Let's assume i have a linear system as:

$-\omega^{2}U^{T}(M+\Delta M)U+U^{T}(K+\Delta K)U=0_M$

Where $\Delta M=diag(\begin{bmatrix}dm1&dm2&dm3&dm4&dm5\end{bmatrix})$

$\Delta K=diag(\begin{bmatrix}dk1&dk2&dk3&dk4&dk5\end{bmatrix})$


$U\in \Re^{5x3}$

$0_M\in \Re^{3x3}$

Note: $diag$ means a matrix with the element on the diagonal.

How can i formulate the problem in an equivalent form:



$A\in \Re^{9x10}$


$B\in \Re^{9x1}$

This is needed in order to find the rank of the system.

  • 1
    $\begingroup$ What does $I_5 [dm1, ..., dm5]^T$ mean? Your matrix is $5 \times 5$ but $[dm1, ..., dm5]^T$ is a $5\times 1$ vector. Also, is $I_5$ an identity matrix? $\endgroup$ – Florian Jun 25 '18 at 10:21
  • $\begingroup$ Yes, i post an edit to clarify $\endgroup$ – iacopo Jun 25 '18 at 10:25
  • 1
    $\begingroup$ Still unclear. If $I_5$ is an identity then $I_5 \cdot [dm1, ..., dm5]^T = [dm1, ..., dm5]^T \in \Re^{5 \times 1}$. However, your $\Delta M$ should be $5\times 5$. $\endgroup$ – Florian Jun 25 '18 at 10:26
  • $\begingroup$ Yes, you are right. fixed $\endgroup$ – iacopo Jun 25 '18 at 11:32

What helps to solve your problem is the rule ${\rm vec}\{A \cdot {\rm diag}(b) \cdot C^T\} = (C \diamond A)\cdot b$, where

  • ${\rm vec}\{X\}$ is the vectorization operator that rearranges the elements of a matrix $X \in \mathbb{R}^{m \times n}$ into a vector $\in \mathbb{R}^{m \cdot n \times 1}$
  • The $\diamond$ operator is the "Khatri-Rao product", also known as the column-wise Kronecker product between two matrices, i.e., for given matrices $A = [a_1, \ldots, a_k] \in \mathbb{R}^{m \times k}$ and $B = [b_1, \ldots, b_k] \in \mathbb{R}^{n \times k}$, the matrix $C = A \diamond B$ is given by $C = [a_1 \otimes b_1, \ldots, a_k \otimes b_k] \in \mathbb{R}^{m \cdot n \times k}$, where $\otimes$ is the Kronecker product.

Let us apply this rule to your problem. First, we rearrange a bit:

$$ \begin{align} -\omega^2U^T(M+\Delta M)U+U^T(K+\Delta K)U&=0_M \\ -\omega^2U^T{\rm diag}\{\Delta m\} U+U^T{\rm diag}\{\Delta k\} U&=\omega^2 U^T M U - U^T K U \end{align} $$ Now, let us vectorize: $$ \begin{align} -(U^T \diamond \omega^2 U^T) \cdot \Delta m +(U^T \diamond U^T) \cdot \Delta k & = {\rm vec} \{ \omega^2 U^T M U - U^T K U\} \\ \left[-U^T \diamond \omega^2 U^T, U^T \diamond U^T\right] \cdot \left[\Delta m^T, \Delta k^T\right]^T & = {\rm vec} \{ \omega^2 U^T M U - U^T K U\}, \end{align} $$ which is in the desired form $A x = b$, where $A = \left[-U^T \diamond \omega^2 U^T, U^T \diamond U^T\right] \in \mathbb{R}^{9 \times 10}$, $x = \left[\Delta m^T, \Delta k^T\right]^T\in \mathbb{R}^{10 \times 1}$, and $b = {\rm vec} \{ \omega^2 U^T M U - U^T K U\} \in \mathbb{R}^{9 \times 1}$.

As a slight simplification, you can rewrite your system matrix $A$ into $$A = \left[-(I_3 \otimes \omega^2) \cdot (U^T \diamond U^T), U^T \diamond U^T\right] = \left[-D \cdot (U^T \diamond U^T), U^T \diamond U^T\right],$$ where $D = {\rm diag}\{\omega_1, \omega_2, \omega_3, \omega_1, \omega_2, \omega_3, \omega_1, \omega_2, \omega_3\} \in \mathbb{R}^{9 \times 9}$, since $\omega$ is diagonal.

Clearly, if your $\omega_i$ are equal, the system has rank at most 5 (not surprisingly). In general, you may be more lucky. I just tried with randomly drawn data and got rank 9. My guess is for randomly drawn $\omega$, $U$ (from a continuous distribution) you get full rank almost surely, but that's a guess only.

  • $\begingroup$ Nice, thank you for the help $\endgroup$ – iacopo Jun 25 '18 at 12:39
  • $\begingroup$ You're welcome, glad to hear it helps. $\endgroup$ – Florian Jun 25 '18 at 12:41

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