$$\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}$$
In a talk on positive definite matrices, Prof. Strang mentions that he remembers the eigenvalues of the matrix above because "These matrices are so important to figure them out.".
In here, he says,
So second derivatives should have a bigger place in calculus. Second, my matrices, which are like the linear algebra version of second derivatives, are in need in linear algebra. Everything comes out right with those guys.
I see that it is a tridiagonal matrix, but I wonder if the $-1$ off-diagonals confer them a last name - are these
$$\begin{bmatrix} a & -1 & 0 \\ -1 & a & -1 \\ 0 & -1 & a \end{bmatrix}$$
arrangements a separate group of tridiagonal matrices? And why are they so important?
Incidentally, he derives them from the eigenvectors, which in turns he deducts by analogy of this difference of second differences to a second order differential equation here.
These are specific forms of band matrices and Toeplitz matrices, and in his book Linear Algebra and Its Applications he describes them as the first of "four families of matrices - simple and useful, absolutely basic" with the nondescript name $K_n.$ Their properties are:
- Symmetric
- Sparse
- Banded (tridiagonal): Rapidly solved.
- Constant diagonals: Shift or time invariance. Can be used as a highpass filter.
- Second difference matrix.
- Invertible
- Positive definite
On page 66 and 67 he explains their value as the matrix formulation of a discretized problem of second derivatives as such:
The first difference is
$$\frac{\Delta u}{\Delta x}=\frac{u(x+\Delta x) - u(x - \Delta x)}{2 \Delta x}$$
and the second difference
$$\small\frac{\Delta^2 u}{\Delta x^2}=\frac{\left(u(x+\Delta x) - u(x)\right) -\left(u(x)-u(x-\Delta x) \right)}{\Delta x^2}=\frac{u(x+\Delta x) - 2u(x) +u(x- \Delta x)}{\Delta x^2}$$
and multiplying by $\Delta x^2,$
$$\small\begin{bmatrix} 2&-1&&\\-1&2&-1&&\\&-1&2&-1\\ &&-1&2&-1&&\\ &&&-1&2&-1&\\ &&&&-1&2 \end{bmatrix} \begin{bmatrix} u_1 \\u_2\\u_3\\u_4\\u_5\\u_6 \end{bmatrix}=h^2 \begin{bmatrix}f(h)\\f(2h)\\f(3h)\\f(4h)\\f(5h)\\f(6h) \end{bmatrix}$$
where $\Delta x= h,$ and the differential equation to solve of the form
$$-\frac{d^2u}{dx^2}=f(x)$$
the temperature distribution in a rod, for example, with ends fixed at 0 degrees and with a heat source $f(x).$
In other words the solution of a Poisson's equation.
