Cfrgtkky

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 $begin{vmatrix}
11end{vmatrix}$, $begin{vmatrix}
11 & 26 \
26 & 66
end{vmatrix}$, $begin{vmatrix}
11 & 26 & 41 \
26 & 66 & 106 \
41 & 106 & 171
end{vmatrix}$, $begin{vmatrix}
11 & 26 & 41 & 56 \
26 & 66 & 106 & 146 \
41 & 106 & 171 & 236 \
56 & 146 & 236 & 326 &
end{vmatrix}$, 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 $begin{vmatrix}
171 & 236 \
236 & 326end{vmatrix}$ make a contribution, even though it is neither utilized by the $A$ nor the $B$ approach?

To test for positive definiteness, as long as the principal submatrices are nested (and we have submatrices of all sizes, of course), they don't need to be the leading ones or the trailing ones. This is because by relabelling the rows and columns, any nested sequence of principal submatrices can be turned into a sequence of leading principal submatrices.

E.g. for any $Ssubseteq{1,2,3,4,5}$, let $A(S)$ denotes the principal submatrix $(a_{ij})_{i,jin S}$. 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 $A=operatorname{diag}(1,-1,-1,1,1)$.

By the way:

• You lack vocabularies. I think you need to look up the meanings of the words "leading/trailing" and principal in your textbook.


• The test has a name. It is called Sylvester's criterion.


• 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.