Prove that a positive $n \times n$ matrix is positive definite

Question

Let constant $0<k<1$. Prove that the $n\times n$ matrix $A$, where its diagonal entries $a_{ii} =1$ and $a_{ij}=k$ for $i \neq j$, is Positive definite.

By definition, I know that a matrix $A$ is positive definite if $x^T A x>0$ for all vectors $x\neq 0$. And I also know that an $n \times n$ matrix is positive definite iff all its eigenvalues are positive. But I am stuck here.

Any help or hint is appreciated. Thanks

This seems false. the matrix is the same as $B-(k-1)I$ where $B$ is that rank $1$ matrix consisting of only $k$'s. It follows that $-(k-1)$ is an eigenvalue with multiplicity $n-1$. Of course if you swap the entries $k$ and $1$ and make $k>1$ then it's true. — Asinomás, Jun 22 '21 at 18:15
No, the matrix with only $1$ as entries (for $k=1)$ is not positive definite for $n\ge 2$. — Dietrich Burde, Jun 22 '21 at 18:15
Generally it is good to try simple examples, in this case a $2 \times 2$ example. — copper.hat, Jun 22 '21 at 18:18
actually it's never positive definite if the size is at least $2$. — Asinomás, Jun 22 '21 at 18:19
This statement will only be true if we are also given that $k < 1$. — Ben Grossmann, Jun 22 '21 at 18:23
The spectrum of the matrix is always ${ [-(k-1)]^{n-1},[nk-(k-1)]^1}$ — Asinomás, Jun 22 '21 at 18:26
Following the ideas from this post, we can see that the eigenvalues of this matrix will be $1-k$ with multiplicity $1$ and $n-1 + k$ with multiplicity $n-1$. — Ben Grossmann, Jun 22 '21 at 18:28
I think $1-k$ has multiplicity $n-1$ and the other eigenvalue is $nk+1-k$ — Asinomás, Jun 22 '21 at 18:33
Write $A= (1-k) I + k e e^T$ where $e$ is the vector of ones. Then the eigenvalues are $1-k +nk$ (eigenspace $\operatorname{sp} { e }$) and $1-k$ (eigenspace ${e}^\bot$). — copper.hat, Jun 22 '21 at 18:38
Thanks for the wonderful comments and observations. @Yorch, and co. This implies that the question isn't correct right? — Pacific, Jun 22 '21 at 18:45
@BenGrossmann thanks again for the observation. I have corrected the question the condition is $0<k<1$. Could you help with the prove. Thanks — Pacific, Jun 23 '21 at 10:40

score 1 · Answer 1 · answered Jun 23 '21 at 12:43

1

Bit of calculation is needed to prove it for a general $n \times n$ matrix. The fact $\left( \sum_{i=1}^n x_i \right)^2 = \sum_{i=1}^n x_i{}^2 + 2 \sum_{1 \leq i < j \leq n} x_i x_j$ is used.

Let $x=\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{pmatrix}$, $x^TAx$ will be of the form \begin{align} \sum_{i=1}^n x_i^2 + 2k \sum_{1 \leq i < j \leq n} x_i x_j. \end{align} If $\sum_{1 \leq i < j \leq n} x_i x_j \ge 0$ we're done. So let $\sum_{1 \leq i < j \leq n} x_i x_j < 0$. Then \begin{align} \sum_{i=1}^n x_i^2 + 2k \sum_{1 \leq i < j \leq n} x_i x_j &= \left(\sum_{i=1}^n x_i \right)^2 -2 \sum_{1 \leq i < j \leq n} x_i x_j +2k \sum_{1 \leq i < j \leq n} x_i x_j \\ &= \left(\sum_{i=1}^n x_i \right)^2+2(1-k) \left(-\sum_{1 \leq i < j \leq n} x_i x_j \right) \\ &>0. \end{align}

answered Jun 23 '21 at 12:43

absolute0

1,027

Thanks @absolute0 Please why did you use the first equality. I thought the goal was to check if it satisfies $x^T A x >0$ – Pacific Jun 25 '21 at 04:34
see the first equality is used in the 3rd last line to show $x^TAx >0$. that's why it's written. – absolute0 Jun 25 '21 at 06:02
oh I see thanks – Pacific Jun 25 '21 at 06:07
you're welcome. – absolute0 Jun 25 '21 at 06:11

score 1 · Answer 2 · answered Jun 23 '21 at 17:53

1

Let $J=\mathbf {11}^T$ be the $n\times n$ ones matrix. Then

$A= (1-k) I_n + k\cdot J\succeq (1-k) I_n \succ \mathbf 0$
Therefore A is PD.
In words: a PD matrix plus a PSD matrix is PD.

answered Jun 23 '21 at 17:53

user8675309

10,034

please why did you use $n \times n$ ones matrix. Since the conditions are $a_{ii} =1$ and $0<a_{ij}<1$. – Pacific Jun 25 '21 at 04:40
Since every off-diagonal entry is the identical, equal to $k$, we can write your matrix as $ D + k \cdot J$ for some diagonal matrix $D$. Furthermore, since all diagonal entries identically are equal to 1, this implies $D=(1-k)I_n$. – user8675309 Jun 25 '21 at 17:29

Prove that a positive $n \times n$ matrix is positive definite

2 Answers2