5.10 Matrices and linear algebra

Commutator is defined as$$[M,N]=MN-NM$$ Where $N,M$ are matrices.

Anti-commutator is when$${[M,N]}_{+}=MN+NM$$ Two matrices commute means $MN-NM=0$. Matrices that commute share an eigenbasis.

Properties of commutators$$\begin{array}{rl}[A+B,C]& =[A,C]+[B,C]\\ [A,B+C]& =[A,B]+[B,C]\\ [A,A]& =0\\ [A^2,B]& =A[A,B]+[A,B]A\\ [AB,C]& =A[B,C]+[A,C]B\\ [A,BC]& =[A,B]C+B[A,C]\end{array}$$

Matrices are generally noncommutative. i.e.  $$MN\ne NM$$ Matrix Inverse$$A^{-1}=\frac{1}{\left|A\right|}{A}_c^T$$ Where $A_c$ is the cofactor matrix.

Matrix inverse satisfies$$A^{-1}A=I=A{A}^{-1}$$ Matrix adjoint is same as Transpose for real matrix. If Matrix is complex, then Matrix adjoint does conjugate in addition to transposing. This is also called dagger.$$A_{ij}^{†}=A_{ji}^{\ast }$$ So dagger is just transpose but for complex, we also do conjugate after transposing. That is all.

If $A_{ij}=A_{ji}$ then matrix is symmetric. If $A_{ij}=-A_{ji}$ then antisymmetric.

Hermitian matrix is one which $A^{†}=A$. If $A^{†}=-A$ then it is antiHermitian.

Any real symmetric matrix is always Hermitian. But for complex matrix, non-symmetric can still be Hermitian. An example is $\left(\begin{array}{cc}1& -i\\ i& 2\end{array}\right)$.

Unitary matrix Is one whose dagger is same as  its inverse. i.e. $$\begin{array}{rl}{A}^{†}& =A^{-1}\\ A^{†}A& =I\end{array}$$

Remember, dagger is just transpose followed by conjugate if complex. Example of unitary matrix is $\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& i\\ i& 1\end{array}\right)$. Determinant of a unitrary matrix must be complex number whose magnitude is $1$.

Also $\left|Av\right|=\left|v\right|$ if $A$ is unitary. This means $A$ maps vector of some norm, to vector which must have same length as the original vector.

A unitary operator looks the same in any basis.

Orthogonal matrix One which satisfies $$\begin{array}{rl}A{A}^T& =I\\ A^{T}A& =I\\ A^{-1}& =A^T\end{array}$$

commute means $\left[MN\right]=MN-NM$. Also $[MN{]}_{+}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$.

Another property is that $det\left({\alpha }_i\right)=-1$. Since they are Hermitian and unitary, then ${\alpha }_i^{-1}={\alpha }_i$.

If $H$ is Hermitian, then $U=e^{iH}$ is unitary.

When moving a number out of a BRA, make sure to complex conjugate it. For example $⟨3v_1|v_2⟩=3^{\ast }⟨v_1|v_2⟩$. But for the ket, no need to. For example $⟨v_1|3v_2⟩=3⟨v_1|v_2⟩$

item $⟨f|\mathrm{\Omega }|g{⟩}^{\ast }=⟨{\left(\mathrm{\Omega }|g\right)}^{\ast }|f⟩=⟨g|{\mathrm{\Omega }}^{†}|f⟩$

item when moving operator from ket to bra, remember to dagger it. $⟨u|Tv⟩=⟨T^{†}u|v⟩$

item  if given set of vectors and asked to show L.I., then set up $Ax=0$ system, and check $\left|A\right|$. If determinant is zero, then there exist non-trivial solution, which means Linearly dependent. Otherwise, L.I.

item if given $A$, then to represent it in say basis $e_i$, we say $A_{ki}^{(e)}=⟨e_k,A{e}_i⟩=⟨e_k|A|e_i⟩$. i.e $A_{1,1}=⟨e_1,A{e}_1⟩$ and $A_{1,2}=⟨e_1,A{e}_2⟩$ and so on.