[linear algebra - Why is determinant a multilinear function? - Mathematics Stack Exchange](https://math.stackexchange.com/questions/1403735/why-is-determinant-a-multilinear-function#:~:text=It%20is%20not%20linear%2C%20or,of%20matrix%20columns%2C%20is%20multilinear.) Examples:: *[[Examples]]* Nonexamples:: *[[Nonexamples]]* Constructions:: *[[Constructions|Used in the construction of...]]* Specializations:: [[determinant as volume of parallelopiped with dimension of the ambient space]] Generalizations:: *[[Generalizations]]* Justifications and Intuition:: *[[Justifications and Intuition]]* ---- > [!definition] Definition. ([[volume of a parallelopiped]]) > Let $k \leq n$. Define $V: \rr ^{n \times k} \to [0, \infty)$ by $V(X) = \sqrt{ \det X^{\top}X }.$ Suppose $x_{1},\dots,x_{k}$ are [[linearly independent]] [[vector|vectors]] in $\rr ^{n}$. Define the **$k$-dimensional volume** of the [[parallelopiped]] $\PP(x_{1},\dots,x_{k})$ by $V(X)$, where $X=\begin{pmatrix} | & | & | \\ x_1 & \dots & x_k \\ | & | & |\end{pmatrix}$. ---- #### > [!justification]- > We need to verify that $V$ is well-defined— that is, we must check that >- $X^{\top}X$ is square (so that the [[determinant]] is well-defined) >- $\det (X^{\top}X) \in [0, \infty)$ (in particular, is real and nonnegative) \ The first is clear from the definition of [[matrix product|matrix multiplication]]. \ The second is clear for the following reason (worded long but conceptually simple): \ The [[matrix product]] is defined to be the [[matrix]] of the [[product of linear maps]]. Since [[hom and mat are isomorphic]], there is exactly one [[linear map]] $ST \in \endo(\rr ^{k})$ whose [[matrix]] (w.r.t. the standard [[basis]]) is $X^{\top} X$. Since [[adjoint matrix w.r.t. orthonormal bases equals conjugate transpose]], it therefore must be the case that $S=T^{\dagger}$. Now, because [[product of linear map and adjoint is a positive semidefinite operator]] and the [[matrix of positive semidefinite operator is positive semidefinite]], $X^{\top}X$ is a [[positive semidefinite matrix]]. By the [[characterization of positive semidefinite operators]] all [[eigenvalue]]s of $SU$ are thus nonnegative. Since [[matrix of an operator shares its eigenvalues]] and the [[determinant]] of an [[linear operator]] is the product of its [[eigenvalue]]s and equal to the [[determinant]] of its [[matrix]] (w.r.t. any [[basis]]), $\det X^{\top}X \geq 0$ and so $\sqrt{ \cdot }$ is well-defined here. > [!intuition] > We can check that this definition is a good one by verifying that it agrees with [[determinant as volume of parallelopiped with dimension of the ambient space|the derived volume of a parallelopiped]] when $k=n$ in general, when $k=1$ and $n$ is general, and when $k=2,n=3$. \ So suppose $k=n$. Then we have $\sqrt{ \det X^{\top}X }=\sqrt{ \det(X^{\top})\det (X) }=\sqrt{ (\det X)^{2} }=|\det X |,$ according with [[determinant as volume of parallelopiped with dimension of the ambient space]]. \ Next suppose $k=1$, $n$ general, so that $X$ is a column vector. We get the [[norm|length formula]], as expected: $\sqrt{ \det X^{\top}X }=\sqrt{ \det X\cdot X }=\sqrt{ \|X\| ^{2}}=\|X\|.$ \ Next suppose $k=2, n=3$. Let $a,b$ be the columns of $X$. We should get the [[cross product]] magnitude $\|a \times b\|= \|a\|\|b\|\sin \theta$: ![[CleanShot 2023-01-08 at 19.53.20.jpg|400]] And indeed, we do: ![[CleanShot 2023-01-08 at 19.53.56.jpg]] ![[MOC volume of a parallelopiped]] ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```