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I have the following matrix , \begin{equation*} \begin{bmatrix} T & -I & 0 & \cdots & \cdots & \cdots & 0 \\ -I & T & -I & \ddots & & & \vdots \\ 0 & \ddots & \ddots & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & & \ddots & \ddots & \ddots & \ddots & 0 \\ \vdots & & & \ddots & \ddots &\ddots & -I \\ 0 & \cdots & \cdots & \cdots & 0 & -I & T \\ \end{bmatrix} \end{equation*} where T = tridiag with 4 at main diagonal , -1 at the sub and super diagonal , I is the identity matrix , i want to prove that is positive definite , i used gershgorin circle theorem , but this theorem provide to me that eigenvalues are non-negative so i want to prove now that $\lambda =0 $ is not an eigevalue $\Leftrightarrow$ A is invertible . If you have an idea how to prove that A is positive definite or why A is invertible it will be helpfull. Thanks in advance

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    $\begingroup$ Let us denote by $m$ the number of times $T$ appears. I guess that $T > I$ always, is that correct? Let $m=2$. Then you can use the Schur complement $T - T^{-1}$ which is p.d. iff the matrix itself is. And if $T > I$, then $T - T^{-1}$ is p.d. (since $t - t^{-1} > 0$ for $t > 1$). So, if the assumption that $T > I$ is right, the matrix is p.d. in the case $m=2$. Maybe you can proceed by induction using the Schur complement. $\endgroup$ – Friedrich Philipp Mar 24 '16 at 18:03
  • $\begingroup$ Sorry what do you mean with $Τ>I$ i don't understand $\endgroup$ – chaviaras michalis Mar 24 '16 at 18:14
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    $\begingroup$ That means $T-I > 0$, i.e., $T - I$ is positive definite. What do you know about the eigenvalues of $T$? Do you know the smallest one? $\endgroup$ – Friedrich Philipp Mar 24 '16 at 18:18
  • $\begingroup$ from gershgorin circle theorem we can see that $T-I$ is positive definite , because $|\lambda - a_{ii}| \leq R_i = \sum_{j=1,j \neq i }^{n} |a_{ij}| $ so $R_i = 1 or 2 $ . All the eigenvalue are on the circle with center 3 and radius 2 or 1 , so all the eigenvalue is positive . Am i wrong ? $\endgroup$ – chaviaras michalis Mar 24 '16 at 18:37
  • $\begingroup$ Ok so i must do the induction now ... but i am still worrying why $T-I$ is p.d induce that $T-T^{-1}$ is p.d $\endgroup$ – chaviaras michalis Mar 24 '16 at 18:40
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Suppose $T$ is $m\times m$ and $A$ has $n$ blocks in each block column (i.e. $A$ is $mn\times mn$). Let $B=T-2I_m$. Then $\det B=m+1$. Hence $B$ is invertible. Clearly $B$ is also weakly diagonally dominant. Therefore $B$ is positive definite. Now $A=B\otimes I_n+I_n\otimes B$. Therefore $A$ is positive definite too.

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  • $\begingroup$ I just saw $\det B = m+1$ by induction, but is there an easier way? $\endgroup$ – Friedrich Philipp Mar 24 '16 at 20:02
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    $\begingroup$ Knowing that $\det(B) = m+1$, its positive definiteness can also be seen from the leading principle minors criterion. $\endgroup$ – Friedrich Philipp Mar 24 '16 at 20:12
  • $\begingroup$ @FriedrichPhilipp Induction (with Laplace expansion) is the only way I know and it's the standard trick to evaluate the determinant of a tridiagonal matrix. $\det B$ is also equal to $U_n(1)$, where $U_n(x)$ is Chebyshev polynomial of the second kind, but $U_n(x)$ is defined recursively like we evaluate $\det B$ by induction. So I don't think the Chebyshev polynomial perspective really adds anything new here. But certainly, as the roots of $U_n(x)$ are known explicitly, if you only want to prove that $B$ is nonsingular, then the Chebyshev polynomial approach is superior. $\endgroup$ – user1551 Mar 24 '16 at 20:19
  • $\begingroup$ user1551 Thanks for this comment. $\endgroup$ – Friedrich Philipp Mar 24 '16 at 20:28

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