<s>[INST] Is there a fast way to prove a tridiagonal matrix is positive definite? I' m trying to prove that 
$$A=\begin{pmatrix}
 4 & 2 & 0 & 0 & 0 \\
 2 & 5 & 2 & 0 & 0 \\
 0 & 2 & 5 & 2 & 0 \\
 0 & 0 & 2 & 5 & 2 \\
 0 & 0 & 0 & 2 & 5 \\  
 \end{pmatrix}$$ 
admits a Cholesky decomposition.
$A$ is symmetric, so it admits a Cholesky decomposition iff it is positive definite. The only methods I know for checking this are:


*

*$X^tAX > 0, \quad \forall X \in \mathbb{K}^n- \{0\}$.

*If $\lambda$ is an eigenvalue of $A$, then $\lambda>0.$


I have failed to prove it using 1 and 2 is taking me so much time. Is there any easier way to do this, given that $A$ is tridiagonal?
 [/INST] Notice $A$ can be rewritten as a sum of 5 matrices. 
$$A =
\left[\begin{smallmatrix}
2 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 3
\end{smallmatrix}\right] + 
\left[\begin{smallmatrix}
2 & 2 & 0 & 0 & 0\\
2 & 2 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 2 & 2 & 0 & 0\\
0 & 2 & 2 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 2 & 2 & 0\\
0 & 0 & 2 & 2 & 0\\
0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right] +
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 2 & 2\\
0 & 0 & 0 & 2 & 2
\end{smallmatrix}\right]
$$
The first matrix 
is diagonal with positive entries on diagonals, so it is positive definite.
The remaining four matrices are clearly positive semi-definite. Being a sum
of a positive definite matrix with a bunch of positive semi-definite matrices,
$A$ itself is positive definite.
</s>