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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?

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Are you only interested in real eigenvalues? –  Jonas Meyer Jun 15 '11 at 4:53

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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

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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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@Jonas: If matrix is symmetric then it has orthogonal eigenvectors. The matrix need not be normal? –  user17762 Jun 14 '11 at 0:47
    
@Sivaram: In $M_n(\mathbb{C})$, normality is equivalent to unitary diagonalizability. If you are working in $M_n(\mathbb{R})$, then the symmetric matrices are orthogonally diagonalizable. Of course, real symmetric matrices are normal. If the OP is thinking of real matrices, then your remarks on symmetric matrices will be more helpful. –  Jonas Meyer Jun 14 '11 at 0:50
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Real symmetric matrices are normal. Complex symmetric matrices need not be normal, or even diagonalizable, e.g. $\left(\matrix{1 & i \cr i & -1\cr}\right)$ –  Robert Israel Jun 14 '11 at 0:54
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@Jonas: In complex setting, the matrices OP is interested in are the Hermitian matrices –  user17762 Jun 14 '11 at 0:56
    
@Sivaram: Bob hasn't said he isn't interested in non-Hermitian normal matrices. Of course, Hermitian matrices are normal. It's also worth noting that a matrix $A$ is normal if and only the Hermitian matrices $\Re(A)=(A+A^*)/2$ and $\Im(A)=(A-A^*)/(2i)$ commute; commuting Hermitian matrices can be simultaneously unitarily diagonalized. –  Jonas Meyer Jun 14 '11 at 0:58

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