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Math55a课程简介
Topics We will spend most of the semester studying (bi)linear algebra, and also lay the groundwork for further topics in abstract algebra, culminating with the main results of either Galois theory or the representation theory of finite groups, if not both. These are fundamental tools for much of modern mathematics. In Math 55b, we will introduce metric topology and combine it with tools from the linear algebra studied in 55a to develop a rigorous treatment of differential and integral calculus, together with an introduction to complex analysis calculus of differentiable functions of one complex variable.
Prerequisites
Abstract algebra is the set of advanced topics of algebra that deal with abstract algebraic structures rather than the usual number systems. The most important of these structures are groups, rings, and fields.
Textbooks We shall base our exposition of linear algebra on Sheldon Axler’s Linear Algebra Done Right (3rd ed., Springer, 2015), though we will routinely go beyond Axler. That text should now be available for purchase at the Coop. ${ }^1$ For the rest of 55 a we’ll likely follow a (necessarily small) selection from Michael Artin’s Algebra.
Math55a Abstract Algebra HELP（EXAM HELP， ONLINE TUTOR）
Solve problem 9 in Axler 10.A (p.305) for any finite-dimensional vector space $V$ over any field $F$. Note the corollary that the trace of $P$ is a nonnegative integer. [As I may have noted in lecture, operators satisfying $P^2=P$ (and thus $P^k=P$ for all $k=1,2,3, \ldots$, so the same as each power of $P)$ are called “idempotent”.]
Let $\$ P \$$ be an idempotent operator on a finite-dimensional vector space $\$ V \$$ over any field \$F\$. We want to show that the trace of \$P\$ is a nonnegative integer.
Since $\$ P \$$ is idempotent, we have $\$ P^{\wedge} 2=P \$$. Let $\$ n=\backslash \operatorname{dim}(V) \$$.
By the Spectral Theorem, $\$ \mathrm{~V} \$$ has an orthonormal basis consisting of eigenvectors of \$P\$. Let $\$ \backslash$ lambda_1,Vdots, \lambda_k $\$$ be the distinct eigenvalues of $\$ P \$$, and let $\$ m_{-} 1, \backslash 1 d o t s, m_{-} k \$$ be their respective multiplicities. Then we have
$\$ m_{-} 1+\backslash c$ dots $+m_{-} k=n \$$, and $\$ P \$$ is diagonalizable with respect to this basis, so we can write $P=\left(\begin{array}{ccc}\lambda_1 I_{m_1} & & \ & \ddots & \ & & \lambda_k I_{m_k}\end{array}\right)$. Since $\$ P \$$ is idempotent, we have $\$ \backslash$ ambda_i^2=\lambda_i $\$$ for all $\$ i=1, \backslash$ dots, $\mathrm{k} \$$. Therefore, $\$ \backslash$ lambda_i $\$$ must be either 0 or 1 .
The trace of $\$ P \$$ is the sum of its diagonal entries, which is \$Vambda_1 $m_{-} 1+\backslash c$ otots + Vambda_k m_k\$. Since $\$ \backslash$ lambda i $\$$ is either 0 or 1 , this sum counts the number of 1 ‘s on the diagonal of $\$ P \$$, which is the rank of \$P\$. By the Rank-Nullity Theorem, we have $\operatorname{rank}(P)+\operatorname{nullity}(P)=\operatorname{dim}(V)=n$, where $\$ \backslash$ operatorname ${$ nullity $}(P) \$$ is the dimension of the null space of $\$ P \$$. Since $\$ P \$$ is idempotent, we have $\$ P V=0 \$$ if and only if $\$ P(P v)=0 \$$, which means $\$ P v=0 \$$ or $\$ P v=v \$$ for all $\$ \mathrm{~V} \backslash$ in $\bigvee \$$. Therefore, $\$ \backslash$ operatorname{nullity} $(P) \$$ is the number of 0 ‘s on the diagonal of $\$ P \$$, which is $\$ n \$$ minus the number of 1 ‘s on the diagonal of $\$ P \$$.
Putting everything together, we have $\operatorname{trace}(P)=\lambda_1 m_1+\cdots+\lambda_k m_k=$ $\operatorname{rank}(P) \geq 0$. Therefore, the trace of $\$ P \$$ is a nonnegative integer, as desired.
Solve problem 6 in Axler 10.B (p.331). This should be easy to do with the $\wedge^n$ definition of the determinant, even though Axler didn’t intend that. Again the ground field is arbitrary.
Let $\$ A \$$ be an $\$ n \backslash$ times $n \$$ matrix over an arbitrary field $\$ F \$$. We want to show that $\$ \backslash \operatorname{det}(\backslash$ peratorname ${\operatorname{adj}}(A))=\backslash \operatorname{det}(A)^{\wedge}{n-1} \$$, where $\$ \backslash$ operatorname ${\operatorname{adj}}$ (A) $\$$ is the adjugate matrix of $\$ A \$$.
By definition, the adjugate matrix of $\$ A \$$ is the transpose of the matrix whose $\$(i, j) \$$ entry is the $\$(j, i) \$$ cofactor of $\$ A \$$. Let $\$ C_{-}{i, j} \$$ denote the $\$(i, j) \$$ cofactor of $\$ A \$$, which is defined as $\$(-1)^{\wedge}{i+j} \$$ times the determinant of the $\$(n-1) \backslash \operatorname{times}(n-1) \$$ matrix obtained by deleting the $\$$ \$ th row and $\$ j \$$ th column of $\$ A \$$.
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