# For which values of $t$ is the subset of $\mathbb{C^2}$ linearly dependent?

The question is given below: 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!

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 $$t\in\mathbb{R}$$.
• 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? – Smart Sep 11 at 18:59
• 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 $ab=1$. – 79037662 Sep 11 at 19:08
• but the matrix I want to reduce does not contain $a$ and $b$ – Smart Sep 11 at 19:17
• @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. – Jan Sep 11 at 20:20
• 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 – Smart Sep 11 at 21:07