Science:Math Exam Resources/Courses/MATH307/April 2005/Question 01 (b)

MATH307 April 2005

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Question 01 (b)
Consider the symmetric matrix $B={\begin{pmatrix}2&-2&0&4\\-2&3&2&1\\0&2&3&3\\4&1&3&2\end{pmatrix}}$ (b) Find the number of positive and negative eigenvalues of B.

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Hint

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Solution

First we note that $B$ has all real entries and is symmetric so $B$ is Hermitian. It follows that the eigenvalues of $B$ must be real. With these conditions, we can apply the fact that the number of positive pivots = the number of positive eigenvalues which is given in by the following paper http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/positive-definite-matrices-and-applications/symmetric-matrices-and-positive-definiteness/MIT18_06SCF11_Ses3.1sum.pdf Hence, since we found in part (a) that row operations yield Failed to parse (unknown function "\begin{pmatrix}"): {\displaystyle B \rightarrow \begin{pmatrix} 2 & -2 & 0 & 4\\ 0 & 1 & 2 & 5\\ 0 & 0 & -1 & -7\\ 0 & 0 & 0 & 18 \end{pmatrix},} since there are 3 positive pivots, we must have 3 positive eigenvalues and 1 negative eigenvalue. We can also check this somewhat by considering that the determinant of a Hermitian matrix is the product of its eigenvalues. Recall that we found in part(a): $B=LDU={\begin{pmatrix}1&0&0&0\\-1&1&0&0\\0&2&1&0\\2&5&7&1\end{pmatrix}}{\begin{pmatrix}2&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&18\end{pmatrix}}{\begin{pmatrix}1&-1&0&2\\0&1&2&5\\0&0&1&7\\0&0&0&1\end{pmatrix}}.$ The determinant of $B$ can be found by using part (a): $det(B)=det(LDU)=det(L)det(D)det(U)=(1)(2*1*-1*18)(1)=-36$. Note that we used the following determinant properties: For any matrices $A,B$, we have $det(AB)=det(A)det(B)$ For upper or lower triangular matrices, the determinant is the product of the diagonal entries Since we found that $det(B)<0$, there must be either 1 or 3 negative eigenvalues. This agrees with our earlier finding that there are 3 positive eigenvalues and 1 negative eigenvalue.