Sign of eigenvalues of a real $3 \times 3$ symmetric matrix

Question

I want to find just the sign of $3$ eigenvalues of a given real symmetric $3 \times 3$ matrix $\mathbf{A}$ without actually calculating the eigenvalues. Is there any way I can do it based on only the matrix entries and the given information?

Answer 1

You may use Sylvester's law of inertia.

In general, you can diagonalise a real symmetric matrix $A$ by congruence (whose computational complexity is in general lower than orthogonal diagonalisation). Then the signs of the resulting diagonal matrix are the signs of the eigenvalues of $A$.

Alternatively, if your matrix happens to contain a nested sequence of nonzero principal minors, you can apply the (lesser known) sign-change version of Sylvester's law of inertia without performing any diagonalisation. More specifically, suppose that there exists a (reversely) nested sequence of principal submatrices $A_0,A_1,A_2,\ldots,A_n$ of $A$ such that each $A_k$ is nonsingular and $k\times k$ (here $A_0$ by convention denotes the empty matrix with determinant $1$ and $A_n=A$) and $A_{k-1}$ is contained in $A_k$. Then the number of sign changes in the numeric sequence $$ \det(A_0)\,(=1),\,\det(A_1),\ldots,\det(A_{n-1}),\,\det(A_n)\,(=\det(A)) $$ is the number of negative eigenvalues of $A$. E.g. consider $$ A=\pmatrix{0&0&1\\ 0&1&2\\ 1&2&-3}. $$ Then $1,\,\det(A(2,2))=1,\,\det(A(2:3,\,2:3))=-7,\,\det(A)=-1$ is a sequence of nonzero nested principal minors. Since there is only one sign change (between $\det(A_1)$ and $\det(A_2)$) in the sequence, we conclude that $A$ has one negative eigenvalue and two positive eigenvalues.

Note that the choice of the nested principal minors does not matter as long as these minors are nonzero. To illustrate, the $A$ in the example above has only three sequences of nonzero nested principal minors:

• $1,\,\det(A(2,2))=1,\,\det(A(2:3,\,2:3))=-7,\,\det(A)=-1$,
• $1,\,\det(A(3,3))=-3,\,\det(A(2:3,\,2:3))=-7,\,\det(A)=-1$,
• $1,\,\det(A(3,3))=-3,\,\det(A(\{1,3\},\,\{1,3\}))=-1,\,\det(A)=-1$.

We see that in every one of them, the number of sign changes remains the same (namely, exactly $1$).

Answer 2

About the example of user1551, we see that it has two positive and one negative eigenvalue. Sylvester says that the eigenvalues must match up with the diagonal elements of $D;$ these may vary depending on choices made while solving $P^THP = D.$

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & - 2 & 1 \\ 1 & - \frac{ 2 }{ 7 } & \frac{ 1 }{ 7 } \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & - \frac{ 2 }{ 7 } \\ 0 & 1 & \frac{ 1 }{ 7 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 7 & 0 \\ 0 & 0 & \frac{ 1 }{ 7 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 0 & - \frac{ 1 }{ 7 } & 1 \\ 1 & 0 & 0 \\ 2 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 7 & 0 \\ 0 & 0 & \frac{ 1 }{ 7 } \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & 2 \\ - \frac{ 1 }{ 7 } & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \\ \end{array} \right) $$

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Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & 0 & 1 \\ 2 & 1 & - 3 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 0 & 1 & 2 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & - 7 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & 0 \\ 0 & 1 & 0 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 7 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & - \frac{ 2 }{ 7 } \\ 0 & 1 & \frac{ 1 }{ 7 } \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrr} 0 & 1 & 2 \\ - \frac{ 1 }{ 7 } & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 7 & 0 \\ 0 & 0 & \frac{ 1 }{ 7 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & - 2 & 1 \\ 1 & - \frac{ 2 }{ 7 } & \frac{ 1 }{ 7 } \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & - 2 & - \frac{ 2 }{ 7 } \\ 0 & 1 & \frac{ 1 }{ 7 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 7 & 0 \\ 0 & 0 & \frac{ 1 }{ 7 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 0 & - \frac{ 1 }{ 7 } & 1 \\ 1 & 0 & 0 \\ 2 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 7 & 0 \\ 0 & 0 & \frac{ 1 }{ 7 } \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 1 & 2 \\ - \frac{ 1 }{ 7 } & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & - 3 \\ \end{array} \right) $$