# Normalizing a matrix via triangular transformations

Consider the following $n\times n$ real-valued matrix:

$$A=\begin{bmatrix} \alpha_1 & \beta_1 & 0 &\cdots & 0 &\gamma_n\\ \gamma_1 & \alpha_2 & \beta_2 & \cdots & \cdots & 0\\ 0 & \ddots & \ddots & \ddots & \cdots & \vdots\\ \vdots & \cdots & \ddots & \ddots & \ddots & 0\\ 0 & \cdots & \cdots & \gamma_{n-2} & \alpha_{n-1} & \beta_{n-1}\\ \beta_n & 0 & \cdots & \cdots & \gamma_{n-1} & \alpha_n \end{bmatrix}$$

There exists a triangular matrix $T$ s.t. $N=T^{-1}AT$ is a normal matrix (that is $N N^\top= N^\top N$)? If so, what is the form of $T$ in terms of $\alpha_i$, $\beta_i$ and $\gamma_i$?

Any help (and/or links to papers which address this or related problems) will be appreciated.

EDIT. The answer is false, if the choice of the coefficients $\alpha_i$, $\beta_i$ and $\gamma_i$ is arbitrary. What can we say if we make the additional assumption that $\alpha_i\neq 0$, $\beta_i\neq 0$ and $\gamma_i\neq 0$ for all $i$?

• If you assume that $T$ also has to be real, and that such a $T$ exists for every choice of $\alpha_i, \beta_i, \gamma_i$, then for $n=2$, and likely for all $n$, this is impossible. A real circulant matrix has to have a real eigenvalue, and we can certainly choose the entries to have no real eigenvalues when $n=2$, and presumably for every $n$ (I didn't bother checking). Maybe the question is supposed to be, if for a fixed choice of entries, $A$ is conjugate to a circulant matrix, then can we say what $T$ would look like? – David Handelman Jan 21 '15 at 1:03
• You are absolutely right. Anyway, my main concern is to find (if there exists) a triangular similarity transformation which normalizes $A$. Now, if we restrict the attention to the subclass of normal matrices which are circulant, this is not possible (for every choice of the parameters). But what can be said about the general case? (I edit the question accordingly.) – Ludwig Jan 21 '15 at 10:05

No, this is a counter-example for $n =2$:
Consider $$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix},\ T = \begin{bmatrix} a & b \\ 0 & c \\ \end{bmatrix}.$$
By hypothesis, $T$ is invertible. Thus, on one hand we have that $a \ne 0$ and $c \ne 0$.
On the other hand, the equality $NN^{T} = N^{T}N$ yields that $$\left(c \left(\frac{1}{a}-\frac{b}{a c}\right)+\frac{b}{a}\right)^2+1 = 1$$ giving the contradiction $c=0$.
• Ok, thank you! How about if we assume that $\alpha_i\neq 0$, $\beta_i\neq 0$, $\gamma_i\neq 0$ for all $i$? – Ludwig Jan 25 '15 at 14:24