0
$\begingroup$

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,

$\endgroup$
3
  • $\begingroup$ Oh in the middle, I did row-reduced echelon form to get it $\endgroup$ Apr 25, 2013 at 9:39
  • 1
    $\begingroup$ $x_1=1,x_2=-\sqrt{3}$ is not a solution for $x_1+\sqrt{3}x_2=0$ $\endgroup$ Apr 25, 2013 at 9:49
  • $\begingroup$ I just corrected it. THanks! $\endgroup$ Apr 25, 2013 at 10:07

2 Answers 2

0
$\begingroup$

$x_1+\sqrt{3}x_2=0\rightarrow x_1=-\sqrt{3}x_2\rightarrow (x_1,x_2)=(-\sqrt{3}x_2,x_2)=(-\sqrt{3},1)x_2\rightarrow \ker(T)=span(-\sqrt{3},1)$

$\endgroup$
3
  • $\begingroup$ How about this ? $$ ker(T) = span\left(\begin{array}{cc} 1\\-\frac{1}{\sqrt{3}}\\\end{array}\right) $$ $\endgroup$ Apr 25, 2013 at 10:06
  • $\begingroup$ Do I have two kernels? $\endgroup$ Apr 25, 2013 at 10:06
  • 1
    $\begingroup$ @user2310042 it is the same , the two vectors $(-\sqrt{3},1)$ and $(1,-\frac{1}{\sqrt{3}})$ differ by the constant $-\frac{1}{\sqrt{3}}$ hence generate the same subspace $\endgroup$ Apr 25, 2013 at 10:17
0
$\begingroup$

The first cast can't be correct. define $A$ the martrix of the linear transformation, and v the row vector $(1; -\sqrt{3})$. It is easy calculate that $Av^t \neq 0$ (in fact it is $(-\frac{1}{2},\frac{\sqrt{3}}{2})$.

In fact you made a mistake calculate the first solution, if $x=t$ then $\sqrt{3}y=-t$ then $y=-\frac{t}{\sqrt{3}}$

$\endgroup$
3
  • $\begingroup$ I just corrected but I still seem to get two kernels... Thanks $\endgroup$ Apr 25, 2013 at 10:05
  • $\begingroup$ you obtain two different representations of the same kernel. the first cast is span$(1,-\frac{1}{\sqrt{3}})$ (i call "v" this vector) wich is equal to the second because $-\sqrt{3}*v=(-\sqrt{3},1)$ so the two spans are the same $\endgroup$
    – Riccardo
    Apr 25, 2013 at 10:09
  • $\begingroup$ remember the definition of span(v) and you will be ok :) $\endgroup$
    – Riccardo
    Apr 25, 2013 at 10:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .