How to judge a special matrix is a positive-definite-matrix

Question

I came across a matrix and I need to prove it is a positive-definite-matrix

the matrix is $(a_{i,j})_{n\times n }$ where $ 0\leq i\leq n-1$ and $ 0\leq j\leq n-1$ $a_{i,j}=\beta_1^{i+j}+\beta_2^{i+j}$ where $1<\beta_1<\beta_2<2$ are fixed.

Answer 1

Consider the matrix $(\beta_k^{i+j})$, $k=1,2$, $0 \le i,j \le n-1$. One can clearly find an explicit form for its eigenvectors, which will be given by $$ \begin{bmatrix} -\beta_k^{n-1} & -\beta_k^{n-2} & \dots & -\beta_k & \frac {1}{\beta_k^{n-1}} \\ 0 & 0 & \dots & 1 & \frac {1}{\beta_k^{n-2}} \\ \vdots & & \ddots & & \vdots \\ 1 & \dots & \dots & & \frac {1}{\beta_k^0} \end{bmatrix}. $$ (The column vectors are the eigenvectors, the first $n-1$ eigenvectors have eigenvalue $0$ and the last one has eigenvalue $n \beta_k^{n-1}$. The diagonal dots should go from bottom left to top right, didn't know how to format them...)

Since its eigenvalues are all non-negative, the matrix is semi-positive definite. Let $v \in \mathbb R^n$. Then if $A = (a_{ij})$ and $B_k = (\beta_k^{i+j})$, then $$ v^{\top} A v = v^{\top} B_1 v + v^{\top} B_2 v \ge 0 $$ by what we have just shown. Assume that $v^{\top} A v = 0$, which means $v^{\top} B_1 v = v^{\top} B_2 v = 0$. By the spectral theorem, we can write $$ V_1^0 \oplus V_1^{n\beta_1^{n-1}} = \mathbb R^n = V_2^0 \oplus V_2^{n\beta_1^{n-1}} $$ where the $V_k^{\lambda}$ are the eigenspaces of $B_k$ for the eigenvalue $\lambda$ and those subspaces are orthogonal. Notice how the kernel of $B_k$ has dimension $n-1$, so $$ \dim( (\ker B_1) \cap (\ker B_2) ) \ge n-2. $$ Therefore, for $n \ge 3$, this intersection has dimension $\ge 1$, so we can find a vector $v \in \mathbb R^n$ such that $B_1 v = B_2 v = 0$, hence $v^{\top} Av = 0$ and $A$ is not positive definite for $n \ge 3$, although it is semi-positive definite.

For $n=2$, the condition $\beta_1 < \beta_2$ ensures that the eigenvectors of $B_1$ from the eigenvalue $n\beta_1^{n-1}$ and of $B_2$ from the eigenvalue $n\beta_2^{n-2}$ are not colinear, hence they give a basis of $\mathbb R^2$ and thus for any $v \in \mathbb R^2 \backslash \{0\}$, we have $$ v^{\top} A v \ge \max\{ v^{\top} B_1 v, v^{\top} B_2 v \} > 0 $$ because $v$ has a non-zero component in one of the two eigenvectors.

P.S. You might as well have assumed $\beta_1, \beta_2 > 0$ and $\beta_1 \neq \beta_2$ instead of $1 < \beta_1 < \beta_2 < 2$.

Hope that helps,