Determinants of tridiagonal matrices

Question

A square matrix $A = [a_{ij}]$ is called ${\bf tridiagonal}$ if $a_{ij}=0$ for $|i-j|>1$. Try to guess a formula for the determinant of tridiagonal matrix, say $a_i = a_{ii}$ for $i=1,...,n$, $b_i = a_{i,i+1}$ and $c_i = a_{i+1,i}$ for $i=1,...,n-1$.

Attempt

So, I was thinking on reducing to smaller matrix. Say for $n=1$, we det A = $a_1$. Not in the case $n=2$, we have just the matrix with rows $[a_1, b_1$] and $[c_1,a_2]$. Thus,

$$ det A = a_1 a_2 - c_1 b_1 $$

Now, for the $n=3$ case, we start to see the zeros appear, but it becomes cumbersome to compute determinant after $n>3$. Is there a way to find closed nice for this problem? Or do I have to keep doing it expressing the actual determinant in terms of the previous as it is evident in the case $n=3$ since if we call $D_n$ to be the determinant on the nth case (for instnace, we saw that $D_2 = a_1 a_2 - c_1 b_1$ so that for the $n=3$ case I see that

$$ D_3 =a_3 D_2 - c_2 b_2 a_1 $$

Is this the right way to approach this problem? Moreover, why are tridiagonal matrices so important? Can someone give intuition into what they do? or in what situations we use them

Answer 1

Recursion is the best way to solve this problem. As a hint, you showed that $$D_3 = a_3D_2-c_2b_2a_1 = a_3D_2 - c_2b_2D_1.$$ Can you generalize this to a formula for $D_n$ in terms of $D_{n-1}$, $D_{n-2}$, and a few of the entries of the matrix?

As for why they are important, many eigenvalue algorithms for symmetric/Hermitian matrices will first use similarity transforms to reduce the matrix to a tridiagonal form, and then find the eigenvalues of a tridiagonal matrix.

Also, tridiagonal matrices come up when solving differential equations via discretization. Suppose we want to solve the $u''(x) = f(x)$ on the interval $[0,1]$. Pick a positive integer $N$, and let $v_n = u(\tfrac{n}{N})$ for $n = 0,1,\ldots,N$. Then, using an approximation of the second derivative, we have $$f(\tfrac{n}{N}) = u''(\tfrac{n}{N}) \approx \dfrac{u(\tfrac{n+1}{N})-2u(\tfrac{n}{N})+u(\tfrac{n-1}{N})}{(\tfrac{1}{N})^2} = N^2(v_{n+1}-2v_n+v_{n-1}).$$ If we do this for all $n = 1, 2, \ldots, N-1$, and then include equations for whatever boundary conditions we might have, we'll get a tridiagonal system of equations. Note, this was a fairly trivial example, but there are more complicated differential equations and PDEs that can be handled this way.