Does there exist a symmetric tridiagonal matrix with zero determinant? I will like to know whether there exists a symmetric tridiagonal matrix with zero determinant? I will refer the definition of a tridiagonal matrix to the one found in Wikipedia: 
"A tridiagonal matrix is a matrix that has nonzero elements only on the main diagonal, the first diagonal below this, and the first diagonal above the main diagonal.".
If yes, kindly provide an example. If not, please illustrate the proof or idea.
Thank you very much.  
 A: The trivial example is
$$
\pmatrix{0 & 0 \\ 0 & 0}.
$$
Or you could consider:
$$
\pmatrix{1 & 1 \\ 1 & 1} \quad\text{or} \quad\pmatrix{1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1}.
$$
Otherwise if you only want non-zero entries of the diagonals, then how about
$$
\pmatrix{1 & a & 0 \\ a & 1 & a \\ 0 & a & 1}
$$
where $a = \frac{1}{\sqrt{2}}$.
Since you ask, I don't know if this might work, but try to consider
$$
\pmatrix{1 & a & 0 & 0\\ a & 1 & a & 0\\ 0 & a & 1& a \\ 0 & 0 & a & 1}
$$
and find the determinant of this matrix. This will be an expression in $a$. Can this ever equal zero? If this doesn't work, maybe this will give you an idea to something that would.
A: If you are interested in a symmetric tridiagonal matrix whose entries on the three diagonals are all nonzero, here's a way to obtain lots of $n \times n$ examples where the determinant is zero.    Let $L_k$ be the determinant of the matrix
formed by the first $k$ rows and columns of the tridiagonal symmetric $A = (a_{ij})$.
Then $L_k = L_{k-1} a_{k,k} - L_{k-2} a_{k-1,k}^2$ for $k \ge 3$.  If $L_{n-1}$ and $L_{n-2}$ and $a_{n-1,n}$ are all nonzero (which is true "generically"), you can take $a_{n,n} = L_{k-2} a_{k-1,k}^2/L_{k-1}$ to get your example. 
Thus for a $4 \times 4$ example, I randomly chose the entries other than the bottom right as
$$ \pmatrix{2 & -1 & 0 & 0\cr -1 & 2 & -1 & 0\cr 0 & -1 & 1 & 1\cr 0 & 0 & 1 & a_{44}\cr}$$
and found that the determinant was $0$ for $a_{44} = 3$.
