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Diagonalizable Vectors

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Suppose v=(x1,x2,...,xn)^t, w=(y1,y2,...,yn)^t, then
vw^t is the matrix below

x1y1 x1y2 ... x1yn
x2y1 x2y2 ... x2yn
xny1 xny2 ... xnyn

Trace(vw^t)=x1y1+x2y2+...+xnyn=(v^t)w, which is not equal to 0.
Then v is not 0. Without the loss of generality, we assume x1 is not 0.
Then the kth row of the matrix vw^t is (xk/x1) times the first row.
i.e. (xky1,xky2,...,xkyn) ...

Solution Summary

Diagonalizable Vectors are investigated.