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Let $A=(1-a)I_n + a J_n$
Find the values of $~a~$ so the Matrix is p.d ?
Note:$~I_n~$ is the identity matrix and $~J_n$ is the $1's$ matrix.

I know that $~~A$ is p.d $~iff~λ_i >0$ so, I need to choose $a$ so that $~λ_i >0$

First I tried to use the definition to find the eigen $det(A-λI_n)=0$
but it real complicated.

any help would be appreciated.

This question from A First Course in Linear Model Theory by Nalini, chapter 2.

Update: Trying to use the hint by @ whuber

By using the idea of inspection, I found the Gershgorin theorem ( first time I heard about it)
https://math.la.asu.edu/~gardner/Gcircle.pdf

By the theorem we conclude that all$~ λi~$ in the disk

$D(1, (n-1)a)~~center~ at~ 1~, ~~and~ radius~ (n-1)a$

so in order to have positive value for $~ λi~$ we need to take

$|(n-1)a|<1~~$ hence, $-1/n-1<a<1/n-1$

@ whuber Is there another idea without using this theorem because we did not cover it in my class? Thanks

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    $\begingroup$ Hint: because $I_n$ and $J_n$ commute, they have the same eigenspaces. This enables you to know the eigenspaces by inspection and allows you to read off the eigenvalues immediately. $\endgroup$
    – whuber
    Commented Feb 9, 2019 at 19:19
  • $\begingroup$ Please add the self-study tag and have a look at its wiki $\endgroup$
    – user20160
    Commented Feb 9, 2019 at 21:43
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    $\begingroup$ Since $A-\lambda I_n=(1-a-\lambda)I_n+a\mathbf1_n\mathbf1_n^\top$, determinant of $A-\lambda I$ can be easily calculated if you want using Matrix determinant lemma. It is important to note that $A$ is symmetric, for only then we can say that $A$ is p.d. iff its eigenvalues are positive. $\endgroup$ Commented Feb 11, 2019 at 7:30

1 Answer 1

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Let's start with a definition and see how far it can take us. $A$ is positive definite if and only if for every nonzero vector $x,$ $x^\prime A x \gt 0.$ Let us therefore lead off with this latter expression and use what we know about $A.$

To avoid interrupting the flow of algebra, let me observe right now that $J_n$ is an outer product:

$$J_n = \mathbf{1}_n \mathbf{1}_n^\prime$$

where $\mathbf{1}_n = (1,1,\ldots,1)^\prime$ is a column vector of ones. Note, too, that

$$x^\prime \mathbf{1}_n = \left(x^\prime \mathbf{1}_n\right)^\prime =\mathbf{1}_n^\prime x$$

because both sides are scalars.

So:

$$\eqalign{ x^\prime A x &= x^\prime\left((1-a)I_n + a J_n\right) x \\ &= (1-a) x^\prime I_n x + a x^\prime \left(\mathbf{1}_n \mathbf{1}_n^\prime\right) x \\ &= (1-a)\|x\|^2 + a \left(x^\prime \mathbf{1}_n\right) \left(\mathbf{1}_n^\prime x\right) \\ &= (1-a)\|x\|^2 + a\left(\mathbf{1}_n^\prime x\right)^2. }$$

At this point there are many ways to proceed: exploit well-known inequalities; use Lagrange multipliers; apply induction; change to a suitable basis; etc. I leave the choice to you and your mathematical taste. If you're still not sure of the answer, read over the replies at Bound for the correlation of three random variables to see what it looks like for $n=3$ and generalize.


Perhaps the easiest and most insightful route to a solution is a geometric one. The geometry is brought out by generalizing the relation $J_n=\mathbf{1}_n\mathbf{1}_n^\prime$ to

$$J = \mathbf{u}\mathbf{u}^\prime$$

for some arbitrary nonzero vector $\mathbf u.$ This vector defines a subspace of dimension $n-1$ via

$$\mathbf{u}^\perp = \{\mathbf x \in\mathbb{R}^n\mid \mathbf{u}^\prime \mathbf x = 0\}.$$

This is the kernel of $J:$ that is, $J$ restricted to $\mathbf{u}^\perp$ is the zero transformation.

When $(\mathbf u_2, \mathbf u_3,\ldots, \mathbf u_n)$ is a basis of $\mathbf{u}^\perp,$ adjoining $\mathbf u_1 = \mathbf u$ gives a basis $\mathcal U = (\mathbf u_1, \mathbf u_2, \ldots, \mathbf u_n)$ for the entire space. Because obviously

$$I_n \mathbf{u}_j = \mathbf{u}_j$$

for all $j=1,2,\ldots, n$ and

$$J \mathbf{u}_1 = J \mathbf{u} = \left(\mathbf{u}\mathbf{u}^\prime\right)\mathbf u = \mathbf{u}\left(\mathbf{u}^\prime \mathbf u\right) = ||\mathbf u||^2 \mathbf{u},$$

$\mathcal U$ is an eigenbasis for both $I_n$ and $J.$ Consequently it is an eigenbasis for any linear combination like $(1-a)I_n + aJ,$ all of whose eigenvalues can be read directly from the coefficients on the right hand sides of the equations

$$((1-a)I_n + aJ)\mathbf{u}_1 = \left((1-a) + a||\mathbf{u}||^2\right)\mathbf{u}_1$$

and

$$((1-a)I_n + aJ)\mathbf{u}_j = (1-a)\mathbf{u}_j$$

for $j=2, 3, \ldots, n.$ Since this linear combination is positive semidefinite if and only if all eigenvalues are nonnegative, we have discovered the necessary and sufficient conditions

$$ (1 - a) + a||\mathbf{u}||^2 \ge 0;\ (1-a) \ge 0.$$

Solving for $a$ in the case $||\mathbf u||^2 \gt 1$ gives

$$ \frac{-1}{||\mathbf u||^2 - 1} \le a \le 1.$$

When $||\mathbf u||^2 \le 1,$ there is no lower bound on $a:$ only the upper bound holds.

For the case $J = J_n$ in the question, $\mathbf u = (1,1,\ldots, 1)$ for which $||\mathbf u||^2 = n.$

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