3.1 KiB
Lecture 38
Chapter VIII Operators on complex vector spaces
Trace 8D
Definition 8.47
For a square matrix A, the trace of A is the sum of the diagonal entries denoted tr(A).
Theorem 8.49
Suppose A is m\times n, B is n\times m matrices, then tr(AB)=tr(BA).
Proof:
By pure computation.
Theorem 8.50
Suppose T\in \mathscr{L}(V) and u_1,...,u_n and v_1,...,v_n are bases of V.
tr(M(T,(u_1,...,u_n)))=tr(M(T,(v_1,...,v_n)))
Proof:
Let A=tr(M(T,(u_1,...,u_n))) and B=tr(M(T,(v_1,...,v_n))), then there exists C, invertible such that A=CBC^{-1},
tr(A)=tr((CB)C^{-1})=tr(C^{-1}(CB))=tr(B)
Definition 8.51
Given T\in \mathscr{L}(V) the trace of T denoted tr(T) is given by tr(T)=tr(M(T)).
Note: For an upper triangular matrix, the diagonal entries are the eigenvalues with multiplicity
Theorem 8.52
Suppose V is a complex vector space such that T\in \mathscr{L}(V), then tr(T) is the sum of the eigenvalues counted with multiplicity.
Proof:
Over \mathbb{C}, there is a basis where M(T) is upper triangular.
Theorem 8.54
Suppose V is a complex vector space, n=dim\ V.$T\in \mathscr{L}(V)$. Then the coefficient on z^{n-1} in the characteristic polynomial is tr(T).
Proof:
(z-\lambda_1)\dots(z-\lambda_n)=z^{n}-(\lambda_1+...+\lambda_n)z^{n-1}+\dots
Theorem 8.56
Trance is linear
Proof:
- Additivity
tr(T+S)=tr(M(T)+M(S))=tr(T)+tr(S) - Homogeneity
tr(cT)=ctr(M(T))=ctr(T)
Theorem/Example 8.10
Trace is the unique linear functional \mathscr{L}\to \mathbb{F} such that tr(ST)=tr(TS) and tr(I)=dim\ V
Proof:
Let \varphi:\mathscr{L}(V)\to \mathbb{F} be a linear functional such that \varphi(ST)=\varphi(TS) and \varphi(I)=n where n=dim\ V. Let v_1,...,v_n be a basis for V define P_{j,k} to be the operator $M(P_{j,k})=\begin{pmatrix}
0&0&0\
0&1&0\
0&0&0
\end{pmatrix}$. Note P_{j,k} form a basis of L(V), now we must show \varphi(P_{j,k})=tr(P_{j,k})=\begin{cases}1\textup{ if }j=k\\0\textup{ if }j \neq k\end{cases}
-
For
j\neq k\varphi(P_{j,j}P_{j,k})=\varphi(P_{j,k})=0\varphi(P_{j,k}P_{j,j})=\varphi(P_{j,k})=0 -
For
j=k
\varphi(P_{k,j},P_{j,k})=\varphi(P_{k,k})=1\varphi(P_{j,k},P_{k,j})=\varphi(P_{j,j})=1
So \varphi(I)=\varphi(P_{1,1}+...+P_{n,n})=\varphi(P_{1,1})+...+\varphi(P_{n,n})=n
Theorem 8.57
Suppose V is finite dimensional vector space, then there does not exists S,T\in \mathscr{L}(V) such that ST-TS=I. (ST-TS is called communicator)
Proof:
tr(ST-TS)=tr(ST)-tr(TS)=tr(ST)-tr(ST)=0, since tr(I)=dim\ V, so ST-TS\neq I
Note: requires finite dimensional.
Chapter ? Multilinear Algebra and Determinants
Determinants ?A
Definition ?.1
The determinant of T\in \mathscr{L}(V) is the product of eigenvalues counted with multiplicity.
Definition ?.2
The determinant of a matrix is given by
det(A)=\sum_{\sigma\in perm(n)}A_{\sigma(1),1}\cdot ...\cdot A_{\sigma(n),n}\cdot sign(\sigma)
perm(\sigma)= all recordings of 1,...,n, number of swaps needed to write \sigma