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I have a matrix of the form $C = I - a a^T$. In this particular case, $a$ is a constant vector with value $1/\sqrt{N}$, where $N$ is the rank of $I$ (number of columns). For $I = eye(5)$, $N$ will be 5.

I am trying to find a matrix $U$ such that $C = U U^T$. I can do this using eigenvalue decomposition.

I have found "Some Modified Matrix Eigenvalue Problems" from Golub and looks very promising, as it shows how to compute eigenvalues for the exact same kind of matrices that I have. It says that when $a$ has all values different than 0, the eigenvalues of C can be computed as the zeros of

$$ 1 - \sum_i^N {\frac{u_i^2}{d_i - \lambda}}$$

Following that formula, in my particular case I can write

$$ 1 - N {\frac{1/N}{1 - \lambda}} = 0$$ which is only true for $\lambda = 0$. Matlab says that I have $N-1$ more eigenvalues that are equal to 1.

Assuming that I have 1 eigenvalues which equals to 0 and $N-1$ which equals to 1, I have tried to compute the eigenvectors using "Rank-One Modification ofthe Symmetric Eigenproblem" from Bunch, Nielsen and Sorensen but I am not sure I can use that. In Section 4 they define $D_i$ as $$D_i = D - \tilde{d}_i I$$ with D being the singular values of the original matrix (called $B$ in this last paper). In my particular case, I can have two different possible values for $D_i$: An indentity matrix, which is associated with the first eigenvalue, which is 0, or a null matrix, associated with the other eigenvalues. They said that "Theorem 1 implies that $D_i$ is nonsingular" which in my particular case is not true at all.

I am not sure if I can use all this theory given that my updated matrix is not rank complete.

Do you have any idea on how could I find a closed form for $U$?

Edit: Fix a typo

Rodrigo de Azevedo
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jjcasmar
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1 Answers1

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For general $a$, such a decomposition exists if and only if $\|a\|_2\leq 1$. This is satisfied in the special case above.

We can compute an eigen-decomposition and the matrix $U$ directly. Clearly, $a$ is an eigenvector of $C$ with the associated eigenvalue $1-a^Ta\geq 0$. The remaining eigenvectors can be taken from the orthogonal complement of $a$ and correspond to repeated eigenvalue $1$ (of multiplicity $N-1$). So if we take an orthonormal matrix $$ V:=\left[\frac{a}{\|a\|_2},\tilde{V}\right] $$ and diagonal $$ D:=\begin{bmatrix}1-a^Ta & 0\\ 0 & I_N\end{bmatrix}, $$ then $A=VDV^T=UU^T$ with $U:=VD^{1/2}$.

There's also a simple way based on Householder reflections which provides an explicit formula for $V$ if the specification above is not explicit enough.

Algebraic Pavel
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