MIT 18.06 习题集 5

This lesson reproduces the Problem Set “MIT 18.06 Problem Set 5” from MIT OpenCourseWare 18.06 Linear Algebra (Spring 2010) (https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/), licensed under CC BY-NC-SA 4.0. Work through 13 problems in order: type your reasoning in the answer box or write it on the handwriting pad, then submit for AI grading.

How the grading works Every problem is graded by an AI mathematics teacher that reads the official solution shipped with the course. It checks your work criterion by criterion, awards partial credit for correct steps, and returns written feedback. Equivalent mathematics and valid alternative methods are accepted — a different route to the same result can still earn full marks.

If your answer is incomplete, the feedback names exactly which steps are missing, so you can correct them and try again.

Every system with no solution is like the one in problem 6. There are numbers $y_1, \dots, y_m$ that multiply the $m$ equations so they add up to $0=1$. This is called Fredholm's Alternative:

Exactly one of these problems has a solution: $A\mathbf{x}=\mathbf{b}$ OR $A^T\mathbf{y}=0$ with $\mathbf{y}^T\mathbf{b}=1$.

If $\mathbf{b}$ is not in the column space of $A$ it is not orthogonal to the nullspace of $A^T$. Multiply the equations $x_1-x_2=1$ and $x_2-x_3=1$ and $x_1-x_3=1$ by numbers $y_1, y_2, y_3$ chosen so that the equations add up to $0=1$.

If $A^T A\mathbf{x}=0$ then $A\mathbf{x}=0$. Reason: $A\mathbf{x}$ is in the nullspace of $A^T$ and also in the $\underline{\hspace{2cm}}$ of $A$ and those spaces are $\underline{\hspace{2cm}}$. Conclusion: $A^T A$ has the same nullspace as $A$. This key fact is repeated in the next section.

The command N=null(A) will produce a basis for the nullspace of $A$. Then the command B=null(N') will produce a basis for the $\underline{\hspace{2cm}}$ of $A$.

Suppose I give you four nonzero vectors $\mathbf{r}, \mathbf{n}, \mathbf{c}, \mathbf{l}$ in $\mathbf{R}^2$.

a. What are the conditions for those to be bases for the four fundamental subspaces $C(A^T), N(A), C(A), N(A^T)$ of a $2 \times 2$ matrix?

b. What is one possible matrix $A$?

Suppose I give you four nonzero vectors $\mathbf{r}_1, \mathbf{r}_2, \mathbf{n}_1, \mathbf{n}_2, \mathbf{c}_1, \mathbf{c}_2, \mathbf{l}_1, \mathbf{l}_2$ in $\mathbf{R}^2$.

a. What are the conditions for those to be bases for the four fundamental subspaces $C(A^T), N(A), C(A), N(A^T)$ of a $2 \times 2$ matrix?

b. What is one possible matrix $A$?

Suppose $A$ is the $4 \times 4$ identity matrix with its last column removed. $A$ is $4 \times 3$. Project $\mathbf{b}=(1, 2, 3, 4)$ onto the column space of $A$. What shape is the projection matrix $P$ and what is $P$?

What linear combination of $(1, 2,-1)$ and $(1, 0, 1)$ is closest to $\mathbf{b}=(2, 1, 1)$?

If $P^2=P$ show that $(I-P)^2=I-P$. When $P$ projects onto the column space of $A$, $I-P$ projects onto the $\underline{\hspace{2cm}}$.

a. Find the projection matrix $P_C$ onto the column space of $A$.

b. Find the $3 \times 3$ projection matrix $P_R$ onto the row space of $A$. Multiply $B=P_C A P_R$. Your answer $B$ should be a little suprising — can you explain it?

In $\mathbf{R}^m$, suppose I give you $\mathbf{b}$ and $\mathbf{p}$, and $n$ linearly independent vectors $\mathbf{a}_1, \dots, \mathbf{a}_n$. How would you test to see if $\mathbf{p}$ is the projection of $\mathbf{b}$ onto the subspace spanned by the $\mathbf{a}$'s?

If $A$ has $r$ independent columns and $B$ has $r$ independent rows, $AB$ is invertible.

Proof: When $A$ is $m$ by $r$ with independent columns, we know that $A^T A$ is invertible. If $B$ is $r$ by $n$ with independent rows, show that $BB^T$ is invertible. (Take $A=B^T$.)

Now show that $AB$ has rank $r$.

With conductances $c_1=c_2=2$ and $c_3=c_4=c_5=3$, multiply the matrices $A^T C A$. Find a solution to $A^T C A \mathbf{x}=\mathbf{f}=(1, 0, 0,-1)$. Where these potentials $\mathbf{x}$ and currents $\mathbf{y}=-CA\mathbf{x}$ on the nodes and edges of the square graph.

Suppose $A$ is a $12 \times 9$ incidence matrix from a connected (but unknown) graph.

a. How many columns of $A$ are independent?

b. What condition on $\mathbf{f}$ makes it possible to solve $A^T \mathbf{y}=\mathbf{f}$?

c. The diagonal entries of $A^T A$ give the number of edges into each node. What is the sum of those diagonal entries?