form of symmetric matrix of rank one The question is:
Let $C$ be a symmetric matrix of rank one. Prove that $C$ must have the form $C=aww^T$, where $a$ is a scalar and $w$ is a vector of norm one.
I think we can easily prove that if $C$ has the form $C=aww^T$, then $C$ is symmetric and of rank one. Why we need $w$ is of norm one?
 A: $C$ is a symmetric matrix and thus has eigen decomposition $C=\sum_{i=1}^{N}\lambda_iu_iu_i^H$. If it is rank one, all eigenvalues except one will be zero, thus $C=\lambda uu^H$ where $\lambda$ is the non-zero eigenvalue and $u$ is the corresponding unit norm eigen vector. 
A: dineshdileep has already provided what is perhaps the natural solution to this problem. Let me present another.
Any matrix has what is called a rank factorization. In the case of a rank $1$ matrix, every row in the matrix is a multiple of some specific row. Let $\mathbf{y}$ be that row and let $\mathbf{x}$ be the vector of multiples associated with $\mathbf{y}$. Then $A$ naturally factors as 
$$A=\mathbf{x}\mathbf{y}^\mathrm{T}$$
We may factor out the magnitudes to get 
$$A = \|\mathbf{x}\|\|\mathbf{y}\|\,\mathbf{\hat{x}}\mathbf{\hat{y}}^\mathrm{T}$$
So far this decomposition holds for any matrix, symmetric or not. Our goal now is to prove that $\mathbf{x} = \pm\mathbf{y}$ in the case of symmetry.
Suppose then that $A$ is symmetric with decomposition $A = c\mathbf{x}\mathbf{y}^\mathrm{T}$ where $\mathbf{x}$ and $\mathbf{y}$ are unit vectors. By symmetry, we get
$$\mathbf{x}\mathbf{y}^\mathrm{T} = \mathbf{y}\mathbf{x}^\mathrm{T}$$
Multiplying on the left by $\mathbf{y}^\mathrm{T}$ and on the right by $\mathbf{x}$, we get
$$1=\|\mathbf{x}\|^2\|\mathbf{y}\|^2 = (\mathbf{x}^\mathrm{T}\mathbf{x})(\mathbf{y}^\mathrm{T}\mathbf{y}) = \mathbf{x}^\mathrm{T}\mathbf{y}\mathbf{x}^\mathrm{T}\mathbf{y} = (\mathbf{x}\cdot\mathbf{y})^2$$
which of course implies
$$|\mathbf{x}\cdot \mathbf{y}| = 1 = \|\mathbf{x}\|\|\mathbf{y}\|$$
This is the equality case of Cauchy-Schwarz, which holds if and only if $\mathbf{x}$ and $\mathbf{y}$ are linearly dependent. The two are both unit vectors so it follows that $\mathbf{x} = \pm\mathbf{y}$. Therefore we have
$$A = \pm c \mathbf{x}\mathbf{x}^\mathrm{T}$$
as required.
