LINEAR ALGEBRA: What is the relationship between ||y||^2 and y . x?

  1. QUESTION (5.1, #30 -> Bretscher, O.):

    Consider a subspace V of [tex]\mathbb{R}^n[/tex] and a vector [tex]\overrightarrow{x}[/tex] in [tex]\mathbb{R}^n[/tex]. Let [tex]\overrightarrow{y}\,=\,proj_v\,\overrightarrow{x}[/tex]. What is the relationship between the following quantities?

    [tex]||\overrightarrow{y}||^2[/tex] and [tex]\overrightarrow{y}\,\cdot\,\overrightarrow{x}[/tex]


    My work so far:

    I only know this for sure...

    [tex]||\overrightarrow{y}||\,\leq\,||\overrightarrow{x}||\,\iff\,\overrightarrow{x}\,\in\,V[/tex]

    But how do I use this to relate the two quantities? Please help, thanks!
     
  2. jcsd
  3. quasar987

    quasar987 4,774
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    I assume [itex]\hat{v}[/itex] is a unit vector? Then [itex]\vec{y}=(\vec{x}\cdot \hat{v})\hat{v}[/itex]. I think this "substitution" is a good place to start.
     
  4. Ok, I use that version and write:

    [tex]\,\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrightarrow{v}\right]\,\cdot\,\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrightarrow{v}\right]\,[/tex]

    [tex]\left[\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)\,\overrightarrow{v}\right]\,\cdot\,\overrightarrow{x}[/tex]

    This is two equations in two variables.

    Is the second one equal to zero?
     
    Last edited: Nov 6, 2006
  5. quasar987

    quasar987 4,774
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    write the squared norm of y as [itex]||\vec{y}||^2=\vec{y}\cdot\vec{y}[/itex].

    and simplify <y,x> a little also.

    the relationship should become clear.

    (Why would you think the second one equals to zero?)
     
  6. The squared norm of y is [tex]||\,\left(\overrightarrow{v}\,\cdot\,\overrightarrow{x}\right)^2\,\overrightarrow{v}^2\,||[/tex]?

    [tex]\overrightarrow{y}\,\cdot\,\overrightarrow{y}\,=\,||\overrightarrow{y}||^2[/tex]

    Does that sound right?
     
    Last edited: Nov 6, 2006
  7. quasar987

    quasar987 4,774
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    As I said, the squared norm (or norm squared if you'd like) of [itex]\vec{y}[/itex] is [itex]||\vec{y}||^2=\vec{y}\cdot\vec{y}[/itex].

    And as I said in post #2, [itex]\vec{y}=(\vec{x}\cdot \hat{v})\hat{v}[/itex]. So

    [tex]||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}[/tex].

    Now use the properties of the scalar product to make that look nicer (e.g. [itex] \vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})[/itex])
     
  8. quasar987

    quasar987 4,774
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    As a general rule, stay away from notational shortcuts like [itex]\vec{y}^2[/itex] in a mathematics course. Always write the ||
     
  9. Like the first equation in post #3?

    Scalar product properties?

    So...

    [tex]\hat{v}\,\left(\overrightarrow{x}\,\cdot\,\hat{v}\right)^2[/tex] is the norm squared?
     
    Last edited: Nov 6, 2006
  10. I am guessing that this might have something to do with the Cauchy-Schwarz inequality since it is in the section and all, right?

    [tex]|\vec{x}\,\cdot\,\vec{y}|\,\leq\,||\vec{x}||\,||\vec{y}||[/tex]

    How do I use the inequality for this problem? Just start substituting?
     
  11. quasar987

    quasar987 4,774
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    The first equation in post #3 doesn't make much sense because the ||a|| notation means "norm of vector a", but in your first equation of post 3, you have a scalar in btw the ||.||. You don't even need absolute values either because [itex]\vec{y}\cdot\vec{y}=y_x^2+y_y^2+y_z^2[/itex] can never be negative.

    So the equation to work with is [itex]||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}[/itex]

    And if that is the norm squared, how did you pass from that to [itex]\hat{v}\,\left(\overrightarrow{x}\,\cdot\,\hat{v}\ \right)[/itex] as the norm squared?! This is not even a scalar!

    You need to re-read your notes carefully I think!
     
    Last edited: Nov 6, 2006
  12. So we have:

    [tex]||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}[/tex]

    But how do you simplify that expression? (without just going back to [tex]||\vec{y}||^2[/tex])
     
    Last edited: Nov 6, 2006
  13. You can simplify it in many ways, but you must compare your RHS
    [tex](\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}[/tex]
    with
    [tex]\overrightarrow{y}\,\cdot\,\overrightarrow{x}[/tex]
     
    Last edited: Nov 6, 2006
  14. quasar987

    quasar987 4,774
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    By the properties of the scalar product (or inner product or whatever you call the [itex]\cdot[/itex] operation). More precesily, use this property: [itex] \vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})[/itex].
     
    Last edited: Nov 6, 2006
  15. [tex]||\vec{y}||^2=(\vec{x}\cdot \hat{v})\hat{v}\cdot(\vec{x}\cdot \hat{v})\hat{v}[/tex]

    [tex]||\vec{y}||^2=(\vec{x} \cdot ||\vec{v}||^2)(\vec{x} \cdot ||\vec{v}||^2)=(||\vec{x}||^2 \cdot ||\vec{v}||^4)[/tex]
     
    Last edited: Nov 6, 2006
  16. quasar987

    quasar987 4,774
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    Sniff. Where do you pull all these equations from? Your last equation doesn't make sense and you should see it! You're taking the dot product of scalars!


    You have the template right here:

    [tex] \vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})[/tex]

    Except in your case, [itex]\alpha = (\vec{x}\cdot \hat{v})[/itex] and [itex]\vec{a}=\vec{b}=\hat{v}[/itex]. And also, both a and b are multiplied by a constant.
     
  17. [tex]||\vec{y}||^2=\alpha(\vec{a}) \cdot \alpha(\vec{a})[/tex]

    But what is the result of a dot product between two identical vectors?
     
    Last edited: Nov 6, 2006
  18. quasar987

    quasar987 4,774
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    You've only rewritten the equation by changing the notation. You're not more advanced.

    Use the property of the dot product that allows you to take the alpha's and pull them in front: [tex] \vec{a}\cdot \alpha\vec{b}= \alpha\vec{a}\cdot \vec{b}= \alpha(\vec{a}\cdot \vec{b})[/tex]
     
  19. [tex]||\vec{y}||^2=\alpha^2 \vec{a} \cdot \vec{a}[/tex]?
     
  20. quasar987

    quasar987 4,774
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    Yes! phew. lol
     
    Last edited: Nov 6, 2006
  21. Now what? Do the same for [tex]\overrightarrow{y}\,\cdot\,\overrightarrow{x}[/tex]?

    [tex](\vec{x}\cdot \hat{v})^2\,\hat{v} \cdot \hat{v}[/tex]

    What qualities does the [tex]\vec{a} \cdot \vec{a}[/tex] part have?
     
    Last edited: Nov 6, 2006
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