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$v_1 = \begin{pmatrix}1\\-a\\1\end{pmatrix}$, $v_2 = \begin{pmatrix}-a\\1\\-a\end{pmatrix}$, $v_3 = \begin{pmatrix}0\\-a\\1\end{pmatrix}$

For which $a \in \mathbb{R}$ is $\{v_1, v_2, v_3\}$ a basis of $\mathbb{R^3}$?

My solution:

$\lambda_1 \cdot v_1 + \lambda_2 \cdot v_2 + \lambda_3 \cdot v_3 = 0$

Linear system:

$\lambda_1 - a\cdot \lambda_2 + \lambda_3 = 0$ (I)

$-a \cdot \lambda_1 + \lambda_2 - a\cdot \lambda_3 = 0$ (II)

$\lambda_1 - a\cdot \lambda_2 + 0 \cdot \lambda_3 = 0$ (III) $\Rightarrow a = \frac{\lambda_1}{\lambda_2}$

Put $a$ in I:

$\lambda_1 - \frac{\lambda_1}{\lambda_2} \cdot \lambda_2 + \lambda_3 = 0 \Rightarrow \lambda_3 = 0$

Put $a$ and $\lambda_3$ in II:

$-\frac{\lambda_1}{\lambda_2} \cdot \lambda_1 + \lambda_2 - \frac{\lambda_1}{\lambda_2} \cdot 0 = 0$

$-\frac{\lambda_1^2}{\lambda_2} + \lambda_2 = 0 \Rightarrow \lambda_1 = \lambda_2$

Here I don't know how to show that $\lambda_1 = \lambda_2 = 0$

Could you please help?

Thanks

John
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  • If you're trying to show that $\lambda_1 = \lambda_2 = \lambda_3 = 0$, then dividing by $\lambda_2$ in the middle of your proof probably isn't a good idea. – got it--thanks Jan 08 '18 at 04:16

2 Answers2

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$$x-ay=0 \\ -ax+y-az=0 \\ x-ay+z=0$$

We want to choose $a$ so that $x$, $y$, and $z$ have to be $0$. First we see from the first equation that $x = ay$. So let's replace $x$ in the other two equations to turn this from a $3\times 3$ system into a $2\times 2$ system:

$$-a(ay)+y-az = (1-a^2)y-az=0 \\ (ay)-ay+z=z=0$$ That last equation tells us that regardless of the value of $a$, $z$ must equal $0$. So let's plug that into the other equation to get $$(1-a^2)y=0$$ At this point we see that either $1-a^2 = 0$ or $y=0$. Since we want $y$ to be equal to $0$, we need to choose $a$ so that $1-a^2\ne 0$. So any value of $a$ other than $\pm 1$ will do. \

Then, just to be sure, let's see what values $x$ can take when $a\ne \pm 1$, $y=0$, and $z=0$: $$x=ay = a\cdot 0 = 0$$ So if $a\ne \pm 1$ then $x=y=z=0$ as we wanted and hence $v_1$, $v_2$, $v_3$ are linearly independent. If you can then show that they span $\Bbb R^3$ then they are indeed a basis.

got it--thanks
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  • Thanks for the quick help.

    At first I had also tried $(1-a^2)y=0$ but stupidly I divided the equation by y, so I couldn't get the solution. Now it makes sense.

    Since there are 3 linear independent vectors from $\mathbb{R}^3$ it's actually clear, that they span $\mathbb{R}^3$ and thus they are a basis of $\mathbb{R}^3$. Am I right?

     – John Jan 08 '18 at 04:39
  • If you've already proven in class that any $n$ linearly independent vectors in an $n$-dimensional space is a basis, then yes -- just make a note of that fact at the end of your work and you're done. – got it--thanks Jan 08 '18 at 04:42
  • Yes we did. Thanks for the notice and confirmation. – John Jan 08 '18 at 04:44
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To show $v_1,v_2,v_3$ are independent we can take them as columns of a matrix and check that whether the resulting matrix is invertible. The resulting matrix is $$\left[\begin{matrix}1&-a&0\\-a&1&-a\\1&-a&1\end{matrix}\right]$$ The determinant of this matrix is $1\cdot(1-a^2)-(-a)\cdot(-a-(-a))=1-a^2\neq0$ for all $a\in\Bbb{R}\setminus\{-1,1\}$. Hence the matrix is invertible, i.e. it has full column rank, and hence $v_1,v_2,v_3$ are independent, if and only if $a\neq1,-1$.

QED
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