# Homework Help: Linear algebra problem

1. Aug 6, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
If A is an n x n matrix over R such that A^3 = A + 1, prove that det(A) > 0 .

2. Relevant equations

3. The attempt at a solution
So, what I've done is factor the expression to get A(A+1)(A-1) = 1, then taking the determinant of both sides, I get det(A)det(A+1)det(A-1) = 1. I thought that maybe I could argue that maybe if det(A) < 0, then this would lead to a contradiction, but that doesn't seem to be going anywhere.
Should I try to solve it without the explicit use of determinants?

2. Aug 6, 2017

### StoneTemplePython

What are the eigenvalues of $\mathbf A$ ?

3. Aug 6, 2017

### Mr Davis 97

Wouldn't any eigenvalue have to satisfy the equation $\lambda^3 = \lambda + 1$? Whose only real solution is approximately 1.3247?

4. Aug 6, 2017

### Staff: Mentor

The decomposition is a good starting point. It tells us that $\det A = 0$ is impossible, so only $\det A < 0$ or $\det A > 0$ are left.
Now assume $\det A < 0$ and consider the real function $f(x)=\det (A-x \cdot 1)$. Is it continuous? Do you know something about the values at certain points? What does this mean for the other points?

Edit: I wouldn't consider eigenvalues, because you cannot be sure whether they are real or not. So eventually they don't exist.

5. Aug 6, 2017

### StoneTemplePython

Indeed. So any possible determinant is a product comprised of this real root (which is positive) raised to some natural number and the product of some conjugate pairs raised to a natural number...

6. Aug 6, 2017

### Mr Davis 97

What can we do with $f(x) = \det (A - xI)$? It doesn't seem to be a very malleable expression. The only two things I see is that f(0) < 0, by supposition. How does this let us conclude things about f(1) or f(-1)?

7. Aug 6, 2017

### Staff: Mentor

Well, you also know $f(0)\cdot f(-1) \cdot f(1) = 1 > 0$ and all are real numbers.

Edit: You can also use @StoneTemplePython's approach with conjugate eigenvalues, if this is easier to you. My suggestion is quasi by foot using the properties of continuous real functions.

8. Aug 6, 2017

### Mr Davis 97

Well if $f(0) < 0$ it would have to be the case that $f(1) \cdot f(-1) < 0$, which is to say that they have different signs, which would mean that we'd have to have a zero of f(x) on the interval $[-1,1]$, right? Is this a contradiction since there are no eigenvalues of A on this interval?

Also, $f(x)$ is continuous since it is an nth degree polynomial.

9. Aug 6, 2017

### Staff: Mentor

Yes, that's right. We have an $x$ with $f(x)=0$. Your conclusion is basically correct. I only would try to demonstrate why this is a contradiction. Where does $f(x)=0$ lead to? Where do the eigenvalues come in here?

10. Aug 6, 2017

### Mr Davis 97

I'm not completely sure where to go with $\det (A - xI)= 0$. Of course, the x's for which this is true are the eigenvalues of A, but I don't see how this is related to the interval [-1,1]

11. Aug 6, 2017

### Staff: Mentor

What does it mean, if a linear function has zero determinant?

12. Aug 6, 2017

### Mr Davis 97

That it's not invertible?

13. Aug 6, 2017

### Staff: Mentor

Yes, and thus it has a non-trivial kernel, i.e. a vector $v$ with $(A-x\cdot 1)(v)=0$. Now you can use $A^3=A+1$ again and apply it on $v$.

14. Aug 6, 2017

### Mr Davis 97

All I get is that $x$ must satisfy the equation $x^3= x+1$

15. Aug 6, 2017

### Ray Vickson

Eigenvalues work perfectly well here. Since $A^3 - A - I = 0$, $p(x) = x^3 - x - 1$ is the "minimal polynomial" of $A$, and that means that its roots are all the eigenvalues of $A$, not counting multiplicity. One of the eigenvalues is $\lambda_1 > 0$, as we can see from the fact that $p(x)$ changes sign between $x = 0$ and large positive $x$. Setting $p'(x) = 0$ implies that there are two stationary points, $x = \pm 1/\sqrt{3}$, and the second-derivative test shows that $x = -1/\sqrt{3}$ is a (local) maximum, while $x = +1/\sqrt{3}$ is a local minimum. Therefore, there can be only one positive root of $p(x) = 0$. Also, $p(-1/\sqrt{3}) < 0$, so $p(x)$ never rises above the 0 for $x < 0$, hence there are no negative real roots. Therefore, the other two eigenvalues are complex conjugates $\lambda_2$ and $\lambda_3 = \overline{\lambda}_2$, hence $\lambda_2 \times \lambda_3 > 0$. The characteristic polynomial $c(x)$ of $A$ must be a multiple of $p(x)$ but with the same roots, hence $c(x) = (x-\lambda_1)^r (x - \lambda_2)^s (x - \lambda_3)^t$. The matrix $A$ is real, so $c(x)$ is a real polynomial and so its complex roots must come in conjugate pairs. This means that we must have have $s = t \geq 1$. Thus, we can, indeed, determine the sign of the product of the eigenvalues.

Last edited: Aug 7, 2017
16. Aug 6, 2017

### Staff: Mentor

Yes, and this is a) the solution and b) as I said, the eigenvalue argument by foot. You get that for some real number $x_0$ between $x=-1$ and $x=1$ that it has to hold $x_0^3-x_0-1 = 0$. Until here you have used, that the function $f(x)$ is continuous, real-valued, that
and that $v \neq 0$.

All which is left is to calculate the real number $x_0$ or at least that it has to be greater than $1$.

17. Aug 6, 2017

### Mr Davis 97

To calculate that real number, is it okay and necessary to use software?

18. Aug 6, 2017

### Staff: Mentor

That's not me to decide it. I also looked it up on the Wolfram page. O.k., I had $A+x\cdot I$ and $x_0 = -1.3...$ but this doesn't make any difference. I added (basically) "or at least outside of $[-1,1]$" to my last post. This way one could do it analytically: calculate inflection points or show that there cannot be a zero between $-1$ and $1$ or even use Cardano's formulas. It would of course make some more work than the software solution, but it's possible.

If we have the polynomial $p(x)=x^3-x-1$ then $p(-1)=-1$ and $p(1)=-1$. Thus if it crossed the $x-$axis in between, it would have to have a maximum $x_m$ anywhere in between with $p(x_m)>0$. But this can easily be calculated by differentiation, solving a quadratic formula and evaluation of $p(x_m)$. So this is one possible way without software.