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