Determinant
Determinants are defined for square matrices and it is a function that maps the matrix into a real number.Determinants are important concepts in linear algebra.It is used in the analysis and solution of system of linear equations. Determinant of a matrix is represented as $det(A)$ or $|A|$.
determinants are used for testing the invertibility. A square matrix $A \in \mathbb{R}^{n \times n}$ is invertable if and only if $det(A) \ne 0$
For a diagonal, upper triangular and lower triangular matrix, the determinant is the product of its diagonal element
Determinant measures volume.If the two sides of the parallelogram is represented as a two columns of a matrix then the absolute value of the determinant represents volume.For a parallelepiped with three sides r,b,g then determinant of the 3x3 matrix $[ r b g ]$ is the volume of the solid.
Laplace expansion can be used to find determinant.It reduces the problem of computing the determinant of $ n \times n$ matrix into computing the determinant of $(n-1) \times (n-1)$ matrices.
Consider a matrix $A \in \mathbb{R}^{n \times n}$ Then Laplace expansion along the column $j$ is
$det(A) = \sum_{k=1}^{n} (-1)^{k+j} a_{kj} det(A_{k,j})$
Expansion along the row is
$det(A) = \sum_{k=1}^{n} (-1)^{k+j} a_{jk} det(A_{j,k})$
Here $A_{k,j} \in \mathbb{R}^(n-1) \times (n-1)$ is the submatrix of $A$ that we obtain when deleting row $k$ and column $j$.
Example 3 x 3 matrix:(Laplace Expansion to find the Determinants)
so $det(A)= -5$
import numpy as np
A=np.array([[1,2,3],[3,1,2],[0,0,1]])
detA=np.linalg.det(A)
print(" det(A)")
print(detA)
o/p
det(A)
-5.000000000000001
Note: determinant can also be computed using Sarru's rule.
Properties of determinants
For $A \in \mathbb{R}^{n \times n}$, the determinant exhibits the following properties.
$1.det(AB)=det(A).det(B)$
$2.det(A)=det(A^T)$
$3.det(A^{-1})=1/det(A)$
4.Similar matrices possess same determinant.So the determinant is invariant to the choice of basis of a linear mapping
5.Adding a multiple of row/column to another does not change $det(A)$
6.Multiplication of a column/row by a scalar $k$ scales the determinant by$ k$
7.Swapping two rows/columns changes the sign of $det(A)$
We can use Gauss elimination to compute $det(A)$ by bringing $A $into row echelon form.For a triangular matrix the determinant is the product of diagonal elements.
A square matrix $A \in \mathbb{R}^{n \times n}$ has $det(A) \ne 0$ if and only if $rank(A)=n$.In other words , A is invertible if and only if it is full rank.
Trace
The trace of a square matrix $A \in \mathbb{R}^{n \times n}$ is defined as
$tr(A)=\sum_{i=1}^{n} a_{ii}$
ie: Trace is the sum of the diagonal elements of A
import numpy as np
A=np.array([[1,2,3],[3,1,2],[0,0,1]])
detA=np.trace(A)
print("A")
print(A)
print("Trace(A)")
print(detA)
o/p
A
[[1 2 3]
[3 1 2]
[0 0 1]]
Trace(A)
3
Properties of trace
$1.tr(A+B)= tr(A)+tr(B)$
$2.tr(kA)=k tr(A)$
$3.tr(I_n)=n$
$4.tr(AB)=tr(BA)$
$5.tr(ABC)=tr(BCA)$ invariant under cyclic permutation
$6.tr(xy^T)=tr(y^Tx)=y^Tx$
$7.$ If A is a Transformation matrix for a linear mapping, for a different base B
$A=S^{-1}AS$
$tr(A)=tr(S^{-1}AS)=tr(SS^{-1}A)=tr(A)$
This shows that trace is independent of the basis in linear mapping.
Characteristic Polynomial
For $\lambda \in \mathbb{R}$ and a square matrix $A \in \mathbb{R}^{n \times n}$
$P_A(\lambda)=det(A-\lambda I )=C_0+C_1\lambda+C_2\lambda^2+\cdots+C_{n-1}\lambda^{n-1}+(-1)^n\lambda^n$
The characteristic polynomial will allow us to compute eigen values and eigen vectors.
In the characteristic polynomial
$C_0=det(A)$
$C_{n-1}=(-1)^{n-1}tr(A)$
Example:
Let $A=\begin{bmatrix}
2 & 1\\
1& 2
\end{bmatrix} $
The characteristic polynomial
$P_A(\lambda)=det(A-\lambda I )=\begin{vmatrix}
2-\lambda & 1\\
1& 2-\lambda
\end{vmatrix}=(2-\lambda)^2-1=3-4\lambda+\lambda^2$
Note that the determinant of the matrix $det(A)=3=C_0$ and the trace of the matrix $tr(A)=4=(-1)^{n-1}C_{n-1}=-4$
Finding characteristic polynomial using sympy
from sympy import Matrix
M=Matrix([[2,1],[1,2]])
M.charpoly().as_expr()
O/P
$\displaystyle \lambda^{2} - 4 \lambda + 3$
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