Are Diagonally dominant Tridiagonal matrices diagonalizable?

Question

I have searched this in the literature but could not find any reference, so I would like to post it here. Hope this is at the research level. Assume that $$ A=\begin{vmatrix} a_1 & b_1 \\ c_1 & a_2 & b_2 \\ & c_2 & \ddots & \ddots \\ & & \ddots & \ddots & b_{n-1} \\ & & & c_{n-1} & a_n \end{vmatrix} $$ where $a_i< 0,\quad 0\leq c_i,b_i\quad \forall i \quad $, $a_1+b_1=0$, $c_{n-1}+a_n=0$ and $c_{i-1}+a_i+b_i=0,\quad 2\leq i\leq n-1$.

My test using matlab with random matrices and they all diagonalizable. This makes me believe that $A$ should be diagonalizable, however, I have not found a simple proof for it.

My question here is : is $A$ diagonalizable ? can we find a "simple" proof for it? Thank you for your time!

Answer 1

Counterexample:

$$ \begin{bmatrix} -1 & 1 & 0 & 0\\ 0 & -1 & 1 & 0\\ 0 & 0 & -2 & 2\\ 0 & 0 & 2 & -2 \end{bmatrix} $$ is defective: its eigenvalues are $-1,-1, 0, -4$ (it is block triangular, so its eigenvalues are those of the $2\times 2$ blocks on the diagonal), but $A+I$ has rank 3.

Strategy to construct it:

(added to the answer on request of @EmilioPisanty)

My first thought was: let's take all the $c_k$ equal to $0$ and make a Jordan block $$ \begin{bmatrix} -1 & 1 \\ & -1 & 1\\ && \ddots & \ddots \\ &&& -1 \end{bmatrix}. $$ This almost works, but the last row has to be adjusted to satisfy the zero-sum condition. My idea was adjusting it by adding rows and columns (but the other natural idea of changing $c_{n-1}$ only can be made to work, too; see below). It is a known fact that if $T_{11}$ and $T_{22}$ have no eigenvalues in common, then $ \begin{bmatrix} T_{11} & T_{12}\\ 0 & T_{22} \end{bmatrix} $ and $ \begin{bmatrix} T_{11} & 0\\ 0 & T_{22} \end{bmatrix} $ are similar (proof: $$ \begin{bmatrix} I & X\\ 0 & I \end{bmatrix} \begin{bmatrix} T_{11} & T_{12}\\ 0 & T_{22} \end{bmatrix} \begin{bmatrix} I & X\\ 0 & I \end{bmatrix}^{-1} = \begin{bmatrix} T_{11} & T_{12}+XT_{22}-T_{11}X\\ 0 & T_{22} \end{bmatrix}, $$ and the Sylvester equation $T_{11}X-XT_{22}=T_{12}$ is solvable whenever $T_{11}$ and $T_{22}$ have disjoint spectra).

So I just took $T_{11}$ a Jordan block, $T_{22}$ any matrix with no eigenvalues in common with it, and I knew that the resulting matrix had to have a Jordan block, too, no matter what $T_{12}$ was.

Alternate idea

Let's adjust the Jordan block by making the minimum possible change: alter $c_{n-1}$: $$A= \begin{bmatrix} -1 & 1 \\ & -1 & 1\\ && \ddots & \ddots \\ &&& -1 & 1\\ &&&& -1 & 1\\ &&&&1 & -1 \end{bmatrix}. $$ This matrix is still block triangular, so its eigenvalues are $n-2$ times $-1$, and then whatever the eigenvalues of $\begin{bmatrix}-1 & 1\\ 1 & -1\end{bmatrix}$ are. Does it still have a Jordan block, or is it diagonalizable? I guess it probably has a Jordan block: matrices with a Jordan block have more degrees of freedom, so I expect them to be "generic" among the matrices with multiple eigenvalues. Let's just throw it into Wolfram Alpha to check and hope it works. Bingo! Oh, wait, in retrospect it is obvious that it works, because $A+I$ has clearly rank $n-1$.

Answer 2

All real tridiagonal matrices with $b_kc_k>0$, are diagonalizable, and their spectra are real and simple.

See, for example, Gantmakher and Krein, Oscillation matrices and kernels..., AMS 2002.

Sketch of the proof. Expanding the determinant $|A-\lambda I|$ write a recurrent formula for characteristic polynomials of truncated matrices. It is seen from this formula that the eigenvalues depend only on $a_k$ and the products $c_kb_k$. Therefore the symmetric matrix with $c_k^\prime=b_k^\prime=\sqrt{c_kb_k}$ has the same spectrum. This shows that the spectrum is real.

To show that all eigenvalues are distinct one applies Sturm's theorem to the sequence of characteristic polynomials.

On the other hand, if you allow all $c_k=0$, for example, you can have a Jordan cell which is not diagonalizable.