Linear Algebra: IV Determinants

Determinants


The determinant of a square matrix $A$, denoted $\det(A)$, describes how the linear transformation $T: \mathbb{R}^n \to \mathbb{R}^n$,


$$ T(\vec{x}) = A\vec{x} $$

scales $n$ dimensional area / volume.


  • If $|\det(A)| > 1$, areas / volumes are expanded.


  • If $|\det(A)| < 1$, areas / volumes are compressed.


  • If $\det(A) = 0$, areas / volumes collapse to lower dimension.


  • The sign of $\det(A)$ indicates whether the transformation preserves or reverses orientation.

Scaling volume, the determinant of a matrix product is the product of determinants:


$$ \det(AB) = \det(A)\det(B) $$

Also, a scalar matrix product scales all $n$ rows / columns by the scalar so:


$$ \det(cA) = c^n \det(A) $$

If a matrix has a row or column with zeros, then $\det(A) = 0$, since the $n$ dimensional volume collapses to zero under a transformation that spans at most an $(n-1)$ dimensional space. If any two rows or columns of $A$ are linearly dependent, then $\det(A) = 0$, since the $n$ dimensional volume collapses to zero also.


If all of the rows or columns of $A$ are linearly independent, then $\det(A) \neq 0$.


Elementary matrices affect determinants as follows:


$$ \det(I) = 1 $$
$$ \det((E^{\text{swap}})_{ij}) = -1 $$
$$ \det((E^{\text{scale}})^k_i) = k $$
$$ \det((E^{\text{shear}})^k_{ij}) = 1 $$

Using Gauss–Jordan elimination, if there exist at most $n^2$ elementary matrices $E_1, \ldots, E_{n^2}$ such that


$$ E_{n^2} \cdots E_1 P A = I $$

Then


$$ A^{-1} = E_{n^2} \cdots E_1 $$

And


$$ A = P^{-1} E^{-1}_1 \cdots E^{-1}_{n^2} $$

A permutation matrix is a product of swapping elementary matrices with determinant $-1$. So $\det(P^{-1}) = \pm 1$.


We observe that $\det(E_i), \ \det(E_i^{-1}), \ \det(P^{-1}) \neq 0$, which means that $A$ has an inverse if $\det(A) \neq 0$.


Also, $\det(A)$ depends only on the swapping and scaling of rows:


$$ \det(A)= \det(P^{-1}) \det(E^{-1}_1) \cdot \ldots \cdot \det(E^{-1}_{n^2}) $$

Scaling a row by $k$ in Gauss-Jordan elimination scales the determinant of $A^{-1}$ by $k$ and scales the determinant of $A$ by $\frac{1}{k}$ (the pivot). Since Gauss-Jordan elimination involves at most $n$ elementary scaling matrices, the determinant of $A$ is the product of the pivots.


Also $\det(A)\det(A^{-1}) = \det(I)$ so:


$$ \det(A^{-1}) = \frac{1}{\det(A)} $$

The determinant of a transpose is equal to the determinant of the original matrix since $\det(A) = 0$ implies $\det(A^\top)=0$ and $\det(A) \neq 0$ implies


$$ A=E^{-1}_1 \cdots E^{-1}_{n^2} $$

And so


$$ A^\top= (E^{-1}_{n^2})^\top \cdots (E^{-1}_1)^\top $$

Where


$$ \det(((E^\text{swap})_{ij})^\top) = \det((E^\text{swap})_{ji}) = -1 $$
$$ \det(((E^\text{scale})^k_{i})^\top) = \det((E^\text{scale})^k_{i}) = k $$
$$ \det(((E^\text{shear})^k_{ij})^\top) = \det((E^\text{shear})^k_{ji}) = 1 $$

Therefore,


$$ \det(A^\top) = \det(A) $$

Comments

Post a Comment

Popular posts from this blog

Linear Algebra: III Square and Inverse Matrices

Square and Inverse Matrices An $n$ by $n$ square matrix has the same number of rows and columns: $$ A = \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} $$ If $A$ is an $n$ by $n$ square matrix and $\vec{x} \in \mathbb{R}^n$, then the product $A\vec{x}$ is also a vector in $\mathbb{R}^n$, defining a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^n$. The identity matrix $I$ is an $n$ by $n$ square matrix with 1s on the main diagonal and 0s elsewhere: $$ I = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} $$ It satisfies: $$ AI = IA = A $$ The inverse of a square matrix $A$, denoted $A^{-1}$, is a matrix such that: $$ A A^{-1} = A^{-1} A = I $$ Which may not exist. If $A^{-1}$ exists, then $A$ is called invertible . If $A$ and $B$ are inve...
Read more

Linear Algebra: VII Orthonormal Matrices

Orthonormal Matrices A square $n$ by $n$ matrix $Q$ is called orthonormal if: $$ QQ^\top = Q^\top Q = I $$ $$ Q^{-1} = Q^\top $$ The columns $\vec{q}_1, \ldots, \vec{q}_n$ of $Q$ (and the rows) form an orthonormal basis of $\mathbb{R}^n$: $$ \vec{q}_i^\top \vec{q}_j = 0 \text{ if } i \neq j, \quad \|\vec{q}_i\| = 1 $$ An orthonormal matrix preserves inner products / dot products: $$ (Q\vec{x})^\top(Q\vec{y}) = \vec{x}^\top Q^\top Q \vec{y} = \vec{x}^\top I \vec{y} = \vec{x}^\top \vec{y} $$ An orthonormal matrix also preserves magnitude: $$ \|Q\vec{x}\|^2 = (Q\vec{x})^\top Q\vec{x} = \vec{x}^\top Q^\top Q \vec{x} = \vec{x}^\top I \vec{x} = \vec{x}^\top \vec{x} = \|\vec{x}\|^2 $$ So an orthonormal matrix preserves angles between vectors: $$ \cos{\theta} = \frac{(Q\vec{x})^\top(Q\vec{y})}{\|Q\vec{x}\|\|Q\vec{y}\|} = \frac{\vec{x}^\top \vec{y}}{\|\vec{x}\|\|\vec{y}\|} $$ There are two special types of orthogonal matrices. First, a...
Read more

Linear Algebra: VI Triangular Matrices and LU Decomposition

Triangular Matrices and LU Decomposition A square matrix $A$ is called upper triangular if: $$ a_{ij} = 0 \quad \text{for all } i > j $$ A square matrix $A$ is called lower triangular if: $$ a_{ij} = 0 \quad \text{for all } i Upper and lower triangular $n \times n$ square matrices both contain at least $$ \frac{n(n-1)}{2} $$ Zeros and $$ \frac{n(n+1)}{2} $$ Other entries. The identity matrix $I$ is both upper triangular and lower triangular. Scaling elementary matrices and their products are also both upper and lower triangular. Shearing elementary matrices are either upper triangular or lower triangular but never both. Swapping elementary matrices are neither upper nor lower triangular. If $A$ and $B$ are upper triangular $n \times n$ square matrices, then $AB$ is also upper triangular. Indeed, for $i > j$: $$ (ab)_{ij} = \sum_{k=1}^n a_{ik} b_{kj} $$ If $i > k$, then $a_{ik} = 0$. If $k \ge i$, then $k > j...
Read more