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  • Trace
  • ​Determinant
  • Invertibility

Matrix Properties

Trace

Trace of a matrix is the sum of its diagonal elements:

Tr(A)=∑iAiiTr(A) = \sum_iA_{ii}Tr(A)=i∑​Aii​

Further,

Tr(A+B)=Tr(A)+Tr(B)Tr(AB)=Tr(BA),evenifAB≠BATr(A)=Tr(AT)Tr(A+B) = Tr(A) + Tr(B) \\ Tr(AB) = Tr(BA), \hspace{0.1cm}even\hspace{0.1cm}if\hspace{0.1cm}AB \neq BA \\ Tr(A) = Tr(A^T)Tr(A+B)=Tr(A)+Tr(B)Tr(AB)=Tr(BA),evenifAB=BATr(A)=Tr(AT)

​Determinant

For a matrix, A=[a11a12a21a22]A = \begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{bmatrix}A=[a11​a21​​a12​a22​​]​, it's determinant is given by:

∣A∣=a11a22−a12a22|A| = a_{11}a_{22} - a_{12}a_{22}∣A∣=a11​a22​−a12​a22​

For a 3×33 \times 33×3 matrix, A=[abcdefghi]A = \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}A=​adg​beh​cfi​​​, it's determinant is given by:

∣A∣=a(ei−fh)−b(di−fg)+c(dh−eg)|A| = a(ei-fh) - b(di-fg) + c(dh - eg)∣A∣=a(ei−fh)−b(di−fg)+c(dh−eg)

​For a 4×44 \times 44×4​ matrix, the determinant is given by:

∣A∣=[a×∣fghjklnop∣]−[b×∣eghiklmop∣]+[c×∣efhijlmnp∣]−[d×∣efgIjkmno∣]|A| = \begin{bmatrix} a \times \begin{vmatrix} f & g & h\\ j & k & l\\ n & o & p \end{vmatrix} \end{bmatrix} - \begin{bmatrix} b \times \begin{vmatrix} e & g & h\\ i & k & l\\ m & o & p \end{vmatrix} \end{bmatrix} + \begin{bmatrix} c \times \begin{vmatrix} e & f & h\\ i & j & l\\ m & n & p \end{vmatrix} \end{bmatrix} - \begin{bmatrix} d \times \begin{vmatrix} e & f & g\\ I& j & k\\ m & n & o \end{vmatrix} \end{bmatrix}∣A∣=​a×​fjn​gko​hlp​​​​−​b×​eim​gko​hlp​​​​+​c×​eim​fjn​hlp​​​​−​d×​eIm​fjn​gko​​​​

Physically, determinant represents volume formed by column vectors.

Invertibility

A square matrix AAA is invertible if and only if det(A)≠0det(A) \neq 0det(A)=0. This automatically means that columns of AAA​ are linearly independent. For example,

A=[v1v2...vn]v1=(a11,a21,...,an1)v2=(a12,a22,...,an2)vn=(a1n,a2n,...,ann)A = \begin{bmatrix} v_1 & v_2 & ... & v_n\\ \end{bmatrix} \\ v_1 = (a_{11}, a_{21}, ... , a_{n1}) \\ v_2 = (a_{12}, a_{22}, ... , a_{n2}) \\ v_n = (a_{1n}, a_{2n}, ... , a_{nn})A=[v1​​v2​​...​vn​​]v1​=(a11​,a21​,...,an1​)v2​=(a12​,a22​,...,an2​)vn​=(a1n​,a2n​,...,ann​)

​Let's presume the following:

vn=α1v1+α2v2+...+αn−1vn−1v_n = \alpha_1v_1 + \alpha_2v_2 + ... + \alpha_{n-1}v_{n-1}vn​=α1​v1​+α2​v2​+...+αn−1​vn−1​

​In such a scenario, the area formed by the (hyper)parallelogram by the "n vectors" (the determinant) would be zero.

Another simpler example, consider a 2×22 \times 22×2​ matrix where v1=2v2v_1 = 2v_2v1​=2v2​​. In such a case, the area of the parallelogram is zero.

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