MIT 18.06 Problem Set 3

This lesson reproduces the Problem Set “MIT 18.06 Problem Set 3” 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 14 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.

The plane $x-3y-z=12$ is parallel to the plane $x-3y-z=0$ in Problem 17. One particular point on this plane is $(12, 0, 0)$. All points on the plane have the form (fill in the first components)

(If possible...) Construct a matrix whose column space contains $(1, 1, 0)$ and $(0, 1, 1)$ and whose nullspace contains $(1, 0, 1)$ and $(0, 0, 1)$.

How is the nullspace $\mathbf{N}(C)$ related to the spaces $\mathbf{N}(A)$ and $\mathbf{N}(B)$, if $C=\begin{bmatrix} A \\ B \end{bmatrix}$?

Kirchhoff's Law says that \emph{current in=current out} at every node. This network has six currents $y_1, \ldots, y_6$ (the arrows show the positive direction, each $y_i$ could be positive or negative). Find the four equations $A\mathbf{y}=0$ for Kirchhoff's Law at the four nodes. Find three special solutions in the nullspace of $A$.

Suppose $A$ and $B$ are $n$ by $n$ matrices, and $AB=I$. Prove from $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$ that the rank of $A$ is $n$. So $A$ is invertible and $B$ must be its two-sided inverse (Section 2.5). Therefore $BA=I$ (which is not so obvious!).

\emph{Neat fact} \textbf{Every} $m$ \textbf{by} $n$ \textbf{matrix of rank} $r$ \textbf{reduces to (}$m$ \textbf{by} $r$\textbf{) times (}$r$ \textbf{by} $n$\textbf{):}

Write the 3 by 4 matrix $A$ in equation (1) at the start of this section as the product of the 3 by 2 matrix from the pivot columns and the 2 by 4 matrix from $R$.

Suppose $R$ is $m$ by $n$ of rank $r$, with pivot columns first:

(a) What are the shapes of those four blocks?

(b) Find a \emph{right-inverse} $B$ with $RB=I$ if $r=m$.

(c) Find a \emph{left-inverse} $C$ with $CR=I$ if $r=n$.

(d) What is the reduced row echelon form of $R^T$ (with shapes)?

(e) What is the reduced row echelon form of $R^T R$ (with shapes)?

Prove that $R^T R$ has the same nullspace as $R$. Later we show that $A^T A$ always has the same nullspace as $A$ (a valuable fact).

Suppose you allow elementary \emph{column} operations on $A$ as well as elementary row operations (which get to $R$). What is the “row-and-column reduced form” for an $m$ by $n$ matrix of rank $r$?

(§3.3, #17 – Optional) (a) Suppose column $j$ of $B$ is a combination of previous columns of $B$. Show that column $j$ of $AB$ is the same combination of previous column of $AB$. Then $AB$ cannot have new pivot columns, so $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$.

(b) Find $A_1$ and $A_2$ so that $\operatorname{rank}(A_1 B)=1$ and $\operatorname{rank}(A_2 B)=0$ for $B=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.

Explain why these are all false:

(a) The complete solution is any linear combination of $\mathbf{x}_p$ and $\mathbf{x}_n$.

(b) A system $A\mathbf{x}=\mathbf{b}$ has at most one particular solution.

(c) The solution $\mathbf{x}_p$ with all free variables zero is the shortest solution (minimum length $\|x\|$). Find a 2 by 2 counterexample.

(d) If $A$ is invertible there is no solution $\mathbf{x}_n$ in the nullspace.

Write down all known relations between $r$ and $m$ and $n$ if $A\mathbf{x}=\mathbf{b}$ has (a) no solution for some $\mathbf{b}$ (b) infinitely many solutions for every $\mathbf{b}$ (c) exactly one solution for some $\mathbf{b}$, no solution for other $\mathbf{b}$ (d) exactly one solution for every $\mathbf{b}$.

Apply Gauss-Jordan elimination to $U\mathbf{x}=0$ and $U\mathbf{x}=\mathbf{c}$. Reach $R\mathbf{x}=0$ and $R\mathbf{x}=\mathbf{d}$:

Solve $R\mathbf{x}=0$ to find $\mathbf{x}_n$ (its free variable is $x_2=1$). Solve $R\mathbf{x}=\mathbf{d}$ to find $\mathbf{x}_p$ (its free variable is $x_2=0$).

Suppose $K$ is the 9 by 9 second difference matrix (2's on the diagonal,-1's on the diagonal above and also below). Solve the equation $K\mathbf{x}=\mathbf{b}=(10, \ldots, 10)$. If you graph $x_1, \ldots, x_9$ above the points $1, \ldots, 9$ on the $x$ axis, I think the nine points fall on a parabola.

Suppose $A\mathbf{x}=\mathbf{b}$ and $C\mathbf{x}=\mathbf{b}$ have the same (complete) solutions for every $\mathbf{b}$. Is it true that $A=C$?