MHB Form of symmetric matrix of rank one

[i_a_n]

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. But what about the opposite direction...that is what we need to prove. How to prove this?)

 

[Opalg]

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. But what about the opposite direction...that is what we need to prove. How to prove this?)

If $C$ has rank 1 then the range space of $C$ is 1-dimensional. Let $w$ be a unit (column) vector in this subspace, then $Cw$ must be a scalar multiple of $w$, say $Cw = aw$. Since $C$ is symmetric, $w^TC = (Cw)^T = aw^T.$ Notice also that $w^Tw = 1$ since $w$ is a unit vector.

For any other column vector $x$, $Cx$ must also be a scalar multiple of $w$, say $Cx = \lambda_xw$, which we could equally well write $Cx = w\lambda_x$ since $\lambda$ is a scalar. Then $$\lambda_x = \lambda_x w^Tw = w^T(\lambda_xw) = w^T(Cx) = (w^TC)x = aw^Tx,$$ hence $Cx = w\lambda_x = w(aw^Tx) = aww^Tx$. That holds for all vectors $x$, therefore $C = aww^T.$