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I have a question about the computation of the norm of a block matrices.

The square block matrix $T_{n}$ is defined as:

\begin{bmatrix} A_{1,1} &. &.&&A_{1,n} \\ A_{2,1} &. &.&.&A_{2,n}\\ . &. &.&.&.\\ . &. &.&.&.\\ . &. &.&.&.\\ A_{n,1} &. &.&.&A_{n,n}\\ \end{bmatrix} where $A_{i,j}$ is a $c \times c$ Toeplitz matrix given as \begin{bmatrix} a_{0} &a_{1} &.&.&a_{c} \\ a_{-1} &a_{0} &a_{1}&.&.\\ . &. &.&.&.\\ . &. &.&.&a_{1}\\ a_{-c} &. &.&a_{-1}&a_{0}\\ \end{bmatrix}

I would like to compute the following norm

$||T_{n}||^{2}=\underset{x}{\max}(x'T_{n}T'_{n}x)/(x'x)$.

In the case of a simple $n \times n$ Toeplitz matrix, it is easy and gives

$||T_{n}||^{2}=\underset{x}{\max}(x'T_{n}T'_{n}x)/(x'x)= \sum \limits_{\underset{}{k=0}}^{n-1}\sum \limits_{\underset{}{j=0}}^{n-1} t_{k-j}x'x /(\sum \limits_{\underset{}{k=0}}^{n-1} |x_{k}|^{2})$

Any ideas or suggestions are welcomed.

  • 0
    To be clear, the $'$ means transpose?2017-02-08
  • 0
    @Semiclassical yes.2017-02-08

0 Answers 0