Tridiagonal Matrix's Rank and Eigenvalues Please proof that a tridiagonal matrix with positive entries on minor diagonals has rank n-1 or n and the eigenvalues of this matrix are real.
http://upload.wikimedia.org/math/3/4/a/34a1edbf99dae57a40b4d66d6d5b1faa.png
 A: Assume the tridiagonal matrix $T$ in this form:
$$
T = \begin{bmatrix}
\alpha_1 & \beta_1  &              &              & \\
\gamma_1 & \alpha_2 & \beta_2      &              & \\
         & \ddots   & \ddots       & \ddots       & \\
         &          & \gamma_{n-2} & \alpha_{n-1} & \beta_{n-1} \\
         &          &              & \gamma_{n-1} & \alpha_{n}
\end{bmatrix}
$$
with $\beta_i>0$ and $\gamma_i>0$ for $i=1,\ldots,n-1$.
If you erase the first row and the last column of $T$, you get an upper triangular matrix with the positive $\gamma$'s on the diagonal (and hence nonsingular). Hence $T$ contains a sub-matrix of the rank $n-1$ and consequently, its rank cannot be smaller than $n-1$.
Take the diagonal matrix $D=\mathrm{diag}(\delta_1,\ldots,\delta_n)$ with positive diagonal entries $\delta_k$ such that
$$
\delta_1=1, \quad \delta_k^2 = \prod_{i=1}^{k-1}\frac{\gamma_i}{\beta_i}, \quad k=2,\ldots,n.
$$
I invite you to show that $D^{-1}TD$ is symmetric, and therefore has real eigenvalues. Since $T$ is similar to $D^{-1}TD$, it has real eigenvalues as well.
A: For the rank: it should be easy to show that the last $n-1$ columns of $T$ must be linearly independent (try row-reduction, or simply note that no two columns have identical support).  This is sufficient to show that the rank of $T$ is at least $n-1$.  Note that this holds even if the diagonal entries are complex.
For the eigenvalues, it suffices to show that the rank of $T - \lambda I$ is $n$ whenever $\lambda$ is non-real.
In fact, because of our earlier work, it suffices to show that the matrix
$$
\pmatrix{
a_1 - \lambda & b_1&0\\
c_1 & a_2 - \lambda & b_2\\
0 & c_2 & a_3 - \lambda
}
$$
Has full rank when $\lambda$ is non-real.
