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威斯康辛大学2210Q 应用线性代数Applied Linear Algebra代考

2210Q. Applied Linear Algebra

3.00 credits

Prerequisites: MATH 1132Q or 1152Q or 2142Q. Recommended preparation: a grade of C- or better in MATH 1132Q. May not be taken out of sequence after passing MATH 2144Q, 3210, 3510, or 3710. Repeat restrictions apply. See advising.uconn.edu/repeat-policy for information.

Grading Basis: Graded

Systems of equations, matrices, determinants, linear transformations on vector spaces, characteristic values and vectors, from a computational point of view. The course is an introduction to the techniques of linear algebra with elementary applications.

威斯康辛大学2210Q 应用线性代数Applied Linear Algebra代考

2210Q 应用线性代数Applied Linear Algebra题目解答

问题 1.

Compute the rank of the linear map $T: \mathbf{R}^{3} \rightarrow \mathbf{R}^{4}$ defined by
$$
T(x, y, z)=(x+y+z, x-y, y-z, z-x) .
$$

证明 .

We let $v_{1}, v_{2}, v_{3}$ be the canonical basis of $\mathbf{R}^{3}$ and compute
$$
T\left(v_{1}\right)=T(1,0,0)=(1,1,0,-1),
$$
thus the first row of the matrix $A$ in the previous discussion is $(1,1,0,-1)$. We do the same with $v_{2}, v_{3}$ and we obtain
$$
A=\left[\begin{array}{cccc}
1 & 1 & 0 & -1 \
1 & -1 & 1 & 0 \
1 & 0 & -1 & 1
\end{array}\right]
$$
Using row-reduction we compute
$$
A_{r e f}=\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & -1
\end{array}\right]
$$
and we deduce that
$$
\operatorname{rank}(T)=3 .
$$

问题 2.

Prove that for all linear transformations $T_{1}, T_{2}: V \rightarrow W$ and all scalars $c \in F$ we have
$$
{ }^{t}\left(T_{1}+c T_{2}\right)={ }^{t} T_{1}+c^{t} T_{2} .
$$

证明 .

We need to prove that if $l$ is a linear form on $W$, then
$$
l \circ\left(T_{1}+c T_{2}\right)=l \circ T_{1}+c l \circ T_{2} .
$$
This follows from the fact that $l$ is linear.

问题 3. a) Let $T_{1}: V_{1} \rightarrow V_{2}$ and $T_{2}: V_{2} \rightarrow V_{3}$ be linear transformations. Prove that
$$
{ }^{t}\left(T_{2} \circ T_{1}\right)={ }^{t} T_{1} \circ{ }^{t} T_{2} .
$$
b) Deduce that if $T: V \rightarrow V$ is an isomorphism, then so is ${ }^{t} T: V^{} \rightarrow V^{}$, and $\left({ }^{t} T\right)^{-1}={ }^{t}\left(T^{-1}\right)$.

证明 .

a) Let $l$ be a linear form on $V_{3}$. Then
$$
\begin{gathered}
{ }^{t}\left(T_{2} \circ T_{1}\right)(l)=l \circ\left(T_{2} \circ T_{1}\right)=\left(l \circ T_{2}\right) \circ T_{1}= \
{ }^{t} T_{1}\left(l \circ T_{2}\right)={ }^{t} T_{1}\left({ }^{t} T_{2}(l)\right)={ }^{t} T_{1} \circ{ }^{t} T_{2}(l) .
\end{gathered}
$$
The result follows.
b) Since $T$ is an isomorphism, there is a linear transformation $T^{-1}$ such that $T \circ$ $T^{-1}=T^{-1} \circ T=$ id. Using part a) and the obvious equality ${ }^{t}$ id $=\mathrm{id}$, we obtain
$$
{ }^{t} T \circ{ }^{t}\left(T^{-1}\right)=\mathrm{id}={ }^{t}\left(T^{-1}\right) \circ t T,
$$
from where the result follows.

问题 4.

Let $A \in M_{2}(\mathbf{C})$ have eigenvalues $\lambda_{1}$ and $\lambda_{2}$. Prove that for all $n \geq 1$ we have
$$
\operatorname{Tr}\left(A^{n}\right)=\lambda_{1}^{n}+\lambda_{2}^{n}
$$
Deduce that $\lambda_{1}^{n}$ and $\lambda_{2}^{n}$ are the eigenvalues of $A^{n}$.

证明 .

Let $x_{n}=\operatorname{Tr}\left(A^{n}\right)$. Multiplying relation (2.1) by $A^{n}$ and taking the trace yields
$$
x_{n+2}-\left(\lambda_{1}+\lambda_{2}\right) x_{n+1}+\lambda_{1} \lambda_{2} x_{n}=0 .
$$
Since $x_{0}=2$ and $x_{1}=\operatorname{Tr}(A)=\lambda_{1}+\lambda_{2}$, an immediate induction shows that $x_{n}=\lambda_{1}^{n}+\lambda_{2}^{n}$ for all $n$.

For the second part, let $z_{1}, z_{2}$ be the eigenvalues of $A^{n}$. By definition, they are the solutions of the equation $t^{2}-\operatorname{Tr}\left(A^{n}\right) t+\operatorname{det}\left(A^{n}\right)=0$. Since $\operatorname{det}\left(A^{n}\right)=(\operatorname{det} A)^{n}=$ $\lambda_{1}^{n} \lambda_{2}^{n}$ and $\operatorname{Tr}\left(A^{n}\right)=\lambda_{1}^{n}+\lambda_{2}^{n}$, the previous equation is equivalent to
$$
t^{2}-\left(\lambda_{1}^{n}+\lambda_{2}^{n}\right) t+\lambda_{1}^{n} \lambda_{2}^{n}=0 \quad \text { or } \quad\left(t-\lambda_{1}^{n}\right)\left(t-\lambda_{2}^{n}\right)=0 .
$$
The result follows.

MATH 2210Q MIDTERM EXAM II PRACTICE PROBLEMS


Date and place: Thursday, November 6th, 2021.
Material: Sections 3.1-3.2-3.3-4.1–4.7. Lecture 13–24, HW5-6-7, and the practice exam and additional practice problems below.
Policy: No calculators will be allowed at the exam.
Format: The actual exam will have the same format as the practice problems. The content will
be reasonably similar to the practice exam and the additional practice problems.

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