Anki cards split by module for the first year of maths at Cambridge
| Front | Theorem. Consider a linear map \(f : U \to V\), where \(U, V\) are vector spaces. Then im(f) is a subspace of V, and ker(f) is a subspace of U |
| Back | Both are non empty since \(f(\bf{0}) = 0\).If \(\mathbf{x,y} \in \text{im }(f), \text{ then } \exists \mathbf{a, b} \in U\) such that \(\mathbf{x} = f(\mathbf{a}), \mathbf{y} = f(\mathbf{b})\). Then \(\lambda x + \mu y = \lambda f(\mathbf{a}) + \mu f(\mathbf{b}) = f(\lambda \mathbf{a} + \mu \mathbf{b})\). Now \(\lambda \mathbf{a} = \mu \mathbf{b} \ in U\) since U is a vector space, so there is an element in U that maps to \(\lambda x + \mu y\), so it is in the image and the image is a subspace of V.Similarly for the kernel except \(f(\lambda \mathbf{x} + \mu \mathbf{y}) = \lambda f(\mathbf{x}) + \mu f(\mathbf{y}) = \lambda \mathbf{0} + \mu \mathbf{0} = \mathbf{0}\). So it is in the kernel and it is a subspace. |
| Front | Theorem. Triangle inequality. \(|z_1 + z_2| \leq |z_1|+|z_2|, ||z_1|-|z_2|| \leq |z_1 - z_2|\) |
| Back | For first inequality use a diagram, shortest path is always a straight line.For second inequality \(z_1 = z_1' - z_2', z_2 = z_2'\), so the first inequality becomes \(|z_1'| \leq |z_1' - z_2'| + |z_2'| \implies |z_1' - z_2'| \geq |z_1'| - |z_2'|\), since order only matters on the RHS we can swap 1 and 2 take the modulus of it and the inequality still holds. |
| Front | Definition. \(\mathbb{C}^n\). It has the same standard basis as \(\mathbb{R}^n\) but the scalar product is defined as \(\bf u \cdot v = \sum u_i^* v_i\).It has the following properties. |
| Back | i) \(\bf u \cdot v = (v \cdot u)^*\)ii) \(\bf u \cdot(\lambda v + \mu w) = \lambda(u \cdot v) + \mu(u \cdot w)\)iii) \(\bf u \cdot u \geq 0\) and \(u \cdot u = 0 \iff u = 0\)iv) \((\lambda \bf u + \mu v) \cdot w = \lambda^* u \cdot w + \mu^* v \cdot w\) |