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Prove some facts about a given set of vectors

  1. Prove that the vectors

        \[ \left( \sqrt{3}, 1,0 \right), \quad \left( 1, \sqrt{3}, 1 \right), \quad \left( 0,1, \sqrt{3} \right) \]

    are linearly independent in \mathbb{R}^3.

  2. Prove that the vectors

        \[ \left( \sqrt{2}, 1, 0 \right), \quad \left( 1, \sqrt{2}, 1 \right), \quad \left( 0,1, \sqrt{2} \right) \]

    are linearly dependent in \mathbb{R}^3.

  3. Find all t \in \mathbb{R} such that the vectors

        \[ (t,1,0), \quad (1,t,1), \quad (0,1,t) \]

    are linearly dependent in \mathbb{R}^3.


  1. Proof. We consider the equation

        \begin{align*}  x \left( \sqrt{3}, 1, 0 \right) + y \left( 1, \sqrt{3}, 1 \right) + z \left( 0, 1, \sqrt{3} \right) = O && \implies && \sqrt{3} x + y &= 0 \\  &&&& x + \sqrt{3} y + z &= 0 \\  &&&& y + \sqrt{3} z &= 0. \end{align*}

    From the first equation we have x = -\frac{1}{\sqrt{3}} y and from the third we have z = -\frac{1}{\sqrt{3}} y. Plugging these into the second we have -\frac{2}{\sqrt{3}} y + \sqrt{3} y = 0 which implies y = 0. Hence, x = y = z = 0, so the vectors are independent. \qquad \blacksquare

  2. Proof. Consider the equation

        \begin{align*}  x \left( \sqrt{2}, 1, 0 \right) + y \left( 1, \sqrt{2}, 1 \right) + z \left( 0,1 \sqrt{2} \right) = O && \implies && \sqrt{2} x + y &= 0 \\ &&&& x + \sqrt{2} y + z &= 0 \\  &&&& y + \sqrt{2} z &= 0. \end{align*}

    The first and third equations implies x = z = -\frac{1}{\sqrt{2}} y. By the second equation we then have

        \[ -\frac{2}{\sqrt{2}} y + \sqrt{2} y = 0 \]

    which implies y is arbitrary (since this will hold for all choices of y). Letting y = 1, we have x = z = -\frac{1}{\sqrt{2}} and we have the equation

        \[ -\frac{1}{\sqrt{2}} (\sqrt{2},1,0) + (1, \sqrt{2}, 1) - \frac{1}{\sqrt{2}} (0,1, \sqrt{2}) = (0,0,0). \]

    Hence, these vectors are linearly dependent. \qquad \blacksquare

  3. For (t,1,0), (1,t,1), (0,1,t) to be independent we must have

        \[ x (t,1,0) + y(1,t,1) + z(0,1,t) = O \]

    for some nontrivial choice of x,y,z. This gives us the equations

        \[ xt + y = 0, \quad x+yt + z = 0, \quad y + zt = 0. \]

    If t = 0 then y = 0 and x = -z can be any value, so the vectors are dependent.
    Assume t \neq 0. From the first equation we have x = -\frac{1}{t} y, and from the third we have z = -\frac{1}{t} y. Therefore, the second equation becomes

        \[ yt - \frac{2}{t} y = 0 \quad \implies \quad yt^2 - 2y = 0. \]

    Thus, we have t = \pm \sqrt{2}.
    Therefore, the vectors are dependent if t = 0, \ \pm \sqrt{2}.

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