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 $V$.

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 $A=CBC^{-1}$,

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

$perm(\sigma)=$ all recordings of $1,...,n$, number of swaps needed to write $\sigma$