Leading Principal Minors and No-Pivot Breakdown

Leading Principal Minors Record the Pivots

Gaussian elimination without row exchanges depends on the entries that appear on the diagonal as elimination proceeds. If one of these pivot entries is zero, the usual multiplier—an entry below the pivot divided by the pivot—is undefined. The determinants of the leading principal submatrices give a way to check these pivots in advance. They record not just whether the whole matrix is singular, but where a no-pivot elimination process first encounters trouble.

Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). For \(1\leq k\leq n\), its leading principal submatrix of order \(k\) is the \(k\times k\) matrix formed from the first \(k\) rows and first \(k\) columns. Its determinant is called the leading principal minor of order \(k\). Write this determinant as \(\Delta_k\), and set \(\Delta_0=1\) by convention. Thus \(\Delta_n=\det\mathbf{A}\), but the earlier \(\Delta_k\) contain information that \(\det\mathbf{A}\) alone does not.

We use the standard determinant facts that replacing a row by itself plus a multiple of another row preserves the determinant, and that the determinant of a triangular matrix is the product of its diagonal entries. In elimination, let \(p_k\) denote the pivot at step \(k\), after the first \(k-1\) columns have been eliminated below their pivots. When the preceding pivots are nonzero, the relationship is

The quotient formula applies only when \(\Delta_{k-1}\ne0\), so the earlier pivots have been defined and are nonzero. The product formula explains why the minors detect the successive pivots.

Proof. First suppose elimination has nonzero pivots \(p_1,\dots,p_n\). For any \(k\), perform the first \(k-1\) elimination steps within the leading \(k\times k\) block. Each step replaces a row by itself plus a multiple of another row, so its determinant remains \(\Delta_k\). After those steps, the leading block is upper triangular, with diagonal entries \(p_1,\dots,p_k\). Its determinant is their product. Therefore \(\Delta_k=p_1\cdots p_k\ne0\). Dividing the product for \(\Delta_k\) by the product for \(\Delta_{k-1}\) gives \(p_k=\Delta_k/\Delta_{k-1}\).

Conversely, suppose every \(\Delta_k\ne0\). We show successively that each pivot exists and is nonzero. For \(k=1\), the first pivot is \(p_1=a_{11}=\Delta_1\ne0\). Now suppose the first \(k-1\) pivots have been obtained and are nonzero. The first \(k-1\) elimination steps can therefore be performed. In the resulting leading \(k\times k\) block, the first \(k-1\) diagonal entries are \(p_1,\dots,p_{k-1}\), all entries below those pivots are zero, and the last diagonal entry is the next pivot candidate \(p_k\). The row operations have preserved the determinant, so \(\Delta_k=p_1\cdots p_{k-1}p_k\). Since \(\Delta_k\ne0\) and the previous pivots are nonzero, \(p_k=\Delta_k/\Delta_{k-1}\ne0\). This completes the induction through \(k=n\). \(\square\)

A Successful No-Pivot Factorization

When a Zero Pivot Stops the Process

If \(\Delta_{k-1}\ne0\) but \(\Delta_k=0\), the quotient formula gives a zero pivot at step \(k\). When \(k<n\), this is an interior step: there are rows below the pivot whose entries might need to be eliminated, but the usual multiplier would require division by zero. Without a row exchange or a different method, that no-pivot elimination step cannot be carried out.

Why the Full Determinant Is Not Enough

The determinant \(\Delta_n\) answers whether the entire matrix is nonsingular. It does not, by itself, tell us whether the intermediate pivots are nonzero. The previous example has \(\det\mathbf{A}=-1\), yet elimination without row exchanges still fails before reaching the last column. To predict the path of elimination, check the leading principal minors in order, not just the determinant of the full matrix.

For an invertible matrix, \(\Delta_n\ne0\). Consequently, if some earlier leading principal minor vanishes, there is a first index \(k<n\) with \(\Delta_k=0\); all the preceding minors are nonzero, and the pivot at step \(k\) is zero. This is the characteristic no-pivot breakdown for an invertible matrix. For a singular matrix, the first vanishing minor might instead be \(\Delta_n\), as in the diagonal example, where the final zero does not require an elimination operation.

Common Pitfall

Wrong claim: “The matrix is nonsingular, so elimination without row exchanges will have nonzero pivots.” Applying that claim to the matrix in the second worked example would predict success because \(\det\mathbf{A}=-1\ne0\). But after the first valid elimination step, the matrix is \(\begin{bmatrix}1&1&0\\0&0&1\\0&1&0\end{bmatrix}\), whose second pivot is \(0\) and whose entry below that pivot is \(1\). The correction is to check each leading principal minor: \(\Delta_1=1\ne0\), but \(\Delta_2=0\). The full determinant tests nonsingularity; the sequence of leading principal minors tests the successive no-pivot pivots.