Condition such that the symmetric matrix has only positive eigenvalues

In summary: If the second element of the diagonal matrix is positive then the first element of the diagonal matrix is positive too.Sorry, I am not sure what you are referring to. Could you please provide more context or clarify?
  • #1
songoku
2,302
325
Homework Statement
Please see below
Relevant Equations
det (A - Iλ) = 0
1681048457077.png


My attempt:
$$
\begin{vmatrix}
1-\lambda & b\\
b & a-\lambda
\end{vmatrix}
=0$$
$$(1-\lambda)(a-\lambda)-b^2=0$$
$$a-\lambda-a\lambda+\lambda^2-b^2=0$$
$$\lambda^2+(-1-a)\lambda +a-b^2=0$$
The value of ##\lambda## will be positive if D < 0, so
$$(-1-a)^2-4(a-b^2)<0$$
$$1+2a+a^2-4a+4b^2<0$$
$$a^2-2a+1+4b^2<0$$
$$(a-1)^2+4b^2<0$$

But the LHS of the inequality is always positive so it is not possible for ##(a-1)^2+4b^2## to be less than 0

Where is my mistake? Thanks

Edit: I just realised my mistake. I will revise my working on post #2
 
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  • #2
The value of ##\lambda## will be positive if:
1.
$$D \geq 0$$
$$(a-1)^2+4b^2 \geq 0 \rightarrow a ~\text{and} ~b \in \mathbb{R}$$

2.
$$\lambda_1 + \lambda_2 > 0$$
$$a+1>0$$
$$a>-1$$

3.
$$\lambda_1 . \lambda_2>0$$
$$a-b^2>0$$
$$a>b^2$$

Intersecting all three conditions, the restriction for a and b is ##a>b^2~,a, b \in \mathbb{R}##

Am I correct? Thanks
 
  • #3
[itex]a > b^2[/itex] is the condition that the determinant must be positive. This is necessary for two positive eigenvalues, but it is not sufficient: A positive determinant is also consistent with two negative eigenvalues. So clearly something further is required.

The characteristic equation of a 2x2 matrix [itex]A[/itex] is [tex]
\lambda^2 - \operatorname{tr}(A)\lambda + \det A = 0.[/tex] For a symmetric matrix [itex]\begin{pmatrix} a & b \\ b & d \end{pmatrix}[/itex] we have [tex]
\operatorname{tr}^2 - 4\det = (a + d)^2 - 4ad + 4b^2 = (a - d)^2 + 4b^2 \geq 0[/tex] showing that the eigenvalues are always real.

Both roots of a quadratic will be positive if the lesser root is, so the condition is [tex]
\operatorname{tr}(A) - \sqrt{\operatorname{tr}(A)^2 - 4\det A} \geq 0.[/tex] Note that if [itex]\det A \geq 0[/itex] (as it must be for two positive roots) then [tex]
0 \leq \sqrt{\operatorname{tr}(A)^2 - 4 \det A } \leq |\operatorname{tr}(A)| [/tex] because the square root function is strictly increasing on the positive reals. It follows that [itex]\operatorname{tr}(A) - \sqrt{\operatorname{tr}(A)^2 - 4 \det A}[/itex] has the same sign as [itex]\operatorname{tr}(A)[/itex].
 
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  • #4
songoku said:
Intersecting all three conditions, the restriction for a and b is ##a>b^2~,a, b \in \mathbb{R}##

Am I correct? Thanks
Hi @songoku. I get the same answer as you though I found your proof difficult to follow.

A symmetric matrix with positive eigenvalues is ‘positive definite’. (Though that's not the definition of positive definiteness.)

If you can use this fact, there are tests for positive definiteness which enable you to answer the question with little effort.

(I’m only a dabbler, so any experts are invited to correct me if this is wrong/misleading.)
 
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  • #5
pasmith said:
[itex]a > b^2[/itex] is the condition that the determinant must be positive. This is necessary for two positive eigenvalues, but it is not sufficient: A positive determinant is also consistent with two negative eigenvalues. So clearly something further is required.

The characteristic equation of a 2x2 matrix [itex]A[/itex] is [tex]
\lambda^2 - \operatorname{tr}(A)\lambda + \det A = 0.[/tex] For a symmetric matrix [itex]\begin{pmatrix} a & b \\ b & d \end{pmatrix}[/itex] we have [tex]
\operatorname{tr}^2 - 4\det = (a + d)^2 - 4ad + 4b^2 = (a - d)^2 + 4b^2 \geq 0[/tex] showing that the eigenvalues are always real.

Both roots of a quadratic will be positive if the lesser root is, so the condition is [tex]
\operatorname{tr}(A) - \sqrt{\operatorname{tr}(A)^2 - 4\det A} \geq 0.[/tex] Note that if [itex]\det A \geq 0[/itex] (as it must be for two positive roots) then [tex]
0 \leq \sqrt{\operatorname{tr}(A)^2 - 4 \det A } \leq |\operatorname{tr}(A)| [/tex] because the square root function is strictly increasing on the positive reals. It follows that [itex]\operatorname{tr}(A) - \sqrt{\operatorname{tr}(A)^2 - 4 \det A}[/itex] has the same sign as [itex]\operatorname{tr}(A)[/itex].
I try to solve this condition but I get more or less same result
[tex]
\operatorname{tr}(A) - \sqrt{\operatorname{tr}(A)^2 - 4\det A} \geq 0.[/tex]
$$1+a \geq \sqrt{(1-a)^2+4b^2}$$
$$(1+a)^2 \geq (1-a)^2+4b^2$$
$$1+2a+a^2 \geq 1-2a+a^2+4b^2$$
$$4a \geq 4b^2$$
$$a \geq b^2$$

Intersecting with previous restriction, the answer will be ##a>b^2##

Am I missing something?

Steve4Physics said:
Hi @songoku. I get the same answer as you though I found your proof difficult to follow.

A symmetric matrix with positive eigenvalues is ‘positive definite’. (Though that's not the definition of positive definiteness.)

If you can use this fact, there are tests for positive definiteness which enable you to answer the question with little effort.

(I’m only a dabbler, so any experts are invited to correct me if this is wrong/misleading.)
Do you mean using REF and restricting the pivot position to be positive?

Thanks
 
  • #6
I also get ## a > b^2 ##
 
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  • #7
songoku said:
Do you mean using REF and restricting the pivot position to be positive?

Thanks
Yes, that's right. For information, another way is to check that all the ‘leading determinants' are positive.

I’m not sure if ‘leading determinants’ is the correct terminology but for an ##n \times n## matrix with elements ##a_{ij}##, the ‘leading determinants’ are:

##\begin{vmatrix}
a_{11}
\end{vmatrix} ##

##\begin{vmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}##

##\begin{vmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{vmatrix}##

etc.

This is simple in your problem: the 1st determinant (single element) is positive by inspection; the requirement that the 2nd determinant is positive gives the required inequality directly.
 
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  • #8
songoku said:
I try to solve this condition but I get more or less same result
[tex]
\operatorname{tr}(A) - \sqrt{\operatorname{tr}(A)^2 - 4\det A} \geq 0.[/tex]
[tex]\begin{split}
1+a &\geq \sqrt{(1-a)^2+4b^2} \\
(1+a)^2 &\geq (1-a)^2+4b^2 \\
1+2a+a^2 &\geq 1-2a+a^2+4b^2\\
4a &\geq 4b^2\\
a &\geq b^2\end{split}[/tex]

Your problem is that although you know that [itex]\det A > 0[/itex], you do not know anything about [itex]\operatorname{tr}(A)[/itex]. Isolating the root and squaring both sides is not helpful, since [itex](\operatorname{tr} A)^2[/itex] cancels, leaving you with [itex]\det A > 0[/itex], which you already knew. This is why I noted that if [itex]\det A > 0[/itex] then [tex]\sqrt{(\operatorname{tr} A)^2 - 4\det A} < |\operatorname{tr} A|.[/tex] The consequence is that if [itex]\operatorname{tr} A > 2\sqrt{\det A} > 0[/itex] then [tex]\operatorname{tr} A - \sqrt{(\operatorname{tr} A)^2 - 4\det A} > 0[/tex] because we are not subrracting enough from [itex]\operatorname{tr} A[/itex] to make this quantity negative.

It is easier, however, to start with general properties of quadratics.

The roots of [itex]\lambda^2 - p\lambda + q = 0[/itex] satisfy [tex]
\begin{split}
\lambda_1 + \lambda_2 &= p, \\
\lambda_1\lambda_2 &= q. \end{split}[/tex] If both roots are positive, then [itex]p > 0[/itex] and [itex]q > 0[/itex]. However, the converse does not hold in general: If [itex]p > 0[/itex] and [itex]q > 0[/itex] then the roots are either real and positive or a complex conjugate pair with positive real part. But here we can rule that case out, because we know that the roots are real. Thus the eigenvalues of a 2x2 symmetric matrix will be positive if and only if both the trace and the determinant are positive.

EDIT: It is correct here that if the determinant is positive, then [itex]a > b^2 \geq 0[/itex], and the trace being positive requires [itex]a > -1[/itex] which is less strict; I;mnot sure how I mised that.)
 
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  • #9
I understand. Thank you very much for the help and explanation pasmith, Steve4Physics, malawi_glenn
 

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