hobz Posted September 9, 2008 Share Posted September 9, 2008 Can somebody post (or direct me to) a proof/explanation that an outer product always has rank 1? Thanks! Link to comment Share on other sites More sharing options...
Dr. Jekyll Posted September 10, 2008 Share Posted September 10, 2008 (edited) 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 Link to comment Share on other sites More sharing options...
hobz Posted September 11, 2008 Author Share Posted September 11, 2008 Thank you. A very concise explanation. Much obliged. Link to comment Share on other sites More sharing options...
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