# 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 $\sigma$ $$