# Symmetric matrix decomposition with orthonormal basis of non-eigenvectors

I like to understand the following transformation found in documentation for deriving Kalman filter.

Abstract Formulation: Given 2 symmetric matrices $A$ ,$B$ $\in$ $\mathbb R^{3,3}$ with $A \ne B$ and a set of orthonormal Eigenvectors ($u_1$, $u_2$, $u_3$) from some other matrix $B$ (not $A$!). Because the matrices are symmetric it is clear that $B$ can be decomposed to $B = U\Lambda U^t$. Now there is stated that A can be written as:

$A$ = ($u_1^t$A$u_1$)$u_1$$u_1^t + (u_2^tAu_2)u_2$$u_2^t$ + ($u_3^t$A$u_3$)$u_3$$u_3^t i.e. with the "foreign" Eigenvectors. Concrete Situation: In the original equation the above mentioned A is defined as H_kP_k^-H_k^t + R_a, where P^- is the a priori estimation error covariance and R_a is the sensor noise error covariance matrix. H_k has 3x9 dimension and contains some "more abstract" content with rotation matrix of a quaternion multiplied with cross product operator of gravity vector (0,0,g). As far as I can see, the term (H_kP_k^-H_k^t + R_a) does not lead to a diagonal matrix and this seems to be irrelevant. What I called B is actually the signal's overall error error covariance named U_k From original paper: Because U_k covariance matrix cannot be obtained at this point in time (a priori estimation), it is approximated by the average of the last M steps i.e. from k-M to k-1. The signal itself might be fluctuating considerably because sometimes there is external acceleration at other times there isn't thus sensor noise is the only thing be measured. Assumption (thanks to Calle's and joriki's comments): The eigendecomposition of U_k is related to PCA Principal component analysis (an easier one here). The most interesting cases are all those measurements with strong accelerations i.e. U_k is much greater than the remaining term. So this decomposition of the 2nd term transforms it (approximatively?) towards the direction of the strongest signal. Thus \lambda - \mu helps to detect these situations respectively distinguish them from phases with no signal aside from noise. • Does this explanation makes sense? • Can this procedure of approximating with "wrong" eigenvectors and -values be applied and compares like with like? • What is the name of this matrix decomposition taking not their own eigenvectors? • What is about the error? Thanks for helping Kay PS: Title changed from "Symmetric matrix multiplied with kind of orthonormal basis" - Seems a bit odd. If A could be written this way, Au_1 = (u_1^T A u_1) u_1, so u_1 should be an eigenvector of A as well. Are you sure there aren't more conditions and the equation is correct? – Calle Apr 20 '11 at 13:24 Is B related to A in any way? – Calle Apr 20 '11 at 14:06 AFAIK that's all about it but I updated the question to get things more clear. – Kay Apr 20 '11 at 14:07 @Kay: I thought your description was clear enough; A and B in my comments referred to just those matrices. My comment about A being diagonal not helping was in reference to your statement that "this product (H_kP_k^-H_k^t+R_a) does not lead to a diagonal matrix". – joriki Apr 20 '11 at 18:15 @moderators: Maybe PCA or more general multivariate-analysis would be a candidate for new tag. – Kay Apr 20 '11 at 22:37 ## 1 Answer Writing a matrix A in terms of a basis that does not diagonalize the matrix A is possible, but it requires a full expansion of all terms, not just the diagonal terms. (If the basis diagonalizes A then all off diagonal terms would be zero). If A is a 3 \times 3 matrix and you write A = U U^{-1} A UU^{-1} with$$U = \begin{bmatrix}\mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3\end{bmatrix}$$then you have the sum along all pairs ij (not just for i=j)$$ A = \sum_{ij}\mathbf{u}_i ( \mathbf{u}_i^\top A \mathbf{u}_j)\mathbf{u}_j^\top$\$

I can not tell you why the off diagonal terms were ignored, perhaps it is by design or perhaps by mistake.

Here is the full expansion:

\begin{align} U^{-1}AU &=\begin{bmatrix}\mathbf{u}_1^\top \\ \mathbf{u}_2^\top \\ \mathbf{u}_3^\top \end{bmatrix} A\begin{bmatrix}\mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3\end{bmatrix} \\ &=\begin{bmatrix}\mathbf{u}_1^\top A \\ \mathbf{u}_2^\top A \\ \mathbf{u}_3^\top A\end{bmatrix} \begin{bmatrix}\mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3\end{bmatrix} \\ &= \begin{bmatrix}\mathbf{u}_1^\top A \mathbf{u}_1 & \mathbf{u}_1^\top A \mathbf{u}_2 & \mathbf{u}_1^\top A \mathbf{u}_3 \\ \mathbf{u}_2^\top A \mathbf{u}_1 & \mathbf{u}_2^\top A \mathbf{u}_2 & \mathbf{u}_2^\top A \mathbf{u}_3 \\\mathbf{u}_3^\top A \mathbf{u}_1 & \mathbf{u}_3^\top A \mathbf{u}_3 & \mathbf{u}_3^\top A \mathbf{u}_3 \\\end{bmatrix} \\ UU^{-1}AUU^{-1} &= \begin{bmatrix}\mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3\end{bmatrix}\begin{bmatrix}\mathbf{u}_1^\top A \mathbf{u}_1 & \mathbf{u}_1^\top A \mathbf{u}_2 & \mathbf{u}_1^\top A \mathbf{u}_3 \\ \mathbf{u}_2^\top A \mathbf{u}_1 & \mathbf{u}_2^\top A \mathbf{u}_2 & \mathbf{u}_2^\top A \mathbf{u}_3 \\\mathbf{u}_3^\top A \mathbf{u}_1 & \mathbf{u}_3^\top A \mathbf{u}_2 & \mathbf{u}_3^\top A \mathbf{u}_3 \\\end{bmatrix}\begin{bmatrix}\mathbf{u}_1^\top \\ \mathbf{u}_2^\top \\ \mathbf{u}_3^\top\end{bmatrix} \\ \hphantom{A} \\ A &= \begin{bmatrix}\mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3\end{bmatrix}\begin{bmatrix}\mathbf{u}_1^\top A \mathbf{u}_1 \mathbf{u}_1^\top + \mathbf{u}_1^\top A \mathbf{u}_2 \mathbf{u}_2^\top + \mathbf{u}_1^\top A \mathbf{u}_3\mathbf{u}_3^\top \\ \mathbf{u}_2^\top A \mathbf{u}_1\mathbf{u}_1^\top + \mathbf{u}_2^\top A \mathbf{u}_2\mathbf{u}_2^\top + \mathbf{u}_2^\top A \mathbf{u}_3\mathbf{u}_3^\top \\\mathbf{u}_3^\top A \mathbf{u}_1\mathbf{u}_1^\top + \mathbf{u}_3^\top A \mathbf{u}_2\mathbf{u}_2^\top + \mathbf{u}_3^\top A \mathbf{u}_3\mathbf{u}_3^\top \\\end{bmatrix} \\ &= \sum_{ij}\mathbf{u}_i ( \mathbf{u}_i^\top A \mathbf{u}_j)\mathbf{u}_j^\top \end{align}

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