#math/linear-algebra We aim to derive fundamental requirements of the determinant rooted in our interpretations in two and three dimensions. We begin first with how it operates. As we saw in [[3A1 - Determinant Preliminaries]], dimensions two and three are determined by our base-times-height rule. We can generalize this in $n$ dimensions as follows: pick one vector, the height is the distance from this vector to the subspace spanned by the remaining, base is the $(n-1)$-dimensional volume. ```tikz \begin{document} \begin{tikzpicture}[>=stealth] \begin{scope}[x={(0.95cm,-0.30cm)},y={(0.55cm,0.40cm)},z={(0cm,0.90cm)}] \coordinate (O) at (0,0,0); \coordinate (A) at (3,0,0); \coordinate (B) at (1,2,0); \coordinate (C) at (4,2,0); \coordinate (P) at (1.2,0.6,0); \coordinate (V) at (1.2,0.6,2.5); % The base plane extends beyond the shaded base parallelogram. \fill[blue!4] (-0.7,-0.4,0) -- (4.1,-0.4,0) -- (5.7,2.8,0) -- (0.9,2.8,0) -- cycle; \draw[blue!25] (-0.7,-0.4,0) -- (4.1,-0.4,0) -- (5.7,2.8,0) -- (0.9,2.8,0) -- cycle; \fill[blue!16] (O) -- (A) -- (C) -- (B) -- cycle; \draw[blue!65!black] (A) -- (C) -- (B); % Translate the base by the slanted vector to show the volume. \fill[blue!5] (V) -- (4.2,0.6,2.5) -- (5.2,2.6,2.5) -- (2.2,2.6,2.5) -- cycle; \draw[gray!65] (V) -- (4.2,0.6,2.5) -- (5.2,2.6,2.5) -- (2.2,2.6,2.5) -- cycle; \draw[gray!65] (A) -- (4.2,0.6,2.5); \draw[gray!65] (C) -- (5.2,2.6,2.5); \draw[dashed,gray!65] (B) -- (2.2,2.6,2.5); \draw[->,very thick,blue!70!black] (O) -- (A) node[below] {$v_1$}; \draw[->,very thick,red!75!black] (O) -- (B) node[above left] {$v_2$}; \draw[->,very thick,teal!70!black] (O) -- (V) node[above left] {$v_3$}; \draw[->,dashed,gray,thick] (O) -- (P); \draw[<->,thick,orange!85!black] (P) -- (V) node[pos=0.65,right] {$h$}; % A right-angle marker: an in-plane direction meets the normal. \draw[orange!85!black] (P) -- (1.44,0.78,0) -- (1.44,0.78,0.2) -- (1.2,0.6,0.2); \fill[orange!85!black] (P) circle (1.5pt); \node[below,yshift=-15pt] at (P) {$p=P_E(v_3)$}; \node[below left] at (O) {$0$}; \node at (2.1,0.9,0) {$A_{\mathrm{base}}$}; \node[below] at (5.95,2.50,0) {$E=\mathrm{span}(v_1,v_2)$}; \end{scope} \end{tikzpicture} \end{document} ``` In the figure above we have $h=\text{dist}(v_{3},E)=\lVert v_{3}-P_{E}v_{3}\rVert$. From which we get $V_3=A_{\mathrm{base}}\,h$ as desired. In $n$ dimensions, $E=\mathrm{span}(v_1,\ldots,v_{n-1})$ and $V_n=V_{n-1}(\mathrm{base})\,\mathrm{dist}(v_n,E)$. ## I. Linearity ### Scaling a Vector when we multiple a vector by a positive number $\alpha$, then the height is multiplied by $\alpha$^[Take $x^{2}+y^{2}=z^{2}$ and multiply both sides by $\alpha^{2}$ and we get $(\alpha x)^{2}+(\alpha y)^{2}=(\alpha z)^{2}.$ Hence all sides, including height, grow by the same factor.]. Formalized we get $ D(\mathbf{v}_1,\ldots,\underset{k}{\alpha\mathbf{v}_k},\ldots,\mathbf{v}_n)=\alpha D(\mathbf{v}_1,\ldots,\underset{k}{\mathbf{v}_k},\ldots,\mathbf{v}_n)\tag{1.1} $ ### Adding Vectors In the first figure imagine if we added to $v_{3}$ either $v_{1}$ or $v_{2}$—the height doesn't change! This makes sense since the directions are perpendicular to the height. Formalized we get $ D(\mathbf{v}_1,\ldots,\underbrace{\mathbf{u}_k+\mathbf{v}_k}_{k},\ldots,\mathbf{v}_n)=D(\mathbf{v}_1,\ldots,\underset{k}{\mathbf{u}_k},\ldots,\mathbf{v}_n)+D(\mathbf{v}_1,\ldots,\underset{k}{\mathbf{v}_k},\ldots,\mathbf{v}_n)\tag{1.2} $ ### Remark Linearity is a great property for a function to have. However, admitting negative scalars, negative heights, and thus negative volumes is a small consequence. In fact, we will see this is not consequence, but a blessing in disguise! ## II. Zero ### Two Vectors Coincide If two of the vectors from our system equal one another. Then the volume should be zero! The makes sense since the height between the equating vectors is zero, which makes the volume zero. So we get $ D(\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n)=0\qquad\text{if }\mathbf{v}_j=\mathbf{v}_k\quad\text{for some }j\ne k. \tag{1.3} $ ### Linear Dependence In fact, if the system is at all linearly dependent, the volume should be zero too. Suppose $v_{n}=a_{1}v_{1}+\ldots+a_{n-1}v_{n-1}$ and observe the following $ \begin{aligned} D(\mathbf{v}_1,\ldots,\mathbf{v}_n) &=D\left(\mathbf{v}_1,\ldots,\mathbf{v}_{n-1},\sum_{j=1}^{n-1}a_j\mathbf{v}_j\right)\\ &=\sum_{j=1}^{n-1}a_jD(\mathbf{v}_1,\ldots,\mathbf{v}_{n-1},\mathbf{v}_j) &&\text{(1.1), (1.2)}\\ &=0 &&\text{(1.3)} \end{aligned} $ ## III. Column Replacement Recall that the column operation of the third type is adding a scalar multiple of one row to another. This is the combination of rules $(1.1)$, $(1.2)$, (1.3)! So column operations do not change the determinant. $ D(\mathbf{v}_1,\ldots,\underbrace{\mathbf{v}_j+\alpha\mathbf{v}_k}_{j},\ldots,\underset{k}{\mathbf{v}_k},\ldots,\mathbf{v}_n)=D(\mathbf{v}_1,\ldots,\underset{j}{\mathbf{v}_j},\ldots,\underset{k}{\mathbf{v}_k},\ldots,\mathbf{v}_n).\tag{1.4} $ ## IV. Antisymmetry When two vectors are interchanged, the determinant sign should change. Recall when we introduced linearity that this was an unexpected, (and for now unintuitive), positive consequence! $ \begin{aligned} &D(\mathbf{v}_1,\ldots,\underset{j}{\mathbf{v}_j},\ldots,\underset{k}{\mathbf{v}_k},\ldots,\mathbf{v}_n)\\ &\quad=D(\mathbf{v}_1,\ldots,\underset{j}{\mathbf{v}_j},\ldots,\underbrace{\mathbf{v}_k-\mathbf{v}_j}_{k},\ldots,\mathbf{v}_n) &&\text{(1.4)}\\ &\quad=D(\mathbf{v}_1,\ldots,\underbrace{\mathbf{v}_j+(\mathbf{v}_k-\mathbf{v}_j)}_{j},\ldots,\underbrace{\mathbf{v}_k-\mathbf{v}_j}_{k},\ldots,\mathbf{v}_n) &&\text{(1.4)}\\ &\quad=D(\mathbf{v}_1,\ldots,\underset{j}{\mathbf{v}_k},\ldots,\underbrace{\mathbf{v}_k-\mathbf{v}_j}_{k},\ldots,\mathbf{v}_n) &&\text{}\\ &\quad=D(\mathbf{v}_1,\ldots,\underset{j}{\mathbf{v}_k},\ldots,\underbrace{(\mathbf{v}_k-\mathbf{v}_j)-\mathbf{v}_k}_{k},\ldots,\mathbf{v}_n) &&\text{(1.4)}\\ &\quad=D(\mathbf{v}_1,\ldots,\underset{j}{\mathbf{v}_k},\ldots,\underset{k}{-\mathbf{v}_j},\ldots,\mathbf{v}_n) &&\text{}\\ &\quad=-D(\mathbf{v}_1,\ldots,\underset{j}{\mathbf{v}_k},\ldots,\underset{k}{\mathbf{v}_j},\ldots,\mathbf{v}_n). &&\text{(1.1)} \end{aligned} $ ## V. Normalization For a standard basis $e_{1}, \ldots{}, e_{n}\in \mathbb{R}^{n}$, the corresponding parallelepiped is the $n\text{-}$dimensional unit cube. Hence $ D(\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n)=1.\tag{1.6} $ In matrix form this is simply $\det(I)=1.$