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Form of symmetric matrix of rank one

ianchenmu

Member
Feb 3, 2013
74
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 $C = aww^T.$