2

The matrix given to me is :

$$A=\begin{pmatrix}1&r&r\\r&1&r\\r&r&1 \end{pmatrix} $$ Find the values of $r$ for which this is positive definite. So,I naturally try to find the determinant of the matrix $$A-xI$$ where $$I$$ is the $3\times 3$ identity matrix.

The determinant is $${(1-x)}^3-2r^2(1-x)+2r^3$$

My idea was that solving this equation $${(1-x)}^3-2r^2(1-x)+2r^3=0$$ for $x$ in terms of $r$ and keeping in mind that each value of $x$ that is the eigen value has to be $\gt 0$ , I could find the required possibilities for $r.$ But the problem here became that I failed to solve this equation.

Please help me solve the equation for $x$ and I think I can finish the rest by myself .

Also, any suggestion of a different approach or different trick is most welcome .

Thank you .

Thomas Andrews
  • 177,126
user118494
  • 5,837
  • (A) given a matrix $K$ with all entries equal to $1,$ what are the eigenvalues? (B) what are the eigenvalues of $sK$ for real $s?$ (C) what are the eigenvalues of $sK + tI?$ Here all are square three by three – Will Jagy Dec 08 '15 at 20:04

2 Answers2

1

Here is a way: Let $e=(1,1,1)^T$, then we see that $e e^T$ is positive semidefinite with one eigenvalue of $3$ (in the direction $e$) and two at zero (on the subspace $\{e\}^\bot$).

Then we have $A = r e e^T + (1-r) I$. The eigenvalues of $A$ are those of $r e e^T$ shifted by $1-r$.

Details:

Since $r e e^T$ has eigenvalues $0, 3r$ we see that $A$ has eigenvalues $1-r, 1+2r$. For these eigenvalues to be positive, we need $1+2r >0$ or $r > -{1 \over 2}$ and $r <1$, so equivalently we need $r \in (-{1 \over 2}, 1)$.

copper.hat
  • 172,524
  • the first one has nonzero eigenvalue $3$ – Will Jagy Dec 08 '15 at 20:07
  • @WillJagy: Thanks for catching that. Bad day. – copper.hat Dec 08 '15 at 20:08
  • I came up with a sort of standard matrix of eigenvectors for the matrix of all ones, all columns eigenvectors and pairwise orthogonal, but integer entries and so not unit length. An orthogonal matrix of eigenvectors can be made just by dividing each column by its length. http://math.stackexchange.com/questions/1414690/matrix-with-all-1s-diagonalizable-or-not/1414925#1414925 You may have seen this before, of course. – Will Jagy Dec 08 '15 at 20:15
  • 1
    @WillJagy: I haven't; that's pretty, curious and pretty neat! (Also slightly similar to a matrix used to show that Gaussian elimination with partial pivoting can have $2^{n-1}$ element growth.) – copper.hat Dec 08 '15 at 20:17
  • don't think I know your Gaussian example. – Will Jagy Dec 08 '15 at 20:20
  • @WillJagy: I misremembered, the matrix used as an example showing the element growth in partially pivoted LU factorisation is of the same form but with all ones on the diagonal, -1 below, zeros above except for the rightmost column of all ones. Still a bad day :-). – copper.hat Dec 08 '15 at 20:28
0

Hint: Before calculating the determinant directly, add the second and the third row to the first row. Then add a multiple of the first row to the second and third rows to bring the matrix into a particularly simple form from which you'll be able to read the determinant immediately.

levap
  • 65,634
  • 5
  • 79
  • 122