Given the vector space $\mathbb{C}^3$ and three vectors $x_1 = (1, 0, 0)$, $x_2 = (0, 1, 0)$, $x_3 = (0, 0, 1)$.

Is it correct that these three vectors are linearly independent if $\mathbb{C}^3$ is defined over the field $\mathbb{R}$, while they are linearly dependent if the field is $\mathbb{C}$?

I'm using the following definiton for linear dependence (Halmos, Finite-Dimensional Vector Spaces, 2e):

A finite set of vectors $\{x_i\}$ is linearly dependent if there exists a corresponding set $\{\alpha_i\}$ of scalars, not all zero, such that

$$ \tag{1}\label{eqn_li}\sum_i \alpha_i x_i = 0, $$

and the reason I'm asking is that $\eqref{eqn_li}$ is satisfied for the scalars $\alpha_1 = i$ , $\alpha_2 = 0$, $\alpha_3 = 0$, hence there is a set of scalars $\{\alpha_i\}$, not all zeros, such that $\eqref{eqn_li}$ holds.

  • $\begingroup$ "...for the scalars..." . What scalars?? $\endgroup$ – DonAntonio Jan 12 at 17:44
  • 5
    $\begingroup$ The given set of scalars doesn't satisfy (1). $\endgroup$ – Thomas Shelby Jan 12 at 17:45
  • 1
    $\begingroup$ Hint: $\sum\alpha_jx_j=(\alpha_1,\alpha_2,\alpha_3)$. $\endgroup$ – David C. Ullrich Jan 12 at 18:03

Those three vectors are linearly independent both over $\mathbb C$ and over $\mathbb R$.

However, $(1,0,0)$ and $(i,0,0)$ are linearly dependent over $\mathbb C$ and linearly independent over $\mathbb R$.


Your assumption that $ix_1+0x_2+0x_3=0$ is wrong. Your three vectors are linearly independent, no matter if we view $\Bbb C^3$ as three-dimensional space over $\Bbb C$, or six-dimensional space over $\Bbb R$ (or e.g., infinite-dimensional space over $\Bbb Q$) in the apparent way.

However, $(1,0,0)$ and $(i,0,0)$ are $\Bbb R$-linearly independant and $\Bbb C$-linearly dependant.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.