(vector times its transpose)^2 proof?

**reddevil14241** · July 6th 2010, 11:39 AM

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,

**MattMan** · July 6th 2010, 12:31 PM

$(vv^T)^2=(vv^T)(vv^T)=v(v^Tv)v^T=v||v||^2v^T=vv^T$ Which is what your answer key has.

**reddevil14241** · July 6th 2010, 12:42 PM

thanks

**jakncoke** · July 6th 2010, 12:44 PM

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.

Thread: (vector times its transpose)^2 proof?

LinkBack

Thread Tools

Display

(vector times its transpose)^2 proof?

Similar Math Help Forum Discussions

Least Squares estimator proof using vector transpose

Proof with transpose - I think this is right

Matrix Calculus - Transpose of Vector Derivatives

Proof for the div (scalar times a vector)

How do I differentiate a vector times vector?

Search Tags