User:Eml4500.f08.ateam.shah/Week 4

Continuation of Lecture from Wednesday October 8, 2008

Note: Consider the case where d4(e) &ne; 0, where d1(e)=d2(e)=d3(e)=0

Multiplying out the equation out results in a 0 matrix as listed below:

f4(e) = k4(e) d4(e) = 0 where f is a 4x1 matrix, k is a 4x4 matrix, d is a 4x1 matrix which creates a 4x1 0 matrix.

The 0 matrix represents the 4th column of the k(e) which is the interpretation of the transverse dof's.

From period 14 slide 3: d4(e) = T4(e) d4(e)

Similarly using the same argument f4(e) = T4(e) d4(e)

Also k4(e) * d4(e) = f4(e)
 * which leads to k4(e) T4(e) d4(e) = T4(e) f4(e)

In the equation listed above the T4(e) is a block diagonal matrix



\begin{bmatrix} R_{2x2}^{(e)} & 0_{2x2}\\ 0_{2x2} & R_{2x2}^{(e)}\\ \end{bmatrix} $$

If T(e) is invertible, then [T(e)-1 k(e) T(e)]d(e) = f(e).

Consider a general block diagonal matrix with $$ A= \begin{bmatrix} D_1& & &  & 0\\ &  .& &  & &\\      &   &  .& & &\\      &   &   & .&&\\      0&   &  &  & D_5\\ \end{bmatrix} $$

Question: What is the inverse of A (A-1? Simple example: Diagonal Matrix

$$ B= \begin{bmatrix} d_{11}& & &  & 0\\ &  d_{12}& &  & &\\ &  &  .& & &\\      &   &   & .&&\\      0&   &  &  & d_{mn}\\ \end{bmatrix} = diag[d_{11}, d_{22}, .... , d_{mn}] $$

thus $$ B^{-1} = Diag \begin{bmatrix} \tfrac{1}{d_{11}},& \tfrac{1}{d_{22}},& .&  .& .& .,&\tfrac{1}{d_{mn}}\\ \end{bmatrix} $$

Assuming that all values of the diagonal coefficients are not equal to 0, i.e. dii &ne; 0 for all values of i=1......n.

For a block diagonal matrix A $$ A = Diag \begin{bmatrix} D_1,& .& .& .&, D_s \end{bmatrix} $$ and $$ A^{-1} = Diag \begin{bmatrix} D_1^{-1}, & .& .& .& .&, D_s^{-1}\\ \end{bmatrix} $$ where D1-1, is the inverse of the matrix D1.

The transport matrix is defined to be
 * T(e)-1=Diag[R(e)-1, R(e)-1]

On Lecture 19 Slide 2 R(e) is defined to be $$ R^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} $$ $$ R^{(e)T} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} $$ Thus multiplying R(e) by its transpose will result in $$ R^{(e)T}R^{(e)} = \begin{bmatrix} l^{(e)} & -m^{(e)}\\ m^{(e)} & l^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)} & l^{(e)} \end{bmatrix} = \begin{bmatrix} 1 & 0\\     0 & l     \end{bmatrix} $$ The resulting matrix represents a 2x2 I or identity matrix.

Thus,
 * R(e)-1=R(e)T
 * T(e)-1=Diag[R(e)T, R(e)T]=(Diag[R(e), R(e)])T

Also
 * T(e)-1=T(e)T