Prove some facts about a given set of vectors

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 $\mathbb{R}^3$ .
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 $\mathbb{R}^3$ .
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 $\mathbb{R}^3$ .

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 $. \qquad \blacksquare$
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 $. \qquad \blacksquare$
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 $t = 0, \ \pm \sqrt{2}$ .

One comment

August 27, 2023 at 2:18 am

Anonymous says:

This is trivial (pointing it out just in case it could help anyone when reading the answer of the exercise) :
I think in c. it should be written:
For (t,1,0), (1,t,1), (0,1,t) to be linearly dependent we must have …

Stumbling Robot

Prove some facts about a given set of vectors

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