I am trying to solve this exericse in Axler's text.
Explain why there does not exist $\lambda \in \mathbb{C}$ such that $$\lambda (2 - 3i, 5 + 4i, -6 + 7i) = (12 - 5i, 7 + 22i, -32 - 9i).$$
This isn't particularly difficult, but my solution seems far too simple. I assume I've made a mistake somewhere.
Solution. Assume, seeking a contradiction, that there exists such a $\lambda$. Scaling the first element by $\lambda$, we have $$(2\lambda - 3\lambda i, 5\lambda + 4\lambda i, -6\lambda + 7\lambda i) = (12 - 5i, 7 + 22i, -32 - 9i).$$ Elements in $\mathbb{C}^4$ are equal if and only if their corresponding entries are equal. Focusing on the first component, we have $$2\lambda - 3\lambda i = 12 - 5i.$$ Elements in $\mathbb{C}$ are equal if and only if their real and imaginary components are equal. Hence, $$2\lambda = 12, \; -3\lambda = -5.$$ The first equality implies that $\lambda = 6$. Plugging into the second equality gives $$-18 = -5,$$ which is absurd. Hence, no such $\lambda$ exists.
This doesn't work because $\lambda$ is complex, so in $2\lambda-3\lambda i = 12- 5i$ we cannot compare real and complex parts!
Just divide: $$\lambda = \frac{12-5i}{2-3i}$$ from the first coordinates.
Can you compute $\lambda$?
Or alternatively: from the second coordinate comparison
$$\lambda= \frac{7+22i}{5+4i}$$ and we might derive a contradiction from
$$\frac{7+22i}{5+4i} = \frac{12-5i}{2-3i}$$ from cross-multiplying:
Is $$(7+22i)(2-3i) = (2-5i)(5+4i) ?$$
If not, we have a contradiction too.