Prove That...¶
C'mon, prove these linear algebra things!
For the content below, I use \(W\leqslant V\) to denote \(W\) is a subspace of \(V\), and \(\exists!\) to denote "there exists a unique".
Group, Ring & Field¶
- \(\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}\) is a field.
- \(\mathbb{Q}\) is the smallest number field (i.e. subfield of \(\mathbb{C}\)).
Linear Space¶
- For \(W\subset V(\mathbf{F})\), \(W\leqslant V\iff W\) is closed under addition and scalar multiplication.
- \(S_1\subset S_2\subset V(\mathbf{F})\Rightarrow\operatorname{span}S_1\subset\operatorname{span}S_2\).
- If \(S_1\) and \(S_2\) are linearly independent, then \(\operatorname{span}S_1=\operatorname{span}S_2\Rightarrow|S_1|=|S_2|\).
- Functions \(\exp(\lambda_1x)\), \(\exp(\lambda_2x)\) and \(\exp(\lambda_3x)\) are linearly independent. (\(\lambda_1\), \(\lambda_2\) and \(\lambda_3\) are distinct)
- For \(W_1, W_2\leqslant V(\mathbf{F})\):
- \(W_1\cup W_2\leqslant V\iff W_1\subset W_2\) or \(W_2\subset W_1\).
- \(W_1 + W_2 = \operatorname{span}(W_1\cup W_2)\).
- \(\dim(W_1 + W_2) = \dim W_1 + \dim W_2 - \dim(W_1\cap W_2)\).
- For \(W_1, W_2\leqslant V(\mathbf{F})\), the following propositions are equivalent:
- \(W_1 \cap W_2 = \{0\}\).
- \(\forall\alpha\in W_1 + W_2\), \(\exists!\alpha_1\in W_1\) and \(\exists!\alpha_2\in W_2\) such that \(\alpha = \alpha_1 + \alpha_2\).
- If \(0 = \alpha_1 + \alpha_2 (\alpha_1 \in W_1, \alpha_2 \in W_2)\), then \(\alpha_1 = \alpha_2 = 0\).
- \(\dim(W_1 + W_2) = \dim W_1 + \dim W_2\).
Inner Product Space¶
- (Cauchy–Schwarz inequality) \(|\langle\alpha,\beta\rangle|\leq \|\alpha\|\cdot\|\beta\|\).
- (Triangle inequality) \(\|\alpha\|+\|\beta\| \geq \|\alpha+\beta\|\).
- (Pythagorean theorem) \(\|\alpha\|^2+\|\beta\|^2 = \|\alpha+\beta\|^2\iff\alpha\perp\beta\iff\angle(\alpha,\beta)=\dfrac{\pi}{2}\).
- (Gram–Schmidt process) Any Euclidean space has an orthonormal basis. (The method to construct it is called Gram–Schmidt process.)
- If \(B = \{\varepsilon_1, \varepsilon_2, \cdots, \varepsilon_n\}\) is an orthonormal basis of \(V(\mathbf{F})\), then \(\forall\alpha\in V\), \(\alpha = \sum\limits_{i=1}^n\langle\alpha,\varepsilon_i\rangle\cdot\varepsilon_i\).
Linear Transformation¶
- \(\sigma:V\to W\) is injective \(\iff\ker\sigma = \{0\}\).
- \(\sigma:V\to W\) is surjective \(\iff\operatorname{im}\sigma = W\).
- If \(B=\{\alpha_1,\alpha_2,\dots,\alpha_n\}\) is a basis of \(V\), then \(\forall \beta_1, \beta_2, \dots, \beta_n\in W\), \(\exists !\sigma:V \to W\) such that \(\sigma(\alpha_i) = \beta_i\).
- For \(\sigma: V \to W\), \(\operatorname{rank}\sigma + \dim\ker\sigma = \dim V\).
- For \(\sigma: V \to W\), if \(\dim V = \dim W = n\), the following propositions are equivalent:
- \(\sigma\) is injective.
- \(\sigma\) is surjective.
- \(\operatorname{rank}\sigma = n\).
- \(V\cong W \iff \dim V = \dim W\).
Matrix¶
- \(\mathbf{M}(\sigma)\) and \(\sigma\) are one-to-one.
- Prove with matrix: if \(\dim V(\mathbf{F})=m\), \(\dim W(\mathbf{F})=n\), then \(\mathcal{L}(V, W)\cong \mathbf{F}^{m\times n}\).
- Prove with matrix: \(\dim\mathcal{L}(V, W)=\dim V\cdot\dim W\).
Last update:
2023-11-13
Created: 2023-11-07
Created: 2023-11-07