\)-th entry is \(x_r\), for \(1\le r\le k\), and whose other entries are zero. Then multiplying out the quadratic forms gives the identity below.

The identity says that the principal submatrix records exactly the values of the full quadratic form on vectors supported in the selected coordinates. Here, a vector is supported in \(I\) when every coordinate outside \(I\) is zero.

Proof. Fix a nonempty index set \(I\), and take any nonzero \(\mathbf{x}\in\mathbb{R}^{|I|}\). Construct \(\mathbf{y}\in\mathbb{R}^n\) by inserting the entries of \(\mathbf{x}\) in the coordinates in \(I\) and zeros elsewhere. Since \(\mathbf{x}\ne\mathbf{0}\), at least one entry placed in \(\mathbf{y}\) is nonzero, so \(\mathbf{y}\ne\mathbf{0}\). Positive definiteness of \(\mathbf{A}\) gives \(\mathbf{y}^{\top}\mathbf{A}\mathbf{y}>0\). By the restriction identity, this is \(\mathbf{x}^{\top}\mathbf{A}[I,I]\mathbf{x}>0\). Also, \(\mathbf{A}[I,I]\) is symmetric because \(\mathbf{A}\) is symmetric. Thus it is positive definite. \(\square\)

This theorem applies to every nonempty coordinate selection, not only to consecutive indices or to the first few indices. In particular, if even one principal submatrix is not positive definite, the full matrix cannot be positive definite: otherwise the theorem would force that submatrix to be positive definite.

Principal Minors Must Be Positive

For a positive definite matrix, the theorem shows that each principal submatrix is positive definite. Its determinant must therefore be positive as well. To justify that claim, use the Spectral Theorem for real symmetric matrices: each real symmetric matrix has an orthogonal diagonalization, and its determinant is the product of its eigenvalues. If \(\mathbf{B}\) is positive definite and \(\mathbf{v}\ne\mathbf{0}\) is an eigenvector with eigenvalue \(\lambda\), then

Since \(\mathbf{v}^{\top}\mathbf{v}>0\), this implies \(\lambda>0\). Thus all eigenvalues of \(\mathbf{B}\) are positive, so their product, \(\det\mathbf{B}\), is positive.

Proof. Let \(I\) be any nonempty index set. By the theorem on restrictions, \(\mathbf{B}=\mathbf{A}[I,I]\) is symmetric positive definite. The Spectral Theorem gives an orthogonal matrix \(\mathbf{Q}\) and real eigenvalues \(\lambda_1,\dots,\lambda_{|I|}\) such that \(\mathbf{B}=\mathbf{Q}\operatorname{diag}(\lambda_1,\dots,\lambda_{|I|})\mathbf{Q}^{\top}\). For each eigenvalue choose a corresponding nonzero eigenvector \(\mathbf{v}\). Then \(\lambda\mathbf{v}^{\top}\mathbf{v}=\mathbf{v}^{\top}\mathbf{B}\mathbf{v}>0\), so \(\lambda>0\). Therefore \(\det\mathbf{B}=\prod_{j=1}^{|I|}\lambda_j>0\). This determinant is the principal minor indexed by \(I\). \(\square\)

Why Proper Submatrices Do Not Decide the Full Matrix

The condition “every principal submatrix is positive definite” is equivalent to positive definiteness of \(\mathbf{A}\) if it includes the full index set \(\{1,\dots,n\}\), because that principal submatrix is \(\mathbf{A}\) itself. The restriction theorem supplies the substantive direction: positive definiteness of the full matrix forces positive definiteness on every coordinate restriction. But if the full matrix is excluded, the reverse inference can fail. In particular, even checking every proper principal submatrix does not guarantee that the quadratic form is positive in all directions.

Common Pitfall: Confusing a Necessary Condition with a Complete Test

A common incorrect argument is: “Every proper principal submatrix of \(\mathbf{C}\) is positive definite, so \(\mathbf{C}\) is positive definite.” The calculation above corrects it: \(\mathbf{x}^{\top}\mathbf{C}\mathbf{x}=-\tfrac32<0\) for \(\mathbf{x}=(1,1,1)^{\top}\). The restriction theorem gives a one-way implication from the full matrix to each principal submatrix. Positive definiteness of selected restrictions does not automatically control vectors with all coordinates active.

Likewise, positivity of a principal minor is not itself a statement that the corresponding submatrix is positive definite. A determinant is one scalar, while positive definiteness requires a quadratic-form inequality for every nonzero vector. The results here give a reliable implication: if \(\mathbf{A}\) is positive definite, then all its principal submatrices are positive definite and all its principal minors are positive. A failed principal-submatrix test disproves positive definiteness; passing tests on only some proper submatrices does not prove it.