The rank of an outer product

[hobz]

Can somebody post (or direct me to) a proof/explanation that an outer product always has rank 1?

 

Thanks!

[Dr. Jekyll]

I take a 2 by 3 case as example and you can work out the general m by n case easily.

 

Let [math]a=

\left[\begin{array}{c}

a_1 \\

a_2\\

\end{array}

\right]

[/math] and [math]b=

\left[\begin{array}{c}

b_1 \\

b_2\\

b_3\\

\end{array}

\right]

[/math]

 

Then [math]ab^T=\left[\begin{array}{ccc}

a_1b_1 & a_1b_2 & a_1b_3\\

a_2b_1 & a_2b_2 & a_2b_3\\

\end{array}

\right]

[/math]

 

Each column vector is on the form

[math]b_i

\left[\begin{array}{c}

a_1 \\

a_2\\

\end{array}

\right]

\ \ i=1,2,3.[/math]

 

Then all column vectors are linear dependent, i.e., column rank is 1.

 

If we discard the knowledge that row and column rank are equal, the same applies to the row rank, as each row vector is on the form

[math]a_i

\left[\begin{array}{ccc}

b_1 & b_2 & b_3\\

\end{array}

\right]

\ \ i=1,2.[/math]

 

All rows are linear dependent and row rank is 1.

Edited September 10, 2008 by Dr. Jekyll

[hobz]

Thank you. A very concise explanation. Much obliged.