What is the square of a column vector?

AI Thread Summary
The discussion revolves around the concept of squaring a column vector in matrix terms. The user inquires if there is a matrix property that allows for this operation, specifically referencing the forms {\bf x}{\bf x}^T and {\bf x}^T{\bf x}. It is clarified that {\bf x}{\bf x}^T results in a 3x3 matrix, while {\bf x}^T{\bf x} produces a 1x1 matrix, which is interpreted as the scalar product of the vector. The square of a vector is generally understood to refer to this scalar product, represented as (x^T)x. Understanding these matrix operations is essential for proper application in linear algebra.
devonho
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Homework Statement



Hi,

Is there a matrix property that can be applied to take the square of a column vector?

Something like:

{\bf x}=\left[x_1,x_2,x_3\right]^T
\left[{\bf x}\right]^2
={\bf x}{\bf x}^T
or
={\bf x}^T{\bf x}?

Thank you.


Homework Equations





The Attempt at a Solution

 
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devonho said:

Homework Statement



Hi,

Is there a matrix property that can be applied to take the square of a column vector?

Something like:

{\bf x}=\left[x_1,x_2,x_3\right]^T
\left[{\bf x}\right]^2
={\bf x}{\bf x}^T
or
={\bf x}^T{\bf x}?

Thank you.
...
What is the size of the matrix: ={\bf x}{\bf x}^T\,?

What is the size of the matrix: ={\bf x}^T{\bf x}\,?
 
Hi SammyS,

{\bf x}{\bf x}^T is 3x3
while
{\bf x}^T{\bf x} is 1x1
 
I'm not sure what you mean by the square of a column vector...
 
Generally the square of a vector refers to the scalar product, which you can crudely think of as a 1 by 1 matrix. Therefore, the answer is (xT)x .
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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