Let $\tau$ be a strictly positive real and $q$ a natural integer greater than or equal to 2. We set $\delta = \frac{1}{q+1}$ and $r = \frac{\tau}{\delta^{2}}$. The numerical scheme imposes, for any natural integer $n$ and any $k \in \llbracket 1, q \rrbracket$: $$\frac{f_{n+1}(k) - f_{n}(k)}{\tau} = \frac{f_{n}(k+1) - 2f_{n}(k) + f_{n}(k-1)}{\delta^{2}}$$ as well as $f_{n}(0) = f_{n}(q+1) = 0$. We set $F_{n} = \left(\begin{array}{c} f_{n}(1) \\ \vdots \\ f_{n}(q) \end{array}\right)$, $I_{q}$ is the identity matrix of order $q$, $B$ is the square matrix of order $q$ with coefficient $(i,j)$ equal to 1 if $|i-j|=1$ and 0 otherwise, and $A = (1-2r)I_{q} + rB$. Show that, for all $n \in \mathbb{N}$, $F_{n+1} = A F_{n}$.
Let $\tau$ be a strictly positive real and $q$ a natural integer greater than or equal to 2. We set $\delta = \frac{1}{q+1}$ and $r = \frac{\tau}{\delta^{2}}$. The numerical scheme imposes, for any natural integer $n$ and any $k \in \llbracket 1, q \rrbracket$:
$$\frac{f_{n+1}(k) - f_{n}(k)}{\tau} = \frac{f_{n}(k+1) - 2f_{n}(k) + f_{n}(k-1)}{\delta^{2}}$$
as well as $f_{n}(0) = f_{n}(q+1) = 0$. We set $F_{n} = \left(\begin{array}{c} f_{n}(1) \\ \vdots \\ f_{n}(q) \end{array}\right)$, $I_{q}$ is the identity matrix of order $q$, $B$ is the square matrix of order $q$ with coefficient $(i,j)$ equal to 1 if $|i-j|=1$ and 0 otherwise, and $A = (1-2r)I_{q} + rB$. Show that, for all $n \in \mathbb{N}$, $F_{n+1} = A F_{n}$.