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I came across the following question that says:

Let $a,b,c$ are positive real numbers such that $b^2+c^2<a<1.$ Consider the $3 \times 3$ matrix $A=\begin{pmatrix} 1 &b &c \\ b &a &0 \\ c &0 &1 \end{pmatrix}$. Now I have to show that all the eigenvalues of $A$ are positive real numbers.

My Attempt: I have taken the column matrix $z=\begin{pmatrix} x &y &z \end{pmatrix}^T$. Now,$z^TAz=x^2+ay^2+z^2+2bxy+2czx.$ I have to show that $z^TAz>0.$ The information I have on my hand is: $a,b,c >0,b^2+c^2<a<1$ and $A$ is symmetric . Now how can I make a conclusion from these that $z^TAz>0 ?$

Can someone point me in the right direction? Thanks in advance for your time.

EDIT: As @DonAntonio commented that $A$ is positive definite symmetric matrix and so all the eigenvalues of $A$ are positive real numbers. But for this I have to show that $z^TAz>0 .$ How Can I do that? Do I have to find out the eigenvalues of $A$ independently and then make a conclusion from it? I am still confused.

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    $\begingroup$ In the link I wrote in my answer there's explained that a matrix is definite positive iff all its leading principal factors are positive, which in this case is rather easy to check. $\endgroup$ – DonAntonio Mar 20 '13 at 9:59
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    $\begingroup$ Thanks a lot sir.I have got your point. $\endgroup$ – learner Mar 20 '13 at 13:22
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For symmetry: $$A=\begin{pmatrix} 1 &b &c \\ b &a &0 \\ c &0 &1 \end{pmatrix}$$ $\Longrightarrow $
$$ A^{T} = \begin{pmatrix} 1 &b &c \\ b &a &0 \\ c &0 &1 \end{pmatrix} = A \; \; \; \checkmark $$


For positive definiteness notice that $A$'s leading principal minors are: $$ 1>0 \; \; \; \checkmark \\ a-b^2> c^2 > 0 \; \; \; \checkmark \\ a + -b\cdot(b) + c\cdot(-ac) = a-b^2 - ac^2> 0 \; \; \; \checkmark $$ Note that a symmetric matrix is just a special case of a hermitian matrix. A hermitian matrix is positive definite if its leading principal minors are all positive. Indeed we observe this, so that the matrix $A$ is positive definite symmetric. Thus the matrix $A$ has all real positive eigenvalues.


Definition

The $k^{\text{th}}$ leading principal minor of a matrix $M$ is the determinant of its upper left $k$ by $k$ sub-matrix.

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  • $\begingroup$ Thanks for the detailed explanation. But I have one question.I can not understand the calculation $"a + -b\cdot(b) + c\cdot(b) = a-b^2 + bc> a-(b^2+c^2) > 0 \; \; \; \checkmark ".$ Is not the third leading principal minor be $|A|=a-b^2-ac^2$? $\endgroup$ – learner Mar 20 '13 at 13:19
  • $\begingroup$ @learner yeah i think i messed up. $\endgroup$ – Rustyn Mar 20 '13 at 14:41
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The matrix is positive definite symmetric and thus all its eigenvalues are real positive.

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