<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] 
I have failed to prove it using 1

Note:
$$\begin{pmatrix}a&b&c&d&e\end{pmatrix}\begin{pmatrix}
 4 & 2 & 0 & 0 & 0 \\
 2 & 5 & 2 & 0 & 0 \\
 0 & 2 & 5 & 2 & 0 \\
 0 & 0 & 2 & 5 & 2 \\
 0 & 0 & 0 & 2 & 5 \\  
 \end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}=\\
\begin{pmatrix}4a+2b&2a+5b+2c&2b+5c+2d&2c+5d+2t&2d+5e\end{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}=\\
4a^2+4ab+5b^2+4bc+5c^2+4cd+5d^2+4de+5e^2=\\
(2a+b)^2+(2b+c)^2+(2c+d)^2+(2d+e)^2+4e^2.$$
</s>