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The question is given below:

enter image description here

My first guess is:

1-for $t = (2n + 1)\pi$, where $n \in \mathbb{Z}$. we always have $\cos(t) = -1$ and $\sin(t) = 0$ so we will get the vectors $(-1, 1)^{t}$ and $(1, -1)^{t}$ which are linearly dependent.

2- for $ t = 2n\pi$, where $n \in \mathbb{Z}$. the two vectors will become the same vector $(1, 1)^t$ and hence are linearly dependent.

But what about the rationals and irrationals and the other values of $t$ in the integers ?

So I started to row reduce the following matrix:

$$\begin{bmatrix} \cos(t) + i\sin(t)&1\\ 1&\cos(t) - i\sin(t)\\ \\ \end{bmatrix}$$

But I am stucked now, could anyone help me please?

Thanks!

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1 Answer 1

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By invertible matrix theorem, the vectors are linearly dependent iff the determinant of the matrix you gave is 0. That is, $$(\cos(t)+i\sin(t))(\cos(t)-i\sin(t))-(1)(1)=0$$ or equivalently $$\cos^2(t)+\sin^2(t)=1$$ It is known that this identity holds for all $t\in\mathbb{R}$.

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  • $\begingroup$ so you are saying is that this subset is linearly independent for all $t \in \mathbb{R}$ correct? .... what about if I want to row reduce the given matrix? $\endgroup$
    – user591668
    Sep 11, 2019 at 18:59
  • $\begingroup$ I'm saying they're linearly DEPendent. To row reduce the given matrix, it might be easier to let $a=\cos(t)+i\sin(t)$ and $b=\cos(t)-i\sin(t)$ and remember that $ab=1$. $\endgroup$
    – 79037662
    Sep 11, 2019 at 19:08
  • $\begingroup$ but the matrix I want to reduce does not contain $a$ and $b$ $\endgroup$
    – user591668
    Sep 11, 2019 at 19:17
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    $\begingroup$ @Smart $a$ and $b$ are abbreviations in order to get the notation simple. Your matrix is $$ \begin{pmatrix} a & 1 \\ 1 & b \end{pmatrix},$$ now row reduce it. $\endgroup$
    – Jan
    Sep 11, 2019 at 20:20
  • $\begingroup$ but what about the case if $t = \pi/2$ ... we will get the two vectors (i , 1)^t and (1, -i)^t ... where t in the power of the vectors means transpose .... those 2 vectors are linearly independent because there is no values that you can multiply with in the reals to make them linearly dependent ..... so $t =\pi/2 $ does not work $\endgroup$
    – user591668
    Sep 11, 2019 at 21:07

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