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What is the condition on x for n by n symmetric matrix with 1 on diagonal and x everywhere else to be positive definite

Let the matrix be

\begin{bmatrix} 1 & x & x & ... &x\\ x & 1 & x & ... &x\\ x & x & 1 & ... &x\\ .. & .. & ..& ...& ..\\ x & x & x & ... &1 \end{bmatrix}

Gerry
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2 Answers2

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The eigenvalues of this matrix are $1-x$ and $1+(n-1)x$. For both to be positive, we must have

$$-\frac1{n-1}\lt x\lt 1\;.$$

joriki
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  • Thanks joriki, could you please elaborate on how you arrived at the fact that eigenvalues of this matrix are $1-x$ and $1 + (n-1)x$ – Gerry May 10 '16 at 22:19
  • @Gerry what are the eigenvalues of a matrix with every entry the same $x?$ For that matter, what are the eigenvalues of a matrix with every entry the same $1?$ What is the *rank* of a matrix with every entry the same $1?$ – Will Jagy May 10 '16 at 22:47
  • @WillJagy Please pardon my ignorance. Could you elaborate. Rank of matrix with every entry 1 is 1. – Gerry May 10 '16 at 23:06
  • @Gerry That's a start. Good. Next, in the $n$ by $n$ case, what does that mean the eigenvalues are? Try $n=2$ and $n=3$ by actually writing out the matrices... – Will Jagy May 11 '16 at 00:05
  • @WillJagy Thanks. For n by n case, eigenvalues are 0, n. So for matrix with every entry same x, it is $nx$. – Gerry May 11 '16 at 00:32
  • @Gerry Finally, what happens when you start with a matrix of all $x$ and add $(1-x)I,$ where $I$ is the $n$ by $n$ identity matrix? If you are not sure, consider some eigenvector $v$ of the matrix of all $x,$ call that $X.$ We know what $Xv$ is, in one case, $nxv,$ in the other cases $0.$ What is $(X + (1-x)I)v?$ – Will Jagy May 11 '16 at 00:38
  • @WillJagy. Ok that would be $(nx + (1-x))v$. $\implies (nx + (1-x))$ would be eigenvector of original matrix. – Gerry May 11 '16 at 00:48
  • @Gerry There you go. One eigenvalue is $1 + (n-1)x,$ the other eigenvalues are all $1-x.$ For what $x$ are all the eigenvalues positive? – Will Jagy May 11 '16 at 00:53
  • @WillJagy Thanks for your time and patience. Once the two eigenvalues are arrived at, the rest follows. Thanks again. – Gerry May 11 '16 at 00:58
  • @Gerry it turns out that we can, well, find a nice pattern for a basis of eigenvectors of the matrix with all entries $1.$ If we do not insist that the columns all have length one (as vectors) we can have all entries integers. I put a copy at http://math.stackexchange.com/questions/1414690/matrix-with-all-1s-diagonalizable-or-not/1414925#1414925 – Will Jagy May 11 '16 at 01:46
  • @WillJagy: That's nice. I guess you mean an orthogonal basis, since just any basis would be easier to specify as all but one of the vectors with $n-1$ ones and one $1-n$. – joriki May 11 '16 at 01:52
  • joriki, good point – Will Jagy May 11 '16 at 03:30
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My long answer would be based on the fact that all the pivots of Positive-Definite matrix are always greater than 0

Now when $n=1$, matrix is $ 1 X 1 $, and pivot is $1$

When $n=2$, row reduced echelon form is \begin{bmatrix} 1 & x\\ 0 & 1-x^2 \end{bmatrix}

and hence its pivots are $1, (1-x^2)$

When $n=3$, row reduced echelon form is, \begin{bmatrix} 1 & x & x \\ 0 & 1-x^2 & x(1-x) \\ 0 & 0 & \frac{(1-x)(1+2x)}{(1+x)} \end{bmatrix} and hence its pivots are $1, (1-x^2), \frac{(1-x)(1+2x)}{(1+x)}$

When $n=4$, row reduced echelon form is, \begin{bmatrix} 1 & x & x & x\\ 0 & 1-x^2 & x(1-x) & x(1-x)\\ 0 & 0 & \frac{(1-x)(1+2x)}{(1+x)} & \frac{(1-x)x}{(1+x)}\\ 0 & 0 & 0 & \frac{(1-x)(1+3x)}{(1+2x)} \end{bmatrix} and hence its pivots are $1, (1-x^2),\frac{(1-x)(1+2x)}{(1+x)}, \frac{(1-x)(1+3x)}{(1+2x)}$

Generalizing the results for $n$, pivots are $1, \frac{(1-x)(1+x)}{(1+0x)},\frac{(1-x)(1+2x)}{(1+x)}, \frac{(1-x)(1+3x)}{(1+2x)}, ... ,\frac{(1-x)(1+(n-1)x)}{(1+(n-2)x)}$

For matrix to be positive-definite, all the pivots should be greater than $0$

Clearly first pivot, $(1)$ is $>0$

for second pivot, $(1-X^2) > 0 \implies -1<x<1$

for third pivot, $\frac{(1-x)(1+2x)}{(1+x)} > 0 \implies \frac{-1}{2}<x<1$

for fourth pivot, $\frac{(1-x)(1+3x)}{(1+2x)} > 0 \implies \frac{-1}{3}<x<1$

Generalizing this way, for n pivots, $\frac{-1}{(n-1)}<x<1$

Gerry
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