Is it always possible to decompose an $n$-by-$n$ matrix $\mathbf{A}$ as $\mathbf{A} = \sum_{i=1}^n \lambda_i \mathbf{v}_i\mathbf{v}_i^{\rm T}$ where $\lambda_i$ is the $i$-th eigenvalue of $\mathbf{A}$ and $\mathbf{v}_i$ its corresponding eigenvector?
Matrix decomposition in eigenvector product
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0Are you only interested in real eigenvalues? – 2011-06-15
2 Answers
If the matrix $\mathbf{A}$ is hermitian then it is possible to write it in the form you have shown i.e. $\mathbf{A} = \sum_{k=1}^{n} \lambda_k \mathbf{v}_k \mathbf{v}_k^*$ The reason why you are able to write it in this form is that for a hermitian matrix the eigenvectors are orthogonal
In general, you can write a matrix as $\mathbf{A} = \sum_{k=1}^{n} \sigma_k \mathbf{u}_k \mathbf{v}_k^*$ where $\sigma_k \geq 0$ are the singular values and $\mathbf{u}_k,\mathbf{v}_k$ are the singular vectors
No it is not. The existence of an orthonormal basis of eigenvectors of a complex $n$-by-$n$ matrix is equivalent to normality; this is the same as unitary diagonalizability. The weaker condition of having a basis of eigenvectors is also not held by all matrices, and is equivalent to diagonalizability.
For example, $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is not diagonalizable, and $\begin{bmatrix}1&1\\0&0\end{bmatrix}$ is diagonalizable but not normal.
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0@Jonas: Sorry for the confusion. My mistake. For some reason I thought, if the eigenvectors are orthonormal, then the matrix is hermitian. – 2011-06-14