Let S = {(0,1,k) , (k,1,0), (1,k,1)} Then, S is a basis for $\mathbb R^3$ for what value of k?

Question

I know that a basis generates the vector space. So I took the linear combination of these $3$ such that $a_1 s_1 + a_2 s_2 + a_3 s_3 = 0$. I got $a_1 = a_2$ but I still wasn't able to figure out the value of $k$. I am really lost here. Thanks in advance for the help!

Answer 1

Sketch:

For the vectors to be linearly independent, you need the only solution to $$ a_1\begin{bmatrix}0\\1\\k\end{bmatrix}+a_2\begin{bmatrix}k\\1\\0\end{bmatrix}+a_3\begin{bmatrix}1\\k\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $$ to be $a_1=a_2=a_3=0$.

By inspection, we see that in order for this equation to be true, $ka_2+a_3=0$ by the first row, $ka_1+a_3=0$ by the third row, and $a_1+a_2+ka_3=0$ by the middle row. Therefore, $a_1=a_2=-\frac{1}{k}a_3$. Plugging this into the second equation gives $-\frac{2}{k}a_3+ka_3=0$ or that $\frac{k^2-2}{k}a_3=0$. Since we need to guarantee that $a_3=0$, the coefficient of $a_3$ should not be zero, so $k\not=\pm\sqrt{2}$. We should also be careful of $k=0$ since that shows up in the denominator.

Therefore, for $k\not=\pm\sqrt{2},0$, we know that $a_3=0$, so $a_1$ and $a_2=0$ as well. Then, you need to check the three remaining cases.

Answer 2

We have: $$a_1\left[\matrix{0\\1\\k}\right] + a_2\left[\matrix{k\\1\\0}\right] + a_3\left[\matrix{1\\k\\1}\right]=\left[\matrix{0\\0\\0}\right]$$

$$\implies\left[\matrix{ a_2k+a_3 \\ a_1+a_2+a_3k \\ a_1k+a_3 }\right]=\left[\matrix{0\\0\\0}\right]$$

$$\implies\left[\matrix{ 0 & k & 1 \\ 1 & 1 & k \\ k & 0 & 1 }\right]\left[\matrix{a_1\\a_2\\a_3}\right]=\left[\matrix{0\\0\\0}\right]$$

This is a homogeneous system, so let's row-reduce the coefficient matrix:

$$\left[\matrix{ 0 & k & 1 \\ 1 & 1 & k \\ k & 0 & 1 }\right] \sim \left[\matrix{ 1 & 1 & k \\ 0 & k & 1 \\ k & 0 & 1 }\right] \sim \left[\matrix{ 1 & 1 & k \\ 0 & k & 1 \\ 0 & -k & 1-k^2 }\right] \sim \left[\matrix{ 1 & 1 & k \\ 0 & k & 1 \\ 0 & 0 & 2-k^2 }\right]$$

That's enough work. It's clear from this form that, if $k=0$, we have no pivot in column $2$, and if $k=\pm\sqrt2$, then we have no pivot in column $3$. In either case, we fail to have linear independence, but any other value of $k$ should give us a basis.