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Math Help - Another Inner Product Question.

  1. #1
    Junior Member pearlyc's Avatar
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    Another Inner Product Question.

    Let A be a real invertible n x n matrix. Show that

    <x,y> ≡ y^T  A^T  Ax = (Ay)^T  (Ax)

    defines an inner product in ℝ^n, where x and y are column vectors in ℝ^n. What happens when A is not invertible? (Note : M^T is the transpose of matrix M, obtained by interchaging the rows and colums of M).
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  2. #2
    Junior Member pearlyc's Avatar
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    Anyone at all :\?
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi

    What has to be shown : ( x,\,y,\,\in\mathbb{R}^n, \alpha\in \mathbb{R})
    • <x,y>\,=\,<y,x> (use ^tA\cdot B=^t( ^tB\cdot A), and remember that two 1 \cdot 1 matrices (=two real numbers) are equal iff their transposes are equal)
    • <x,y+z>\,=\,<x,y>+<x,z>
    • <\alpha x,y>\,=\,<x,\alpha y>\,=\,\alpha<x,y>
    • <x,x>\,\geq \, 0 and <x,x>\,=\,0 \Leftrightarrow x=0. To show both, use <x,x>\,=\,^t(Ax)Ax and the definition of the product of matrices which involves a sum : (Ax)_{i}=\sum_{k=1}^n a_{i,k}x_k
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  4. #4
    Junior Member pearlyc's Avatar
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    Hmm, I still have no idea how to approach this :\ Can you demonstrate how to do the first property and the last property for me to see? Perhaps I could understand from there. Thanks!
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Let x,\,y\in\mathbb{R}^n.

    First one :
    • <x,y>\,=\,^ty ^tAAx\,=\, ^t(^t(Ax)Ay)
    • <y,x>\,=\,^tx ^tAAy\,=\, ^t(Ax)Ay
    But <x,y> is a real numbers (i.e. 1\cdot 1 matrix) hence ^t<x,y>\,=\,<x,y> and we get what we want.

    Last one :
    Denote Ax=(Ax_i)=\left(\begin{array}{c} Ax_1\\ \vdots \\Ax_n\end{array}\right), x=\left(\begin{array}{c} x_1\\ \vdots \\x_n\end{array}\right) and A=(a_{i,j}).
    • <x,x>\,=\,^t(Ax)Ax=\sum_{i=1}^n(Ax)_i^2\geq 0
    • If x=0, we check that <x,x>\,=\,0 because (Ax)_i=\sum_{k=1}^na_{i,k}x_k=0
    • If <x,x>\,=\,0\,=\,\sum_{i=1}^n(Ax)_i^2, hence, for all i such that 1\leq i \leq n, (Ax)_i=0. This exactly means that x is in the nullity of A. As A is invertible, x=0.
    Last edited by flyingsquirrel; May 3rd 2008 at 11:03 PM.
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  6. #6
    Junior Member pearlyc's Avatar
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    Hmm, sorry but I still don't quite understand

    Do I have to put like,

    x = [ x1 x2 and y = [ y1 y2
    x3 x4] y3 y4]

    and insert them into whatever you provided above and work from there?

    This is such a tough question
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  7. #7
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by pearlyc View Post
    Do I have to put like,

    x = [ x1 x2 and y = [ y1 y2
    x3 x4] y3 y4]

    and insert them into whatever you provided above and work from there?
    No, there is no need to add this. (at least, I don't know why it would be interesting to introduce this)
    This is such a tough question
    Tell me what you don't understand and I'll try to explain it or to make the post #5 clearer.
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  8. #8
    Junior Member pearlyc's Avatar
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    Hmm, I don't know how did we progress from the question,

    <x,y> ≡ =

    to



    Can you show me how to do the other two as well ? Properties 2 & 3.
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  9. #9
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by pearlyc View Post
    Hmm, I don't know how did we progress from the question,

    <x,y> ≡ =

    to

    There may have a problem of notation : I write ^tA the transpose of A. Anyway :

    <x,y>\,=\,^ty ^tAAx (that's the definition)

    <x,y>\,=\,^t(Ay)Ax because ^t(BC)=^tC\cdot  ^tB for any two matrices B and C.

    hence <x,y>\,=\,^t(^t(Ax)Ay) because ^t( ^t(Ax)Ay)= ^t(Ay) \cdot^t( ^t(Ax))\,=\, ^t(Ay)Ax (previous rule with B=^t(Ax) and C=Ay)

    Can you show me how to do the other two as well ? Properties 2 & 3.
    No. These ones are really basic questions that you should be able to answer by yourself. The only things you need to know is that A\cdot (B+C)= A\cdot B+A\cdot C and that (\lambda B)\cdot C=B\cdot(\lambda C)=\lambda (B\cdot C) with A,\,B,\,C\in\mathcal{M}_n(\mathbb{R}) and \lambda\in\mathbb{R}. Try to do it and, if you want, post your answer, I'll tell you if it's correct or not.
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  10. #10
    Junior Member pearlyc's Avatar
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    Okay, I have a slightly clearer view of it now considering the confusion was about the notation. Hm, to prove Property 1 only require that two steps ?

    I don't know how to explain how did it go from <x,y> to <y,x>.

    I am quite doomed right now, this is my assignment question which dues tomorrow and I am pretty much still quite blur about it
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  11. #11
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by pearlyc View Post
    I don't know how to explain how did it go from <x,y> to <y,x>.
    We got <x,y>\,=\, ^t(^t(Ax)Ay) and we also know that <x,y> is a real number so it can be seen as the one-one matrix \begin{pmatrix}<x,y>\end{pmatrix}. This gives us ^t<x,y>\,=<br />
\,^t\begin{pmatrix}<x,y>\end{pmatrix}<br />
=\begin{pmatrix}<x,y>\end{pmatrix}<br />
=\,<x,y>.
    Hence  <x,y>\,=\,^t<x,y>\,=\, ^t\left( ^t( ^t(Ax)Ay)\right)= ^t(Ax)Ay=<y,x>.

    I am quite doomed right now, this is my assignment question which dues tomorrow and I am pretty much still quite blur about it
    Does your assignment only depend on one question ?
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  12. #12
    Junior Member pearlyc's Avatar
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    Unfortunately, it does depend on this question.

    So stressed out! I still don't know how to do.
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  13. #13
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by pearlyc View Post
    Unfortunately, it does depend on this question.
    I know, you said it. I wanted to know it it *only* depends on it. (as you said "*my* assignment question")

    So stressed out! I still don't know how to do.
    What don't you know how to do ? The two remaining properties ? Show me what you've tried and I'll help you.
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  14. #14
    Junior Member pearlyc's Avatar
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    Sigh. It is quite bad shape now.

    I am working on another assignment now. Assignment overload! Sad.

    Why we don't have to put x = a matrix this time yea? Cause in my lectures, they always do that!
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  15. #15
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by pearlyc View Post
    Why we don't have to put x = a matrix this time yea? Cause in my lectures, they always do that!
    What for ? x is a vector that's to say a n\cdot 1 matrix. What would it change to say "x=a matrix" ?
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