# Positive/Negative Definite/Semidefinite Test Generality

A test to determine whether a matrix is positive definite, negative definite, positive semidefinite, negative semidefinite, or none of the above, is to calculate the determinant of every cascading submatrix inclusively between the $$1$$x$$1$$ top-left entry and the matrix itself. For example, $$\begin{bmatrix} 11 & 26 & 41 & 56 & 71 \\ 26 & 66 & 106 & 146 & 186 \\ 41 & 106 & 171 & 236 & 301 \\ 56 & 146 & 236 & 326 & 416 \\ 71 & 186 & 301 & 416 & 531 \end{bmatrix}$$ is positive definite if and only if $$\begin{vmatrix} 11\end{vmatrix}$$, $$\begin{vmatrix} 11 & 26 \\ 26 & 66 \end{vmatrix}$$, $$\begin{vmatrix} 11 & 26 & 41 \\ 26 & 66 & 106 \\ 41 & 106 & 171 \end{vmatrix}$$, $$\begin{vmatrix} 11 & 26 & 41 & 56 \\ 26 & 66 & 106 & 146 \\ 41 & 106 & 171 & 236 \\ 56 & 146 & 236 & 326 & \end{vmatrix}$$, and $$\begin{vmatrix} 11 & 26 & 41 & 56 & 71 \\ 26 & 66 & 106 & 146 & 186 \\ 41 & 106 & 171 & 236 & 301 \\ 56 & 146 & 236 & 326 & 416 \\ 71 & 186 & 301 & 416 & 531 \end{vmatrix}$$ are each positive. I have been informed that an acceptable alternative is to consider the submatrices between the $$1$$x$$1$$ bottom-right entry and the full matrix, i.e., $$\begin{vmatrix} 531 \end{vmatrix}$$, $$\begin{vmatrix} 326 & 416 \\ 416 & 531 \end{vmatrix}$$, $$\begin{vmatrix} 171 & 236 & 301 \\ 236 & 326 & 416 \\ 301 & 416 & 531 \end{vmatrix}$$, $$\begin{vmatrix} 66 & 106 & 146 & 186 \\ 106 & 171 & 236 & 301 \\ 146 & 236 & 326 & 416 \\ 186 & 301 & 416 & 531 \end{vmatrix}$$, and $$\begin{vmatrix} 11 & 26 & 41 & 56 & 71 \\ 26 & 66 & 106 & 146 & 186 \\ 41 & 106 & 171 & 236 & 301 \\ 56 & 146 & 236 & 326 & 416 \\ 71 & 186 & 301 & 416 & 531 \end{vmatrix}$$. Does mixing these approaches work?

Suppose we label the subdeterminants from the first approach $$A_n$$, and those from the second approach $$B_n$$, where $$n$$ is the number of rows/columns in that subdeterminant. For example, $$B_2$$ represents $$\begin{vmatrix} 326 & 416 \\ 416 & 531 \end{vmatrix}$$. If I find an arbitrary combination of positive subdeterminants from the $$A$$ and $$B$$ approaches, such that all five subscripts are accounted for, i.e., $$A_1$$, $$B_2$$, $$A_3$$, $$A_4$$, and $$B_5$$, can I conclude that the original matrix is positive definite?

If so, is it possible to generalize the approach even further? For example, can the subdeterminant $$\begin{vmatrix} 171 & 236 \\ 236 & 326\end{vmatrix}$$ make a contribution, even though it is neither utilized by the $$A$$ nor the $$B$$ approach?

E.g. for any $$S\subseteq\{1,2,3,4,5\}$$, let $$A(S)$$ denotes the principal submatrix $$(a_{ij})_{i,j\in S}$$. Then $$A(\{4\}),\,A(\{2,4\}),\,A(\{1,2,4\}),\,A(\{1,2,4,5\})$$ and $$A$$ form a nested sequence of principal submatrices, and $$A$$ is positive definite if and only if the determinants of these submatrices are all positive.
In your example, $$A_1=A(\{1\},\,B_2=A(\{4,5\}),\,A_3=A(\{1,2,3\}),\,A_4=A(\{1,2,3,4\})$$ and $$B_5=A$$ do not form a nested sequence of principal submatrices. Even if their determinants are all positive, $$A$$ is not necessarily positive definite. For a counterexample, consider $$A=\operatorname{diag}(1,-1,-1,1,1)$$.
• You seem to have misunderstood how to use Sylvester's criterion for semidefiniteness. To verify that a Hermitian matrix is positive semidefinite, you need to show that all principal minors are nonnegative. It does not suffice to show that all leading principal minors are nonnegative. For a counterexample, consider $$A=\operatorname{diag}(0,1,-1,0)$$. All leading principal minors of this $$A$$ are zero (hence nonnegative), but $$A$$ is indefinite.