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Thread: Linear Dependence

  1. #1
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    Linear Dependence

    Set $\displaystyle \alpha = \sqrt{2}$ and $\displaystyle \beta= \sqrt{3} $. Note that $\displaystyle \alpha^2 = 2$ and $\displaystyle \beta^2 = 3$. $\displaystyle Q(\alpha, \beta) $ is the 4 dimensional vector space of R having basis $\displaystyle \{1, \alpha, \beta, \alpha \beta \}$.

    (a) Show that real numbers $\displaystyle (\alpha, \beta)^2$ and $\displaystyle (\alpha, \beta)^4$ are both in space $\displaystyle Q(\alpha, \beta) $ by expanding the powers and expressing them as a linear combination of the basis vectors.

    (b) Show that the vectors $\displaystyle \{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent.

    My Attempt:

    (a) $\displaystyle (\alpha, \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2 = 2+ 2 \alpha \beta +3$, which can be written as:

    $\displaystyle 2.1 + 2 \alpha \beta +3 . 1$ (since $\displaystyle 1 \in Q(\alpha, \beta)$ and $\displaystyle 2,3 \in R$)

    Now, $\displaystyle (\alpha, \beta)^4 = \alpha^4 + 4 \alpha^3 \beta + 6 \alpha^2 \beta^2 + 4 \alpha \beta^3 + b^4$

    $\displaystyle (2.2).1+(4.2) \alpha \beta + (6.2.3).1+(3.4)\alpha \beta + (4).1$ $\displaystyle = (4).1+(8) \alpha \beta + (46).1+(12) \alpha \beta + (4).1$

    Is this right?

    (b) This is the HARD part! I know that the vectors $\displaystyle \{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent if there are $\displaystyle \lambda_1, \lambda_2, \lambda_3$ in reals, NOt all 0, such that

    $\displaystyle \lambda_1 (\alpha + \beta)^4), \lambda_2 (\alpha+ \beta)^2,\lambda_3 1 = 0$

    But I cant' figure out how to prove/represent this. Can anyone show me?
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  2. #2
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    Quote Originally Posted by Roam View Post
    Set $\displaystyle \alpha = \sqrt{2}$ and $\displaystyle \beta= \sqrt{3} $. Note that $\displaystyle \alpha^2 = 2$ and $\displaystyle \beta^2 = 3$. $\displaystyle Q(\alpha, \beta) $ is the 4 dimensional vector space of R having basis $\displaystyle \{1, \alpha, \beta, \alpha \beta \}$.

    (a) Show that real numbers $\displaystyle (\alpha, \beta)^2$ and $\displaystyle (\alpha, \beta)^4$ are both in space $\displaystyle Q(\alpha, \beta) $ by expanding the powers and expressing them as a linear combination of the basis vectors.

    (b) Show that the vectors $\displaystyle \{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent.

    My Attempt:

    (a) $\displaystyle (\alpha, \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2 = 2+ 2 \alpha \beta +3$, which can be written as:

    $\displaystyle 2.1 + 2 \alpha \beta +3 . 1$ (since $\displaystyle 1 \in Q(\alpha, \beta)$ and $\displaystyle 2,3 \in R$)

    Now, $\displaystyle (\alpha, \beta)^4 = \alpha^4 + 4 \alpha^3 \beta + 6 \alpha^2 \beta^2 + 4 \alpha \beta^3 + b^4$

    $\displaystyle (2.2).1+(4.2) \alpha \beta + (6.2.3).1+(3.4)\alpha \beta + (4).1$ $\displaystyle = (4).1+(8) \alpha \beta + (46).1+(12) \alpha \beta + (4).1$

    Is this right?

    (b) This is the HARD part! I know that the vectors $\displaystyle \{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent if there are $\displaystyle \lambda_1, \lambda_2, \lambda_3$ in reals, NOt all 0, such that

    $\displaystyle \lambda_1 (\alpha + \beta)^4), \lambda_2 (\alpha+ \beta)^2,\lambda_3 1 = 0$

    But I cant' figure out how to prove/represent this. Can anyone show me?
    You just said that $\displaystyle (\alpha+ \beta)^2= 5+ \alpha\beta$ and that $\displaystyle (\alpha+ \beta)^4= 20+ 20\alpha\beta$. (I didn't check to see if you were right!)

    So you need to determine if there exist $\displaystyle \lambda_1$, $\displaystyle \lambda_2$, and $\displaystyle \lambda_3$ such that $\displaystyle \lambda_1(5+ \alpha\beta)+ \lambda_2(20+ 20\alpha\beta)+ \lambda_3= (5\lambda_1+ 20\lambda_2+ \lambda_3)+ (\lambda_1+ 20\lambda_2)\alpha\beta= 0$. That gives you two equations for $\displaystyle \lambda_1$, $\displaystyle \lambda_2$, and $\displaystyle \lambda_3$.

    That gives you two equations
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    You just said that $\displaystyle (\alpha+ \beta)^2= 5+ \alpha\beta$ and that $\displaystyle (\alpha+ \beta)^4= 20+ 20\alpha\beta$. (I didn't check to see if you were right!)
    No. I didn't say that in my last post.

    So you need to determine if there exist $\displaystyle \lambda_1$, $\displaystyle \lambda_2$, and $\displaystyle \lambda_3$ such that $\displaystyle \lambda_1(5+ \alpha\beta)+ \lambda_2(20+ 20\alpha\beta)+ \lambda_3= (5\lambda_1+ 20\lambda_2+ \lambda_3)+ (\lambda_1+ 20\lambda_2)\alpha\beta= 0$. That gives you two equations for $\displaystyle \lambda_1$, $\displaystyle \lambda_2$, and $\displaystyle \lambda_3$.

    That gives you two equations
    I can't see where to get the two equations from (and those equations probably wouldn't be linear)?
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  4. #4
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    What I meant was that $\displaystyle (\alpha+ \beta)^2$ simplifies to $\displaystyle 5+2 \alpha\beta$ and also $\displaystyle (\alpha, \beta)^4 = \alpha^4 + 4 \alpha^3 \beta + 6 \alpha^2 \beta^2 + 4 \alpha \beta^3 + \beta^4$ simplifies to $\displaystyle 4+8 \alpha \beta + 36 +12 \alpha \beta +9$ $\displaystyle = 49+20 \alpha \beta$.

    Now I want to show that for $\displaystyle \lambda_1, ... \lambda_n \in R$ not all zeros,

    $\displaystyle
    \lambda_1 (5+2 \alpha\beta) + \lambda_2 (49+20 \alpha \beta) = 0
    $

    I don't know how this can be proved.
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