3
$\begingroup$

I am trying to understand how to determine the linear dependence/independence of a set of complex vectors. I know the process is the same as determining linear dependence/independence of a set of real vectors, but I am a little confused on how to augment the sets of vectors. For instance if I have $$ \begin{bmatrix} 2 + i \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 - i \end{bmatrix} $$ I know when expanded out into the real and complex parts I have $$ \begin{bmatrix} 2 \\ 1 \end{bmatrix} + i\begin{bmatrix} 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 1 \end{bmatrix} + i\begin{bmatrix} 0 \\ -1 \end{bmatrix} $$ For the augmented matrix do I just use the original set of vectors like the following? $$ \left[\begin{array}{cc|c} 2 + i & 1 & 0\\ 1 & 1 - i &0\end{array}\right] $$ or do I some how have to use the expanded set of vectors?

Any help would be greatly appreciated.

$\endgroup$
3
  • $\begingroup$ Your augmented matrix looks fine. $\endgroup$
    – user137731
    Mar 18, 2015 at 4:06
  • $\begingroup$ Just like for real vectors, one can check the linear independence of vectors ${\bf x}_1, \ldots, {\bf x}_n \in \mathbb{C}^n$ by checking whether the determinant $$\det \begin{pmatrix} {\bf x}_1 & \cdots & {\bf x}_n\end{pmatrix}$$ is nonzero. $\endgroup$
    – Travis Willse
    Mar 18, 2015 at 4:07
  • $\begingroup$ Oh, I forgot all about the determinant. I always found linear independence by putting the matrix into RREF to see if there were any free variables. If I were to go about doing it my way how would I perform row operations on the augmented matrix? In both columns only one value has a real and imaginary part so if I tried to cancel out the real part I am still left with an imaginary part as opposed to 0. $\endgroup$
    – steveclark
    Mar 18, 2015 at 4:12

2 Answers 2

Reset to default
2
$\begingroup$

Check the determinant just like you do for real numbers. If an n x n determinant A such that det(A) = 0, it's rank r < n. This implies that at least one row or column is a linear combination of the rest.

$\endgroup$
0
$\begingroup$

$$\left|\begin{array}\ 2 + i & 1 \\ 1 & 1 - i\end{array}\right| = -\left|\begin{array}\ 1 & 1 - i \\ 2 + i & 1\end{array}\right| = -\left|\begin{array}\ 1 & 1 - i \\ 0 & -2+i\end{array}\right| = 2-i\ne 0$$

Thus your two vectors are linearly independent.

Alternatively (because you asked about Gaussian elimination):

$$\left[\begin{array}{cc|c} 2 + i & 1\\ 1 & 1 - i\end{array}\right]_{R_1 \leftrightarrow R_2} \sim \left[\begin{array}{cc|c} 1 & 1 - i \\ 2 + i & 1\end{array}\right]_{R_2 \rightarrow R_2-(2+i)R_1} \sim \left[\begin{array}{cc|c} 1 & 1 - i \\ 0 & -2+i\end{array}\right]_{\begin{matrix}R_2 \rightarrow R_2/(-2+i) \\ R_1 \rightarrow R_1-(1-i)R_2\end{matrix}} \sim \left[\begin{array}{cc|c} 1 & 0 \\ 0 & 1\end{array}\right]$$

Thus your two vectors are linearly independent.

$\endgroup$
5
  • $\begingroup$ I don't understand how you went from $$ \begin{bmatrix} 1 & 1 - i \\ 2 + i & 1 \end{bmatrix} $$ to $$ \begin{bmatrix} 1 & 1 - i \\ 0 & -2 + i \end{bmatrix} $$ I don't think we learned about those kinds of transformations. $\endgroup$
    – steveclark
    Mar 18, 2015 at 4:27
  • $\begingroup$ I've added in the steps I used. $\endgroup$
    – user137731
    Mar 18, 2015 at 4:28
  • $\begingroup$ Ahh, I understand now. Would you recommend using this approach? Or using the determinant like everyone else is suggesting? $\endgroup$
    – steveclark
    Mar 18, 2015 at 4:32
  • 1
    $\begingroup$ For small matrices, I don't think it really matters. For large matrices, use a computer. $\endgroup$
    – user137731
    Mar 18, 2015 at 4:32
  • $\begingroup$ Sorry for the necropost: Is there a way of extending the idea of the determinant as the volume of a parallelepiped, with maybe the imaginary part used as orientation or something? $\endgroup$
    – MSIS
    Jan 9, 2022 at 20:32

You must log in to answer this question.

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