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I have a matrix \begin{bmatrix} 6I &-4I&0&0&......0\\ -4I &6I& -4I & 0&......0\\ 0 & -4I & 6I & -4I &......0\\ .\\ .\\ .\\ 0&0&0&........-4I& 6I \end{bmatrix}

where $I$ is a $2 \times 2$ matrix. I want to show that the above matrix is a positive definite.

This matrix is symmetric and I need to show that all of the Eigen values of the matrix are positive. To find all the Eigen values by using the definition looks little hard to me.

Is there any other easy way to show that the above mentioned matrix is positive definite?

User124356
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  • Which $2\times 2$ matrix is $I$? – Arthur Mar 26 '22 at 22:28
  • If $n$ is number of rows/columns of $I$, above matrix isn't positive definite for $n \ge 4$. – achille hui Mar 26 '22 at 23:55
  • Let $u$ be the $n \times 1$ column vector with all $1$. If a real symmetric $n \times n$ matrix $A = (a_{ij})$ is positive definite, then sum of all its entries $\sum_{i,j} a_{ij} = u^T A u > 0$. Your matrix start to fail at $n \ge 4$. – achille hui Mar 27 '22 at 00:07
  • Assume your $I$ is the $2\times 2$ identity matrix, its eigenvalues will be $6 - 8\cos\left(\frac{k\pi}{n+1}\right)$ for $k = 1,\ldots,n$. up to some sign, you can find a proof of that here. – achille hui Mar 27 '22 at 00:14

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