Consider $A: n \times n$ a symmetric matrix.
a) Explain the main steps to perform on Orthogonal transformations so as to
to obtain a matrix $T$, tridiagonal, orthogonally similar to $A$.b) Obtained the tridiagonal matrix $T$, orthogonally similar to $A$, shows that:
if $T ^{(0)} = T$, then the matrix $T^{(k)}$ obtained in each iteration of the QR algorithm is also tridiagonal.
My references are Watkins, Fundamentals of Matrix Computations, third edition and Trefethen and Bau, Numerical Linear Algebra.
I understand the item (a). I can conclude that in a process of Householder with less modification we can get a product
$$
Q_{k-2}^{H}Q_{k-3}^{H}\ldots Q_{2}^{H}Q_{1}^{H}AQ_1Q_2 \ldots Q_{k-3}Q_{k-2} = H
$$
where $H$ is an upper Hessenberg matrix.
With you write $Q_1Q_2\ldots Q_{k-2} = Q$, we have the compact form $Q^HAQ=H$ and if $A$ is a symmetric matrix, we can put $H = T$ and showed that $T$ is tridiagonal matrix orthogonally similar to $A$.
Now, the part (b), I need to show the process follow:
If I put $T^{(0)} = T_{0}$. I have for the first iteration ($k=1$):
$$
T_0 = Q_0R_0 \Longrightarrow R_0 = Q_{0}^{T}T_0.
$$
So,
$$
T_1 = R_0Q_0 \Longrightarrow T_1 = Q_{0}^{T}T_0Q_0.
$$
I need to show that $T_1$ is tridiagonal matrix. But I can't do it.
Someone could help me this.
Best Answer
If you need to show that $T_1$ is a tridiagonal matrix in an iterative QR process, you put $T_0 = Q_0R_0 \Longrightarrow R_0 = Q_0^{T}T_0$. So, $$ T_1 = R_0Q_0 \Longrightarrow T_1 = Q_0^{T}T_0Q_0. $$ Now you calculate $T_1^{T}$ and conclude that $T_1$ is a symmetric matrix. I think that you can use that $T_0$ is symmetric by first part of your question. Then, you write $$ T_1 = R_0Q_0 \Longrightarrow T_1 = R_0T_0R_0^{-1}. $$ Now we have $T_1$ a product of an upper triangular matrix $R_0$, a tridiagonal matrix $T_0$ and an upper triangular matrix $R_0^{-1}$. This implies that $T_1$ is an upper Hessenberg matrix. After that, you can see that $T_1^{T}$ is in the same way that is a lower Hessenberg matrix. How $T_1$ is a symmetric matrix, we can conclude that $T_1$ is tridiagonal.