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I have a matrix of the form: $$ \begin{bmatrix} 0 & q & 0 & 0 & 0 & 0 & \cdots \\ p & 0 & q & 0 & 0 & 0 & \cdots \\ 0 & p & 0 & q & 0 & 0 & \cdots \\ 0 & 0 & p & 0 & q & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix} $$ Now, does there exist a similarity transformation that turns this into a symmetric matrx. If yes then how to find it.

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    $\begingroup$ What have you tried? $\endgroup$ – Monadologie Aug 18 at 8:55
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This obviously isn't always possible. E.g. if the underlying field is real, $p=0$ and $q=1$, we have a non-zero nilpotent matrix that cannot possibly be similar to any real symmetric matrix.

However, if $p$ and $q$ are non-zero, $\frac pq=r^2$ for some scalar $r$ in the underlying field and $D=\operatorname{diag}(1,r,r^2,\ldots,r^{n-1})$, then $$ D^{-1}AD=\pmatrix{0&s\\ s&0&s\\ &s&\ddots&\ddots\\ &&\ddots&0&s\\ &&&s&0} $$ where $A$ is your matrix and $s=qr=\frac{p}{r}$.

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  • $\begingroup$ [+1] very short solution ! $\endgroup$ – Jean Marie Aug 19 at 7:51
  • $\begingroup$ Another example where this transformation $M \to D^{-1}MD$ is used with the same matrix $D$ : math.stackexchange.com/q/2817767 $\endgroup$ – Jean Marie Aug 19 at 16:17
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Let $r=\sqrt{pq}$ (we assume sign$(p)$=sign$(q)$).

The given tridiagonal matrix $A$ is similar to

$$B:=\begin{bmatrix} 0 & r & 0 & 0 & 0 & 0 & \cdots \\ r & 0 & r & 0 & 0 & 0 & \cdots \\ 0 & r & 0 & r & 0 & 0 & \cdots \\ 0 & 0 & r & 0 & r & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix}$$

Why that ?

Let $P_n=\det(A-\lambda I_n)$, resp. $Q_n=\det(B-\lambda I_n)$ be the characteristic polynomial of $A$, resp. $B$.

Expanding the first determinant along its first column gives the recurrence relationship :

$$P_n=-\lambda P_{n-1}-pq P_{n-2} \ \ \text{with} \ \ P_1=-\lambda \ \ \text{and} \ \ P_2=\lambda^2-pq \tag{1}$$

(this is a classical way to compute the characteristic polynomial of a tridiagonal matrix).

Doing a similar expansion for the second determinant, one obtains the same relationship as (1). Therefore $A$ and $B$ have the same determinant.

As a symmetric matrix is diagonalisable, one can write : $D=P^{-1}AP$ and $D=Q^{-1}BQ$ with the same diagonal matrix $D$ (diagonal entries being the common eigenvalues).

From $P^{-1}AP=Q^{-1}BQ$, one deduces

$$A=(QP^{-1})^{-1}B(QP^{-1})$$

Thus $A$ and $B$ are similar.

Important remark : In the case $r = 1$, the characteristic polynomial of $B$ is easily shown to be $U_n(-2x)$ where $U_n$ is the n-th Chebyshev polynomial of the second kind (see https://en.wikipedia.org/wiki/Chebyshev_polynomials) ; as a consequence, in the odd case $n=2m+1$ ; the eigenvalues of $A$ and $B$ are $2 \sqrt{pq} \sin(k \pi/(n+1))$ for $k=-m...m$.

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    $\begingroup$ $P_n=Q_n$ doesn't imply that $A$ and $B$ are similar. Take $$A=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$and $$B=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$ $\endgroup$ – Mostafa Ayaz Aug 18 at 9:49
  • $\begingroup$ @MostafaAyaz Actually you are true, similarity of characteristic polynomials does not suffice similarity of two. I was being ignorant. $\endgroup$ – user539586 Aug 18 at 10:08
  • $\begingroup$ It is OK. I explained it in my own answer. Hope it help! $\endgroup$ – Mostafa Ayaz Aug 18 at 10:09
  • $\begingroup$ Following the remark of @Mostafa Ayaz, I have modified my answer. $\endgroup$ – Jean Marie Aug 18 at 10:12
  • $\begingroup$ See added remark. $\endgroup$ – Jean Marie Aug 18 at 12:32

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