Complex Numbers and Linear Algebra

Question

Explain why there does not exist a $\lambda $ in the Complex Field such that

$$\lambda \left(2-3i, 5+4i, -6+7i \right) = \left(12-5i, 7+22i, -32-9i \right)$$

Can someone help me figure out how to go about this problem?

Answer 1

Let $\lambda = a + bi$ then $$\lambda(2-3i,5+4i,-6+7i) = ((a+bi)(2-3i), (a+bi)(5+4i), (a+bi)(-6+7i)) = (12-5i,7+22i,-32-9i)$$ thus looking at the first coordinate and performing multiplication we see $$(2a+3b)+(2b-3a)i = 12-5i$$ and this is a system of 2 equations with 2 unknowns $a$ and $b$. Similarly the other coordinates give 2 more equations in the same 2 unknowns $a$ and $b$. This will be an over determined system and hence no such $\lambda$ exists.

Answer 2

If your tuples are ordered, then you just have to divide element-wise and check whether the ratio is the same for all three coordinates. If it is then that ratio is your $λ$. If it is not, then there cannot be a $λ$ because $\mathbb{C}$ is a field.

Answer 3

Consider that, geometrically, the product of two complex numbers multiplies the respective lengths and adds the respective arguments. Then $\lambda$ should rotate each of the three complex numbers $(2-3i,5+4i, -6+7i)$ by the same amount, to get to each of $(12-5i, 7+22i, -32-9i)$ respectfully. Is this possible?