Is result of vector inner product retained after matrix multiplication?

  • #1
Master1022
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Homework Statement
Let us imagine that we have two vectors ## \vec{a} ## and ## \vec{b} ## and they point in similar directions, such that the inner-product is evaluated to be a +ve number. If we now multiply both of the vectors by a matrix ## W ## which has real entries, will the inner product of the 'transformed' vectors also be positive?
Relevant Equations
Inner product
Hi,

I was thinking about the following problem, but I couldn't think of any conclusive reasons to support my idea.

Question:
Let us imagine that we have two vectors ## \vec{a} ## and ## \vec{b} ## and they point in similar directions, such that the inner-product is evaluated to be a +ve number. If we now multiply both of the vectors by a matrix ## W ## which has real entries, will the inner product of the 'transformed' vectors also be positive?

Attempt:

Intuitively I think along the lines of: if we imagine the operation as transforming a vector in some way, then the two vectors ## \vec{a}## and ## \vec{b}##, which were similar, should be transformed to similar vectors?

Mathematically, I can write the following:
[tex] <W \vec{a} , W \vec{b} > = (W \vec{a}) \cdot (W \vec{b}) = (W \vec{a})^{T} (W \vec{b}) = \vec{a}^{T} W^{T} W \vec{b} [/tex]

- ## W^{T} W ## is positive semi-definite.
- I suppose if ## \vec{a} ## and ## \vec{b} ##, then if/how positive the outcome will depend on some type of sensitivity of ## W ##? That is, we could view ## \vec{b} = \vec{a} + \vec{\epsilon} ## and think about that?

Any help is greatly appreciated.
 
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  • #2
If ##\vec a## and ##\vec b## are parallel, then the assertion follows directly from ##W^T W## being positive semi-definite (as long as you change the statement to be non-negative instead of positive - it should be pretty obvious that the result can be zero if ##W^T W## happens to have eigenvalues that are zero, such as when ##W = 0##).

However, consider what happens when you let ##\vec a## and ##\vec b## be linearly independent and ask yourself the following questions:
  • Taking any set of two vectors ##\vec c## and ##\vec d##, can you find a linear transformation ##W## such that ##W\vec a = \vec c## and ##W \vec b = \vec d##?
  • What does this mean for your assertion?
 
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  • #3
Thanks for the reply @Orodruin !
Orodruin said:
If ##\vec a## and ##\vec b## are parallel, then the assertion follows directly from ##W^T W## being positive semi-definite (as long as you change the statement to be non-negative instead of positive - it should be pretty obvious that the result can be zero if ##W^T W## happens to have eigenvalues that are zero, such as when ##W = 0##).
Agreed

Orodruin said:
However, consider what happens when you let ##\vec a## and ##\vec b## be linearly independent and ask yourself the following questions:
  • Taking any set of two vectors ##\vec c## and ##\vec d##, can you find a linear transformation ##W## such that ##W\vec a = \vec c## and ##W \vec b = \vec d##?
I think so... I could imagine forming some matrix equation like:
[tex] W [ \vec{a} \quad \vec{b}] = [\vec{c} \quad \vec{d}] [/tex]
and if ## \vec{a} ## and ## \vec{b} ## are linearly independent, then the matrix is full-rank and that can be inverted to find ## W ##.

Orodruin said:
  • What does this mean for your assertion?
That might suggest that the outcome of the transformed inner product is still positive? This doesn't necessarily seem correct, but I am confused still...
 
  • #4
Master1022 said:
That might suggest that the outcome of the transformed inner product is still positive? This doesn't necessarily seem correct, but I am confused still...
It took me about 30 seconds to find a counterexample. Hint. Let the vectors be ##(1,0)## and ##(1, 1)##. Look for a 2 x 2 matrix that maps ##(1, 0)## to itself and ##(1, 1)## to something in approximately the opposite direction.
 
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  • #5
Master1022 said:
That might suggest that the outcome of the transformed inner product is still positive? This doesn't necessarily seem correct, but I am confused still...
So you are free to choose any ##\vec c## and ##\vec d##. Are there any such pairs with a negative inner product?
 
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  • #6
PeroK said:
It took me about 30 seconds to find a counterexample. Hint. Let the vectors be ##(1,0)## and ##(1, 1)##. Look for a 2 x 2 matrix that maps ##(1, 0)## to itself and ##(1, 1)## to something in approximately the opposite direction.
Okay thanks @PeroK ! That makes sense, and yes I can see how the answer to my question was 'not necessarily'.
 
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