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Math Help - (vector times its transpose)^2 proof?

  1. #1
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    (vector times its transpose)^2 proof?

    Simplify as far as possible: (v*v^T)^2, assuming v is a column vector of R^n with ||v|| = 1.

    This was a problem on a recent quiz I took and I am having trouble understanding it. The short answer key given after the quiz says that it can be simplified to just (v*v^T), no square needed. Can anyone please help with this proof?

    Thanks,
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  2. #2
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    (vv^T)^2=(vv^T)(vv^T)=v(v^Tv)v^T=v||v||^2v^T=vv^T Which is what your answer key has.
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    thanks
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  4. #4
    Senior Member jakncoke's Avatar
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    so  (v*v^T)^2 which is just  (v*v^T)(v*v^T) The dot product two vectors a, b can be computed  a \cdot b = a^Tb knowing that.  v^Tv = v \cdot v so our equation turns into  (v \cdot v \cdot v)*v^T since the magnitude  ||v|| = 1 the magnitude is  \sqrt{a_1^2 + a_2^2 + ....} where  a_1, a_2, ... are components of v. we know the value under the square root must be 1 as  \sqrt{1} = 1 and so  v \cdot v = a_1^2 + a_2^2 + ... = 1. So  v \cdot v = 1 \cdot vv^T = vv^T . Hope this helps.
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