Q. I think I find the kernel but several... which is correct? Seems like depending on which variable I put as kernel, I can get several kernels. Correct?
T is the transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ whose matrix is A $$A = \left(\begin{array}{cc} \frac{1}{4}&\frac{\sqrt{3}}{4}\\\frac{\sqrt{3}}{4}&\frac{3}{4}\\\end{array}\right)$$
I need to find the kernel of T. And the following is my solution. $$\left(\begin{array}{cc} \frac{1}{4}&\frac{\sqrt{3}}{4}\\\frac{\sqrt{3}}{4}&\frac{3}{4}\\\end{array}\right)\left(\begin{array}{cc} x_{1}\\x_{2}\\\end{array}\right) = \vec{O}$$
So I have
$$\left(\begin{array}{cc} \frac{1}{4}&\frac{\sqrt{3}}{4}\\\frac{\sqrt{3}}{4}&\frac{3}{4} \end{array}\;\middle\vert\;\begin{array}{cc}\\0\\0\end{array}\right) ---rref--- \left(\begin{array}{cc} 1&\sqrt{3}\\0&0 \end{array}\;\middle\vert\;\begin{array}{cc}\\0\\0\end{array}\right)$$
Then I get the following equation in which I get really confused.
$$ x_{1} + \sqrt{3}x_{2} = 0 $$
Which variable should I put t as a free variable?
1st cast: $$ \left(\begin{array}{cc} x_{1}\\x_{2}\\\end{array}\right) = \left(\begin{array}{cc} t\\-\frac{1}{\sqrt{3}}t\\\end{array}\right) = \left(\begin{array}{cc} 1\\-\frac{1}{\sqrt{3}}\\\end{array}\right)t ---> ker(T) = span\left(\begin{array}{cc} 1\\-\frac{1}{\sqrt{3}}\\\end{array}\right) $$
2nd cast: $$ \left(\begin{array}{cc} x_{1}\\x_{2}\\\end{array}\right) = \left(\begin{array}{cc} -\sqrt{3}t\\t\\\end{array}\right) = \left(\begin{array}{cc} -\sqrt{3}\\1\\\end{array}\right)t ---> ker(T) = span\left(\begin{array}{cc} -\sqrt{3}\\1\\\end{array}\right) $$
Which is correct? I am so confused.
Help me. Thanks,
