# Determinant of Block Tridiagonal Matrix

I found in the following paper: Comments on ‘‘A note on a three-term recurrence for a tridiagonal matrix’’ that we can compute the determinant of a block tridiagonal matrix A via a recursion.

In my particular case A is $4n\times4n$,

$$\textbf{A}=\begin{pmatrix} \textbf{B}_L-h\textbf{R} & J\space\textbf{R} & \textbf{0} & \cdots & \textbf{0} \\ J\space \textbf{R} & -h\textbf{R} & J\space\textbf{R} & & \textbf{0} \\ \textbf{0} & J\space \textbf{R} & -h\textbf{R} &\ddots &\vdots \\ \vdots & &\ddots &\ddots & J\space\textbf{R}\\ \textbf{0} & \textbf{0} & \cdots & J \space\textbf{R} & \textbf{B}_R-h\textbf{R} \end{pmatrix}$$

where $$\textbf{R}=\begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0\\ \end{pmatrix},\space \textbf{B}_{L,R}=\begin{pmatrix} 0&\frac{i}{2}\Gamma_{+}^{\text{L,R}}&-\frac{i}{2}\Gamma_{-}^{\text{L,R}}&\frac{1}{2}\Gamma_{-}^{\text{L,R}}\\ -\frac{i}{2}\Gamma_{+}^{\text{L,R}}&0&\frac{1}{2}\Gamma_{-}^{\text{L,R}}&\frac{i}{2}\Gamma_{-}^{\text{L,R}}\\ \frac{i}{2}\Gamma_{-}^{\text{L,R}}&-\frac{1}{2}\Gamma_{-}^{\text{L,R}}&0&\frac{i}{2}\Gamma_{+}^{\text{L,R}}\\ -\frac{1}{2}\Gamma_{-}^{\text{L,R}}&-\frac{i}{2}\Gamma_{-}^{\text{L,R}}&-\frac{i}{2}\Gamma_{+}^{\text{L,R}}&0\\ \end{pmatrix}$$ and $$J,h,\Gamma_{+}^{\text{L,R}},\Gamma_{-}^{\text{L,R}} \in \mathbb{R}$$

Now let me state the recursion mentioned in the above paper,

$$\text{det}(\textbf{A})=\prod_{k=1}^{n}\text{det}(\Lambda_{k})\space\space\space\space(1)$$

where (in my case),

$$\Lambda_{1} = \textbf{B}_L-h\textbf{R}\\ \Lambda_{k} = -h\textbf{R}-J^{2}\textbf{R}\Lambda_{k-1}^{-1}\textbf{R}\\ \Lambda_{n}=\textbf{B}_R-h\textbf{R}-J^{2}\textbf{R}\Lambda_{n-1}^{-1}\textbf{R}$$ Now according to (1), the set of eigenvalues of $\textbf{A}$ should contain the eigenvalues of $\Lambda_{1} = \textbf{B}_L-h\textbf{R}$ since applying (1) to $\textbf{A}-\lambda I_{4n}$ gives $\Lambda_{1}^{'} = \textbf{B}_L-h\textbf{R}-\lambda I_{4}$.

Now here comes my issue.

I computed the spectrum of A in Mathematica for $n=50$ for the values $h=1,J=1.5,\Gamma_{+}^{\text{L}}=1.6,\Gamma_{+}^{\text{R}}=1.3,\Gamma_{-}^{\text{L}}=-0.4,\Gamma_{-}^{\text{R}}=-0.7$

I found that the spectrum did not contain the eigenvalues of $\textbf{B}_{L}-h\textbf{R}$. My question is: Is this recursion not applicable to my case, or am I incorrect in stating that the set of eigenvalues of A should contain the eigenvalues of $\Lambda_{1}$?

• From the dimensional argument, I suspect that the upper bound for k in (1) is n, not 4n (this doesn't solve the question, but is a prerequisit for further actions). Also, the second $I_{4n}$ is probably $I_4$. – colt_browning Jan 20 '18 at 0:27
• From what dimensional argument? A is 4nx4n. Also I made the appropriate change to the identity – user1058860 Jan 20 '18 at 6:54
• $\det(\Lambda_k)$ is a 4th degree homogeneous function of the elements of A. det(A) is a 4n-th degree homogeneous function of the elements of A. So det(A) can be a product of n $\det(\Lambda_k)$'s, but not 4n. – colt_browning Jan 20 '18 at 7:47
• ah, yes. i shall make the change – user1058860 Jan 20 '18 at 14:41

The recursion in the paper is a generalization of the classical determinant relation for block matrices (see Wiki): if $A$ is invertible then $$\det\begin{bmatrix}A & B\\C & D\end{bmatrix}=\det A\cdot\det(D-CA^{-1}B).$$ The block matrix does not share eigenvalues with the $A$ block in general, since $\lambda I-A$ becomes not invertible for an eigenvalue $\lambda$ of $A$, hence, the formula cannot be applied.
Edit: there is an alternative formula for evaluation of the determinant of a tridiagonal matrix here (look Theorem 2). It uses only inversions of off-diagonal blocks that do not contain $\lambda$ in your case.
• What if I computed when $\text{det}(\Lambda_{n}^{'})=0$?(where by adding the prime I mean take $h\textbf{R}-\lambda\text{I}_{4}, \textbf{B}_{R}-h\textbf{R}-\lambda\text{I}_{4}, \textbf{B}_{L}-h\textbf{R}-\lambda\text{I}_{4}$) Shouldn't the solutions to that equation give eigenvalues of $\textbf{A}$? – user1058860 Jan 20 '18 at 15:27
• @user1058860 If you calculate with polynomials in $\lambda$ then you will get a rational function with lots of cancellations. If you manage to cancel everything you should get a polynomial, which is the characteristic polynomial of $A$ (in your notation). But it is almost an impossible task numerically due to round off errors. Maybe symbolically for some relatively low dimensions is ok. – A.Γ. Jan 20 '18 at 21:01
Second. Indeed, $\Lambda'_2$ is already not a polynomial in $\lambda$; moreover, it becomes undefined when $\lambda$ is an eigenvalue of $\Lambda_1$.