# Showing that given matrix does not have negative eigenvalues without using the knowledge that it is positive definite.

Let $a,b,c$ be a positive real number such that $b^2+c^2<a<1$. Let $A=\begin{bmatrix} 1&b&c\\ b&a & 0\\ c & 0 & 1\end{bmatrix}$. Then

Consider the above matrix. I want to comment about the nature of eigenvalues of the matrix in the sense that, they are all positive, all negative, mix of positive or negative, non zero, real or non real etc..

My efforts

We look at the matrix first and see if it looks like one of the familiar type introduced in standard Linear Algebra Text.

We can see, this matrix is symmetric.

As soon we hear the term "symmetric matrix" and there is already the term "eigenvalue" in the question. We go the next standard result which says that a real symmetric matrix is diagonalizable with all eigenvalues real.

Conclusion so far The given matrix has only real eigenvalues.

Another standard result is the sum of eigenvalues is equal to the trace of the matrix. Trace is positive here due to the conditions specified.

So not all eigenvalues can be negative.

So we are left with two choices

1. all eigenvalues are positive

2. Eigenvalues of A are either positive or negative.

I know this matrix is positive definite(I have already proved it, by showing that all sub determinant are positive) so all eigenvalues are positive.

My aim is to show that there are no negative eigenvalues without going into the theory of positive definite matrices.

• Assume the opposite of what you're trying to prove and demonstrate that that assumption leads to a contradiction. – phdmba7of12 Sep 14 '18 at 14:28
• Suppose it has some negative eigenvalue how can I effectively say that sum of all eigenvalues will negative. @phdmba – StammeringMathematician Sep 14 '18 at 14:32
• Reproduce a proof of the claim for positive definite matrices. I wonder whether you can simplify it (in your particular case). – metamorphy Sep 14 '18 at 14:40
• As far I know if we add two matrices eigenvalues need not add up. Right? @amsmath – StammeringMathematician Sep 14 '18 at 14:55
• Forget it. The matrix $A_1$ is not positive semi-definite. I was mistaken. – amsmath Sep 14 '18 at 14:59

The characteristic polynomial of your matrix is $$p(x) = (a-x)(1-x)^2-c^2(a-x)-b^2(1-x).$$ Now, if $x < 0$, then \begin{align*} p(x) &> (a-x)(1-x)^2-c^2(1-x)-b^2(1-x) = (1-x)\cdot[(a-x)(1-x)-c^2-b^2]\\ &> (1-x)\cdot[x^2-(a+1)x] = x(1-x)(x-a-1) > 0. \end{align*} Therefore, $p$ cannot have zeros in $(-\infty,0)$. Also, $p(0) = \det A > 0$. Thus, the eigenvalues of $A$ are positive.

You may use Sylvester's Law of Inertia. Here is a matrix of determinant $1$

$$R = \left( \begin{array}{ccc} 1& -b & \frac{-ac}{a-b^2} \\ 0&1 & \frac{bc}{a-b^2} \\ 0&0 &1 \\ \end{array} \right)$$ and a "congruence diagonalization" $R^T AR = D$ $$\left( \begin{array}{ccc} 1&0 &0 \\ -b&1 &0 \\ \frac{-ac}{a-b^2}& \frac{bc}{a-b^2}&1 \\ \end{array} \right) \left( \begin{array}{ccc} 1&b &c \\ b& a&0 \\ c& 0&1 \\ \end{array} \right) \left( \begin{array}{ccc} 1& -b & \frac{-ac}{a-b^2} \\ 0&1 & \frac{bc}{a-b^2} \\ 0&0 &1 \\ \end{array} \right) = \left( \begin{array}{ccc} 1&0 &0 \\ 0&a-b^2 &0 \\ 0&0 & \frac{a-ac^2 - b^2}{a-b^2}\\ \end{array} \right)$$

I found $R$ using a fairly clean algorithm, see http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr

Let's see, I have written a C++ program for this congruence diagonalization, but it accepts only matrices with integer entries (and square and symmetric). In my program, the matrix of determinant $1$ uses only rational entries. For your problem, with symbol entries, back to one step at a time. However, it is exactly the same algorithm. The bottom line is that you do not need to calculate the eigenvalues themselves, this process tells you how many are positive, how many are exactly zero, how many are negative.

parisize = 4000000, primelimit = 500000
? h = [ 1,b,c; b,a,0; c,0,1]
%1 =
[1 b c]

[b a 0]

[c 0 1]

? ht = mattranspose(h)
%2 =
[1 b c]

[b a 0]

[c 0 1]

? h-ht
%3 =
[0 0 0]

[0 0 0]

[0 0 0]

? r1 = [ 1,-b,-c; 0,1,0; 0,0,1]
%4 =
[1 -b -c]

[0  1  0]

[0  0  1]

? r1t = mattranspose(r1)
%5 =
[ 1 0 0]

[-b 1 0]

[-c 0 1]

? r1t * h * r1
%6 =
[1        0        0]

[0 -b^2 + a     -c*b]

[0     -c*b -c^2 + 1]

? r2 = [ 1,0,0; 0,1,(b * c)/(a - b^2); 0,0,1]
%7 =
[1 0              0]

[0 1 c*b/(-b^2 + a)]

[0 0              1]

? r2t = mattranspose(r2)
%8 =
[1              0 0]

[0              1 0]

[0 c*b/(-b^2 + a) 1]

? r2t * r1t * h * r1 * r2
%9 =
[1        0                                0]

[0 -b^2 + a                                0]

[0        0 (-b^2 + (-a*c^2 + a))/(-b^2 + a)]

? r
%10 = r
? r = r1 * r2
%11 =
[1 -b -a*c/(-b^2 + a)]

[0  1  c*b/(-b^2 + a)]

[0  0               1]

? rt = mattranspose(r)
%12 =
[              1              0 0]

[             -b              1 0]

[-a*c/(-b^2 + a) c*b/(-b^2 + a) 1]

? h
%13 =
[1 b c]

[b a 0]

[c 0 1]

? r
%14 =
[1 -b -a*c/(-b^2 + a)]

[0  1  c*b/(-b^2 + a)]

[0  0               1]

? rt * h * r
%15 =
[1        0                                0]

[0 -b^2 + a                                0]

[0        0 (-b^2 + (-a*c^2 + a))/(-b^2 + a)]

? d = rt * h * r
%16 =
[1        0                                0]

[0 -b^2 + a                                0]

[0        0 (-b^2 + (-a*c^2 + a))/(-b^2 + a)]

? q = d[3,3]
%17 = (-b^2 + (-a*c^2 + a))/(-b^2 + a)
? p
%18 = p
? p = q * (a - b^2)
%19 = -b^2 + (-a*c^2 + a)
?


Your argument is very constructive. $$A=\begin{bmatrix} 1&b&c\\ b&a & 0\\ c & 0 & 1\end{bmatrix}$$ $$\det A = c(-ac) +(a-b^2) = a - b^2 -c^2+c^2 -ac^2 = \underbrace{a -(b^2+c^2)}_{>0} +c^2(\underbrace{1-a}_{>0})$$ Now you know that both the trace and determinant are positive. That leaves you with two choices:

1) Either all three eigenvalues are positive.

2) Or Two are negative and one is positive.

The eigenvalue/eigenvector relation give ($Av = \lambda v$) \begin{align} v_1 + bv_2 + cv_3 &= \lambda v_1\\ bv_1 + av_2 &= \lambda v_2\\ cv_1 + v_3 &= \lambda v_3 \end{align} which gives $$v_3 = \frac{c}{\lambda -1}v_1$$ $$v_2 = \frac{b}{\lambda - a}v_1$$ $$\frac{b^2}{\lambda - a}+ \frac{c^2}{\lambda-1} = \lambda -1$$ If $\lambda < 0$, then $\lambda - 1<-1$. But $$\frac{b^2}{\lambda - a}+ \frac{c^2}{\lambda-1} > \frac{b^2}{\lambda - 1}+ \frac{c^2}{\lambda-1} = \frac{b^2 + c^2}{\lambda - 1} > -(b^2 + c^2) > -a > - 1$$ This means that $$\underbrace{\frac{b^2}{\lambda - a}+ \frac{c^2}{\lambda-1}}_{>-1} = \underbrace{\lambda -1 }_{< -1}$$ CONTRADICTION

The characteristic polynomial is:

$$p(λ) = λ^3 - (a + 2) λ^2 + (2 a - b^2 - c^2 + 1) λ + a c^2 - a + b^2$$

Using that $\,b^2+c^2 < a < 1\,$, the coefficients turn out to have alternating signs:

$$-(a+2) \lt 0 \\ 2 a - b^2 - c^2 + 1 \gt 2a - a + 1 = a+1 \gt 0 \\ a c^2 - a + b^2 \lt c^2+b^2 - a \lt 0$$

By Descartes' rule of signs $\,p(\lambda)\,$ has no negative roots, and either $\,1\,$ or $\,3\,$ positive roots. Also:

$$\require{cancel} p(0) \lt 0 \\ p(1) = \cancel{1} -(\bcancel{a}+\cancel{2})+(\bcancel{2a}-\xcancel{b^2}-c^2+\cancel{1})+ac^2-\bcancel{a}+\xcancel{b^2} = (a-1)c^2 \lt 0 \\$$

Let $\,\lambda = x+1\,$, then:

$$q(x) = p(x-1) = x^3 + \underbrace{(1 - a)}_{\gt \,0} x^2 \underbrace{- (b^2 + c^2)}_{\lt \,0} x + \underbrace{(1-a)}_{\gt \,0}c^2$$

Again by the rule of signs, $\,q(x)\,$ has (at least) one positive root, and (at least) one negative root, therefore all roots of $\,q(x)\,$ are real, and so must be the roots of $\,p(x)\,$. Moreover, since $\,p(x)\,$ is negative at $\,0\,$ and $\,1\,$, it follows that two of the roots are in $\,(0,1)\,$ and the third one in $\,(1, \infty)\,$.