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How will I be able to do this question?

Find a vector C $\in\mathbb{C}^3$ which is orthogonal to both A and B. Where $A = [2,0,i]^T$ and $B = [i,-1,2]^T$.

The answer is $C =c[i/2, 5/2, 1]^T$ , but how did they get that? And does $\mathbb{C}^3$ mean a continuous function and doesn't '$i$' = square root of $-1$ ?

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3 Answers 3

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Looks like $\mathbb{C}^3$ means the 3-dimensional complex vector space and (as you guessed) $i$ is the square root of $-1$.

As usual, your $a\times b$ rule also applies here ($a\times b$ is vector orthogonal to $a$ and $b$): So, a possible vector is $$\left|\begin{array}{rrr} \hat{e_1}&\hat{e_2}&\hat{e_3}\\2 &0&i\\i&-1&2\end{array}\right|$$ which is $(i,5,2)$. As we are free to multiply with any constant, the answer could be written as $c(i/2,5/2,1)$.

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Understand now, thank you very much!! –  Q.matin Sep 25 '12 at 5:17
Sorry, there are sign errors, as pointed out by @Julien S-R: the first component should have a $-$ sign. –  Tapu Sep 25 '12 at 5:24
yea i caught the mistake too when I was solving it your way. I got that the orthogonal vector to be [i,-5,-2] –  Q.matin Sep 25 '12 at 5:43
Correct,...now multiply with $-c$. As I said, you can multiply with any constant - it will not affect the orthogonality. –  Tapu Sep 25 '12 at 5:45

The set of all complex numbers $a+bi$ is denoted by $\mathbb{C}$ (or sometimes just by $\textbf{C}$). Your problem is referring to $\mathbb{C}^3$, which is the set of vectors $(z_1,z_2,z_3)$ where $z_1$, $z_2$, and $z_3$ are complex numbers.

Many things work analogously when you deal with complex vectors vs. regular vectors. If you take the cross product of two vectors in $\mathbb{R}^3$, you get back a new vector orthogonal to both of the original vectors. The same works for complex numbers, and you can check this yourself. (What tests of orthogonality do you know?)

Finally, it is important to realize that there is not just one answer to this question. In fact, there are infinitely many answers. If $v$ is in $\mathbb{C}^3$, and is perpendicular to both $A$ and $B$, then so will be $2v$, $(1+3i)v$, or any complex multiple of $v$. Hopefully you will understand this phenomenon better as you go on with your linear algebra class!

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Yup, I understand a bit more clearly now with your help. Thank you!! –  Q.matin Sep 25 '12 at 5:44

First of all, that's almost certainly not the answer. Unless I've made a mistake in my calculations, there's a sign error. As already stated, the vector resulting from the cross product between $A$ and $B$ is orthogonal to both. Another useful fact: the dot product of two orthogonal vectors is $0$. This can be used to check your answers quickly, so it's very useful when you're tackling this sort of question.

As for $\mathbb{C}^3$, this is analogous to $\mathbb{R}^3$, the 3-D space you're used to working with, except the components of your vectors are allowed to be complex numbers. For a more complete answer, read up on the cartesian product.

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Thank you very much on the great feedback!! –  Q.matin Sep 25 '12 at 5:42

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