2
$\begingroup$

I need to prove, that the matrix $ \begin{pmatrix} A &B \\ B & C \end{pmatrix} $ has at least one positive eigenvalue, if known that $ A+4B+5C > 0 $.

I was told to show that $ \begin{pmatrix} 1 &2 \\ 2 & 5 \end{pmatrix} $ is positive definite. But I don't know what to do with that hint, and what is the connection between the inequality and and the matrix.

Two things I know and think that might be usefull is that the sum of eigenvalues is the trace of the matrix, and their product is the determinent

any usefull hints/directions?

big thank you

  • 0
    Yes they are scalars, not blocks.2012-06-25

2 Answers 2

2

Here's a way to do it without the hint.

The characteristic polynomial of the matrix is $\lambda^2-(A+C)\lambda+AC-B^2\tag1$ so the eigenvalues are ${A+C\pm\sqrt{(A-C)^2+4B^2}\over2}\tag2$ First, note that what's under the square root sign is non-negative, so the eigenvalues are real (this could also have been deduced from the theorem that says that the eigenvalues of a symmetric matrix are real). The larger eigenvalue is the one with the plus sign, so we just have to prove that if $A+4B+5C\gt0$ then $A+C+\sqrt{(A-C)^2+4B^2}\gt0\tag3$ This is certainly true if $A+C\ge0$, so we may assume $A+C\lt0$. Then (3) is equivalent to $(A-C)^2+4B^2\gt(A+C)^2\tag4$ which is equivalent to $B^2\gt AC\tag5$ Now from $A+4B+5C\gt0$ we get $4B\gt-A-5C\ge2\sqrt{5AC}\tag6$ where we have used the inequality of the arithmetic and geometric means. But from (6) we get $B^2\gt(5/4)AC\gt AC\tag7$ which is (5), and we're done.

  • 0
    fantastic! big thank you. I also think that I found a way to prove it by contradiction using Sylvester's criterion, but I still need to look into it. thank you!2012-06-26
0

With the hint by Cocopuffs, $A + 4B + 5C = \mathrm{tr} (\begin{pmatrix} A & B \\ B & C \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix})$. You only need to say $\begin{pmatrix} A & B \\ B & C \end{pmatrix}$ having two negative eigenvalues is not possible.

Fact: Let $X, Y$ be $n\times n$ positive semidefinite matrices, then $tr XY\ge 0$.

  • 0
    thank you for your help. but I dont get it, why is that true? Its been a long time since Ive learnt linear algebra... is there somekind of an assumption I need to know about trace of a product of matrices ?2012-06-25