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I encounter the following $mn\times mn$ tridiagonal block matrix

$$\displaystyle A=\left[ {\begin{array}{*{20}{c}} T & {\frac{1}{{{{k}^{2}}}}{{I}_{n}}} & {} & {} & {} \\ {\frac{1}{{{{k}^{2}}}}{{I}_{n}}} & T & {\frac{1}{{{{k}^{2}}}}{{I}_{n}}} & {} & {} \\ {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {\frac{1}{{{{k}^{2}}}}{{I}_{n}}} & T & {\frac{1}{{{{k}^{2}}}}{{I}_{n}}} \\ {} & {} & {} & {\frac{1}{{{{k}^{2}}}}{{I}_{n}}} & T \end{array}} \right]$$

where

$$\displaystyle T=\left[ {\begin{array}{*{20}{c}} {-2\left( {\frac{1}{{{{h}^{2}}}}+\frac{1}{{{{k}^{2}}}}} \right)} & {\frac{1}{{{{h}^{2}}}}} & {} & {} & {} \\ {\frac{1}{{{{h}^{2}}}}} & {-2\left( {\frac{1}{{{{h}^{2}}}}+\frac{1}{{{{k}^{2}}}}} \right)} & {\frac{1}{{{{h}^{2}}}}} & {} & {} \\ {} & \ddots & \ddots & \ddots & {} \\ {} & {} & {\frac{1}{{{{h}^{2}}}}} & {-2\left( {\frac{1}{{{{h}^{2}}}}+\frac{1}{{{{k}^{2}}}}} \right)} & {\frac{1}{{{{h}^{2}}}}} \\ {} & {} & {} & {\frac{1}{{{{h}^{2}}}}} & {-2\left( {\frac{1}{{{{h}^{2}}}}+\frac{1}{{{{k}^{2}}}}} \right)} \end{array}} \right]$$

is an $n\times n$ matrix. How can I compute the eigenvalues and eigenvectors of $A$?

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1 Answer 1

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In general, every $N\times N$ real symmetric tridiagonal Toeplitz matrix $\mathbf T(a,b)$ (with main diagonal entries equal to $a$ and sub/super-diagonal entries equal to $b$) can be orthogonally diagonalisable as $$ \mathbf T(a,b) = Q\operatorname{diag}(\lambda_1,\ldots,\lambda_N)\,Q^\top, $$ where the eigenvalues are given by $\lambda_i = a+2b\cos\left(\frac{i\pi}{N+1}\right)$ for each $i$ and the (real orthogonal) eigenvector matrix is given by $q_{ij}=\sqrt{\frac2{N+1}}\sin\left(\frac{ij\pi}{N+1}\right)$.

Now, let $T=Q\Lambda Q^\top$ be such a diagonalisation for your $T$. Then $A=(I_m\otimes Q)\ \widetilde{A} (I_m\otimes Q^T)$, where $\widetilde{A}$ is the block-Toeplitz matrix obtained by replacing each $T$ in $A$ by $\Lambda$. Obviously, $\widetilde{A}$ is permutation-similar to $T_1\oplus T_2\oplus\cdots\oplus T_m$, where each diagonal sub-block $T_i$ is the $m\times m$ symmetric Toeplitz matrix $$ T_i=\begin{bmatrix} \lambda_i & \frac1{k^2}\\ \frac1{k^2} & \ddots & \ddots\\ &\ddots & \ddots &\frac1{k^2}\\ &&\frac1{k^2} & \lambda_i \end{bmatrix} $$ and $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $T$. In fact, if $K^{(m,n)}$ denotes the commutation matrix that maps $\operatorname{vec}(A)$ to $\operatorname{vec}(A^\top)$ for a generic $A$, then $$ A=(I_m\otimes Q)\ K^{(m,n)}(T_1\oplus T_2\oplus\cdots T_n)\ {K^{(n,m)}} (I_m\otimes Q^T). $$

Now, for each $i$, using the formula mentioned in the first paragraph, we see that the eigenvalue matrix $\Lambda_i=\operatorname{diag}(\lambda_{i1},\lambda_{i2},\ldots,\lambda_{im})$ of $T_i$ is given by the formula $$ \lambda_{ij}=-2\left(\frac1{h^2}+\frac1{k^2}\right)+\frac{2}{h^2}\cos\left(\frac{i\pi}{n+1}\right)+\frac{2}{k^2}\cos\left(\frac{j\pi}{m+1}\right) $$ for $j=1,2,\ldots,m$. Let $T_i=Q_i\Lambda_iQ_i^\top$ be an orthogonal diagonalisation. It follows that

$$ U=(I_m\otimes Q)\ K^{(m,n)}(Q_1\oplus Q_2\oplus\cdots Q_m), $$ an orthogonal diagonalisation of $A$ is given by $$ A=U(\Lambda_1\oplus\Lambda_2\oplus\cdots\oplus\Lambda_n)U^\top. $$ That is, the eigenvalues of $A$ are those $\lambda_{ij}$s for $i\in\{1,2,\ldots,n\}$ and $j\in\{1,2,\ldots,m\}$, and their corresponding eigenvectors are the columns of $U$.

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  • $\begingroup$ I appreciate for your help. Futhermore, I also need a set of eigenvectors of $A$, too. Can you give me one? Thank in advanced. $\endgroup$ Nov 18, 2016 at 15:52
  • $\begingroup$ @QuảnBáHồngNguyễn Done. See the edit. $\endgroup$
    – user1551
    Nov 18, 2016 at 18:30
  • $\begingroup$ Thanks for your help. I really need this computation. $\endgroup$ Nov 20, 2016 at 3:05

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