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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

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Are $A$, $B$, and $C$ scalars, or are they blocks (matrices themselves)? –  Arturo Magidin Jun 25 '12 at 18:20
    
The hint is likely that $A + 4B + 5C = \mathrm{tr} (\begin{pmatrix} A & B \\ B & C \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix})$ –  Cocopuffs Jun 25 '12 at 18:26
    
Yes they are scalars, not blocks. –  YNWA Jun 25 '12 at 18:34

2 Answers 2

up vote 2 down vote accepted

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.

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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! –  YNWA Jun 26 '12 at 9:57

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 $tr XY\ge 0$.

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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 ? –  YNWA Jun 25 '12 at 20:28

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