复习的方式

最后一讲没有重新讲定义，而是拿旧考题现场推理。每道题都对应课程的一条主线；把它们放在一起，就是一张完整的知识地图。

下面按题型整理，重点不是答案，而是看到条件后应该先问什么。

时间轴

题型一：由可解性反推 m、n、r

题目

设 $A$ 是 $3\times n$ 矩阵。已知：

• $Ax=(1,0,0)^{\mathsf T}$ 无解；
• $Ax=(0,1,0)^{\mathsf T}$ 恰有一个解。

问 $m,n,r$ 的关系。

推理

1. 右边是三维向量，所以 $m=3$。
2. 存在某个右端无解，说明消元会出现零行，因此 $r<m$，即 $r<3$。
3. 「恰有一个解」说明零空间只有零向量，因此列线性无关，$r=n$。

合起来：

$$r=n<3.$$

例子

取

$$A=\begin{bmatrix}0\\1\\0\end{bmatrix}.$$

它是 $3\times1$，$r=n=1$。列空间是 $y$ 轴，$(0,1,0)$ 在其中且解唯一；$(1,0,0)$ 不在其中，无解。

若想用两列，可以取

$$A=\begin{bmatrix}0&0\\1&0\\0&1\end{bmatrix},$$

列空间是 $yz$ 平面，同样满足条件。

题型二：关于 $A^{\mathsf T}A$ 与 $AA^{\mathsf T}$ 的真假判断

在上面的设定下（$r=n<3$），判断三条命题。

命题一：$\det(A^{\mathsf T}A)=\det(AA^{\mathsf T})$

假。两个矩阵尺寸不同：$A^{\mathsf T}A$ 是 $n\times n$，$AA^{\mathsf T}$ 是 $3\times3$。

只有当 $A$ 是方阵时，$\det(BA)=\det(AB)$ 才成立，并且此时两者都等于 $(\det A)^2$。

命题二：$A^{\mathsf T}A$ 可逆

真。因为 $r=n$，列线性无关，$A^{\mathsf T}A$ 是 $n\times n$ 且满秩。

命题三：$AA^{\mathsf T}$ 正定

假。$AA^{\mathsf T}$ 是 $3\times3$，但秩只有 $r=n<3$，必然奇异，最多半正定。

对应的例子中 $AA^{\mathsf T}$ 的第一行全为零，显然不正定。

题型三：证明 $A^{\mathsf T}y=c$ 总有解

$A^{\mathsf T}$ 是 $n\times3$。因为 $A$ 的列线性无关，$A^{\mathsf T}$ 的行线性无关，所以

$$\operatorname{rank}(A^{\mathsf T})=n,$$

行满秩意味着对任意右端都有解。

至于为什么有无穷多解：$A^{\mathsf T}$ 的零空间维数是

$$m-r=3-r>0,$$

所以只要有一个解，就能加上零空间中的任意向量，得到无穷多个解。

这道题把「行满秩 $\Rightarrow$ 总有解」与「零空间维数 $=m-r$」两条结论合在一起，是典型的综合题。

题型四：矩阵乘法与零空间

设 $A$ 的列是 $v_1,v_2,v_3$。

解 $Ax=v_1-v_2+v_3$

按列组合的定义，

$$A\begin{bmatrix}1\\-1\\1\end{bmatrix} =1\cdot v_1+(-1)\cdot v_2+1\cdot v_3 =v_1-v_2+v_3.$$

所以

$$x=\begin{bmatrix}1\\-1\\1\end{bmatrix}.$$

这是矩阵乘法最基本的事实：$Ax$ 是列向量的线性组合，系数就是 $x$ 的分量。

若 $v_1-v_2+v_3=0$

那么上面这个 $x$ 就在 $A$ 的零空间中。于是对任意解，都可以加上它而结果不变：

$$A(x+kc)=Ax.$$

所以解不唯一。唯一性等价于零空间只有零向量。

题型五：正交向量与最近点

设 $v_1,v_2,v_3$ 标准正交。

问：$v_1,v_2$ 的哪个线性组合最接近 $v_3$？

因为 $v_3$ 与 $v_1,v_2$ 都正交，它在 $\operatorname{span}(v_1,v_2)$ 上的投影是零。所以最接近的组合是

$$0\cdot v_1+0\cdot v_2=0.$$

用标准基想象：$v_1,v_2,v_3$ 就是 $x,y,z$ 轴，$xy$ 平面上离 $z$ 轴最近的点是原点。

题型六：Markov 矩阵与稳态

矩阵与特征值

课堂上给出的矩阵满足三个结构性质：

• 每一列元素非负且和为 $1$（Markov 性质）；
• 第一列加第二列等于第三列的两倍，因此列线性相关，矩阵奇异；
• 迹为 $0.8$。

由此直接读出三个特征值：

• 因为 Markov，$1$ 一定是特征值；
• 因为奇异，$0$ 是特征值；
• 因为迹等于特征值之和，第三个是 $0.8-1-0=-0.2$。

补充：录音中的矩阵数字有噪声。一个满足上述全部条件的具体矩阵是 $A=\begin{bmatrix}0.2&0.4&0.3\\0.4&0.2&0.3\\0.4&0.4&0.4\end{bmatrix}.$ 它的列和为 $1$，第一列加第二列等于 $(0.6,0.6,0.8)=2(0.3,0.3,0.4)$，迹为 $0.8$。

稳态

设初值 $u_0=(0,10,0)$，考察 $u_k=A^ku_0$。把 $u_0$ 展开成特征向量组合：

$$u_k=c_1\lambda_1^kx_1+c_2\lambda_2^kx_2+c_3\lambda_3^kx_3.$$

当 $k\to\infty$，$\lambda_1=0$ 与 $\lambda_3=-0.2$ 的项都趋于零，只剩 $\lambda_2=1$ 的项：

$$u_\infty=c_2x_2.$$

求 $\lambda=1$ 的特征向量，解 $(A-I)x=0$ 得

$$x_2=\begin{bmatrix}3\\3\\4\end{bmatrix}.$$

系数为什么是 1

Markov 矩阵保持总人口不变：

$$\mathbf 1^{\mathsf T}Au=\mathbf 1^{\mathsf T}u.$$

对特征向量 $x_i$，若 $\lambda_i\neq1$，则

$$\mathbf 1^{\mathsf T}Ax_i=\lambda_i\mathbf 1^{\mathsf T}x_i=\mathbf 1^{\mathsf T}x_i,$$

所以 $(\lambda_i-1)\mathbf 1^{\mathsf T}x_i=0$，即 $\mathbf 1^{\mathsf T}x_i=0$。也就是说，非 $1$ 特征向量的分量和为零。

初值的分量和是 $10$，$x_2$ 的分量和也是 $10$，因此 $c_2=1$。稳态是

$$u_\infty=\begin{bmatrix}3\\3\\4\end{bmatrix}.$$

题型七：按性质构造矩阵

第三题的要点：$B^{\mathsf T}B$ 永远对称，所以任何非对称矩阵都不可能具有这种形式。

第四题的要点：正交特征向量不只属于对称矩阵。反对称矩阵与正交矩阵同样满足 $AA^{\mathsf T}=A^{\mathsf T}A$，它们的特征向量也可以选成正交（可能是复向量）。

题型八：最小二乘的四个追问

题目

$$A=\begin{bmatrix}1&0\\1&1\\1&2\end{bmatrix},\qquad b=\begin{bmatrix}3\\4\\1\end{bmatrix}.$$

1. 最小二乘解

$$A^{\mathsf T}A=\begin{bmatrix}3&3\\3&5\end{bmatrix},\qquad A^{\mathsf T}b=\begin{bmatrix}8\\6\end{bmatrix}.$$

解正规方程得

$$\hat x=\begin{bmatrix}11/3\\-1\end{bmatrix}.$$

2. 投影是什么

投影就是最优组合：

$$p=A\hat x=\frac{11}{3}\begin{bmatrix}1\\1\\1\end{bmatrix} -\begin{bmatrix}0\\1\\2\end{bmatrix} =\begin{bmatrix}11/3\\8/3\\5/3\end{bmatrix}.$$

3. 对应的直线

$A$ 的两列分别对应「常数项」和「斜率」，所以这是在拟合

$$y=\frac{11}{3}-t.$$

在 $t=0,1,2$ 处，直线给出 $11/3,\,8/3,\,5/3$，与实际值 $3,4,1$ 的误差平方和最小。

4. 什么时候最小二乘解为零

要找非零的 $b$，使得最优组合是零向量。这等价于 $b$ 与整个列空间正交。

列空间由 $(1,1,1)$ 与 $(0,1,2)$ 张成。取

$$b'=\begin{bmatrix}1\\-2\\1\end{bmatrix},$$

它满足

$$\begin{bmatrix}1&1&1\end{bmatrix}b'=0,\qquad \begin{bmatrix}0&1&2\end{bmatrix}b'=0.$$

所以 $b'$ 与两列都正交，投影为零，最小二乘解为零。

整门课的一条主线

把这些题放在一起，可以看到同一条思路反复出现：

1. 秩决定一切：可解性、唯一性、逆的类型，都由 $r$ 与 $m,n$ 的关系决定。
2. 四个子空间是坐标系：零空间管唯一性，列空间管可解性，行空间与左零空间管投影与最小二乘。
3. 矩阵乘法就是列的线性组合：这是所有计算的起点。
4. 特征值描述长期行为：微分方程看指数，Markov 过程看幂。
5. 特殊矩阵有特殊谱：对称（实、正交特征向量）、反对称（纯虚）、正交（模长 $1$）、投影（$0$ 或 $1$）、Markov（$1$ 与不超过 $1$）。
6. SVD 把一般矩阵化归为对角：伪逆、最小二乘、四个子空间的基，都从它得到。

理解检验

1. 由「某右端无解、另一右端唯一解」如何推出 $r=n<m$？

参考答案：右端是三维向量，所以 $m=3$。存在一个右端 $Ax=(1,0,0)^{\mathsf T}$ 无解，说明 $A$ 不能对每个 $\mathbf b\in\mathbb R^3$ 都求解，即列空间不是整个 $\mathbb R^3$，于是 $r<m$，也就是 $r<3$。另一方面 $Ax=(0,1,0)^{\mathsf T}$ 恰有一个解，说明零空间只有零向量（否则可加上零空间向量得到更多解），所以列线性无关，$r=n$。合并得 $r=n<3$。注意这里 $n$ 可以是 $1$ 或 $2$，但必须等于 $r$。

2. 为什么 $A^{\mathsf T}A$ 可逆而 $AA^{\mathsf T}$ 不正定？

参考答案：$A$ 是 $3\times n$ 且 $r=n$，列线性无关，所以 $A^{\mathsf T}A$ 是 $n\times n$ 且满秩，可逆；由 $x^{\mathsf T}A^{\mathsf T}Ax=\lVert Ax\rVert^2>0$（$x\ne0$）它还正定。而 $AA^{\mathsf T}$ 是 $3\times3$，秩只有 $r=n<3$，所以奇异，必有零特征值，最多半正定，不可能正定。两者尺寸和秩都不同，不能混用；只有 $A$ 为方阵且可逆时，两者才都可逆并互为转置关系。

3. 若 $v_1-v_2+v_3=0$，$Ax=v_1$ 的解是否唯一？为什么？

参考答案：不唯一。条件说明向量 $c=(1,-1,1)^{\mathsf T}$ 满足 $Ac=v_1-v_2+v_3=0$，即 $c\in N(A)$，且 $c\ne0$，所以零空间非平凡。$Ax=v_1$ 显然有解 $x=e_1=(1,0,0)^{\mathsf T}$（取第一列为 $v_1$ 的系数）；于是对任意 $k$，

$$A(e_1+kc)=v_1+k\cdot0=v_1,$$

得到无穷多个解。唯一性等价于零空间只有零向量，而这里不满足。

4. Markov 矩阵的非 $1$ 特征向量为什么分量和为零？这如何帮助求稳态系数？

参考答案：Markov 矩阵每列和为 $1$，即 $\mathbf 1^{\mathsf T}A=\mathbf 1^{\mathsf T}$。对特征向量 $Ax_i=\lambda_ix_i$，两边左乘 $\mathbf 1^{\mathsf T}$：

$$\lambda_i\mathbf 1^{\mathsf T}x_i=\mathbf 1^{\mathsf T}Ax_i=\mathbf 1^{\mathsf T}x_i,$$

所以 $(\lambda_i-1)\mathbf 1^{\mathsf T}x_i=0$。当 $\lambda_i\ne1$ 时必有 $\mathbf 1^{\mathsf T}x_i=0$，即分量和为零。求稳态系数时把 $u_0=\sum c_ix_i$ 两边取分量和：非 $1$ 项贡献为零，只剩 $c_2\mathbf 1^{\mathsf T}x_2=\mathbf 1^{\mathsf T}u_0$，于是 $c_2=\mathbf 1^{\mathsf T}u_0/\mathbf 1^{\mathsf T}x_2=10/10=1$。这避免了真正解方程组求系数。

5. 最小二乘解为零的充要条件是什么？如何构造这样的 $b$？

参考答案：最小二乘解满足正规方程 $A^{\mathsf T}A\hat x=A^{\mathsf T}b$，所以 $\hat x=0$ 当且仅当 $A^{\mathsf T}b=0$，即 $b$ 与 $A$ 的每一列都正交，也就是 $b\in N(A^{\mathsf T})$（左零空间），等价于 $b\perp C(A)$。构造方法是取一个非零的左零空间向量：本题 $A$ 的列是 $(1,1,1)^{\mathsf T}$ 与 $(0,1,2)^{\mathsf T}$，解 $b_1+b_2+b_3=0$、$b_2+2b_3=0$，取 $b_3=1$ 得 $b_2=-2$、$b_1=1$，即 $b'=(1,-2,1)^{\mathsf T}$。此时投影 $p=A\hat x=0$，最小二乘解 $\hat x=0$。

结语

线性代数的力量在于：不同的题目背后是同一组结构。秩、子空间、特征值、分解——掌握这四件事，具体计算只是它们的展开。

How the review works

The last lecture does not restate definitions. It works old exam problems live. Every problem maps to one thread of the course; together they form a complete map.

Below, the problems are grouped by type. The point is not the answer but what to ask first when a condition appears.

Timeline

Type one: recover m, n, r from solvability

Problem

Let $A$ be $3\times n$. Suppose:

• $Ax=(1,0,0)^{\mathsf T}$ has no solution;
• $Ax=(0,1,0)^{\mathsf T}$ has exactly one solution.

Find the relation among $m,n,r$.

Reasoning

1. The right side is a three-dimensional vector, so $m=3$.
2. Some right side has no solution, so elimination produces a zero row, hence $r<m$, that is $r<3$.
3. "Exactly one solution" means the null space is trivial, so the columns are independent and $r=n$.

Therefore

$$r=n<3.$$

An example

$$A=\begin{bmatrix}0\\1\\0\end{bmatrix}.$$

It is $3\times1$ with $r=n=1$. Its column space is the $y$-axis, which contains $(0,1,0)$ uniquely but not $(1,0,0)$.

With two columns, take

$$A=\begin{bmatrix}0&0\\1&0\\0&1\end{bmatrix},$$

whose column space is the $yz$-plane and which satisfies the same conditions.

Type two: true or false about $A^{\mathsf T}A$ and $AA^{\mathsf T}$

Under $r=n<3$, decide three statements.

Statement one: $\det(A^{\mathsf T}A)=\det(AA^{\mathsf T})$

False. The matrices have different sizes: $A^{\mathsf T}A$ is $n\times n$, while $AA^{\mathsf T}$ is $3\times3$.

The identity $\det(BA)=\det(AB)$ requires square matrices, and then both equal $(\det A)^2$.

Statement two: $A^{\mathsf T}A$ is invertible

True. Since $r=n$, the columns are independent and $A^{\mathsf T}A$ is $n\times n$ of full rank.

Statement three: $AA^{\mathsf T}$ is positive definite

False. $AA^{\mathsf T}$ is $3\times3$ but has rank $r=n<3$, so it is singular and at most semidefinite.

In the example, the first row of $AA^{\mathsf T}$ is entirely zero, so it certainly is not positive definite.

Type three: proving $A^{\mathsf T}y=c$ is always solvable

$A^{\mathsf T}$ is $n\times3$. Since the columns of $A$ are independent, the rows of $A^{\mathsf T}$ are independent, so

$$\operatorname{rank}(A^{\mathsf T})=n.$$

Full row rank means every right side is solvable.

Why infinitely many? The null space of $A^{\mathsf T}$ has dimension

$$m-r=3-r>0.$$

So once one solution exists, adding any null-space vector gives another.

This problem combines two facts: full row rank implies solvability, and the left null space has dimension $m-r$.

Type four: matrix multiplication and null spaces

Let the columns of $A$ be $v_1,v_2,v_3$.

Solve $Ax=v_1-v_2+v_3$

By the definition of multiplication by columns,

$$A\begin{bmatrix}1\\-1\\1\end{bmatrix} =v_1-v_2+v_3.$$

So

$$x=\begin{bmatrix}1\\-1\\1\end{bmatrix}.$$

This is the most basic fact about matrix multiplication: $Ax$ is a combination of the columns, with the entries of $x$ as coefficients.

If $v_1-v_2+v_3=0$

Then this $x$ lies in the null space of $A$. Any solution can be shifted by it:

$$A(x+kc)=Ax.$$

So solutions are not unique. Uniqueness is exactly the statement that the null space is trivial.

Type five: orthonormal vectors and the nearest point

Suppose $v_1,v_2,v_3$ are orthonormal.

Which combination of $v_1,v_2$ is closest to $v_3$?

Since $v_3$ is orthogonal to both, its projection onto $\operatorname{span}(v_1,v_2)$ is zero. The closest combination is therefore

$$0\cdot v_1+0\cdot v_2=0.$$

Picture the standard basis: the point in the $xy$-plane closest to the $z$-axis is the origin.

Type six: Markov matrices and steady state

The matrix and its eigenvalues

The lecture's matrix has three structural properties:

• every column is nonnegative and sums to $1$ (Markov);
• column one plus column two equals twice column three, so the columns are dependent and the matrix is singular;
• the trace is $0.8$.

The eigenvalues follow immediately:

• Markov forces $1$ to be an eigenvalue;
• singularity forces $0$;
• the trace is the sum, so the third is $0.8-1-0=-0.2$.

Background: the recorded digits are noisy. One matrix satisfying every stated property is $A=\begin{bmatrix}0.2&0.4&0.3\\0.4&0.2&0.3\\0.4&0.4&0.4\end{bmatrix}.$ Its columns sum to $1$, column one plus column two equals $(0.6,0.6,0.8)=2(0.3,0.3,0.4)$, and its trace is $0.8$.

Steady state

Start from $u_0=(0,10,0)$ and iterate $u_k=A^ku_0$. Expand the initial vector in the eigenbasis:

$$u_k=c_1\lambda_1^kx_1+c_2\lambda_2^kx_2+c_3\lambda_3^kx_3.$$

As $k\to\infty$, the terms with $\lambda_1=0$ and $\lambda_3=-0.2$ vanish, leaving only the $\lambda_2=1$ term:

$$u_\infty=c_2x_2.$$

Solving $(A-I)x=0$ gives

$$x_2=\begin{bmatrix}3\\3\\4\end{bmatrix}.$$

Why the coefficient is one

A Markov matrix preserves total population:

$$\mathbf 1^{\mathsf T}Au=\mathbf 1^{\mathsf T}u.$$

For an eigenvector $x_i$ with $\lambda_i\neq1$,

$$\mathbf 1^{\mathsf T}Ax_i=\lambda_i\mathbf 1^{\mathsf T}x_i=\mathbf 1^{\mathsf T}x_i,$$

so $(\lambda_i-1)\mathbf 1^{\mathsf T}x_i=0$, hence $\mathbf 1^{\mathsf T}x_i=0$: eigenvectors for other eigenvalues have zero component sum.

The initial vector sums to $10$, and $x_2$ also sums to $10$, so $c_2=1$. The steady state is

$$u_\infty=\begin{bmatrix}3\\3\\4\end{bmatrix}.$$

Type seven: construct a matrix with given properties

The third point: $B^{\mathsf T}B$ is always symmetric, so any non-symmetric matrix is impossible.

The fourth point: orthogonal eigenvectors are not exclusive to symmetric matrices. Skew-symmetric and orthogonal matrices also satisfy $AA^{\mathsf T}=A^{\mathsf T}A$, and their eigenvectors can be chosen orthogonal, possibly over the complex numbers.

Type eight: four questions about one least-squares problem

The setup

$$A=\begin{bmatrix}1&0\\1&1\\1&2\end{bmatrix},\qquad b=\begin{bmatrix}3\\4\\1\end{bmatrix}.$$

1. The least-squares solution

$$A^{\mathsf T}A=\begin{bmatrix}3&3\\3&5\end{bmatrix},\qquad A^{\mathsf T}b=\begin{bmatrix}8\\6\end{bmatrix}.$$

Solving the normal equations:

$$\hat x=\begin{bmatrix}11/3\\-1\end{bmatrix}.$$

2. The projection

The projection is the optimal combination:

$$p=A\hat x=\frac{11}{3}\begin{bmatrix}1\\1\\1\end{bmatrix} -\begin{bmatrix}0\\1\\2\end{bmatrix} =\begin{bmatrix}11/3\\8/3\\5/3\end{bmatrix}.$$

3. The corresponding line

The two columns correspond to an intercept and a slope, so this fits

$$y=\frac{11}{3}-t.$$

At $t=0,1,2$ the line gives $11/3,\,8/3,\,5/3$, minimizing the squared error against $3,4,1$.

4. When is the least-squares solution zero

Find a nonzero $b$ whose optimal combination is the zero vector. This is equivalent to $b$ being orthogonal to the whole column space.

The column space is spanned by $(1,1,1)$ and $(0,1,2)$. Take

$$b'=\begin{bmatrix}1\\-2\\1\end{bmatrix},$$

which satisfies

$$\begin{bmatrix}1&1&1\end{bmatrix}b'=0,\qquad \begin{bmatrix}0&1&2\end{bmatrix}b'=0.$$

So $b'$ is orthogonal to both columns, its projection is zero, and the least-squares solution is zero.

One thread through the whole course

Put together, the problems repeat a single pattern:

1. Rank governs everything: solvability, uniqueness, and the kind of inverse all follow from how $r$ compares with $m$ and $n$.
2. The four subspaces are the coordinate system: the null space controls uniqueness, the column space controls solvability, and the row and left null spaces control projection and least squares.
3. Matrix multiplication is a combination of columns: this is where every computation starts.
4. Eigenvalues describe long-term behavior: exponentials for differential equations, powers for Markov chains.
5. Special matrices have special spectra: symmetric (real, orthogonal eigenvectors), skew-symmetric (purely imaginary), orthogonal (unit modulus), projection ($0$ or $1$), Markov ($1$ and no larger).
6. SVD reduces any matrix to diagonal form: pseudoinverses, least squares, and bases for all four subspaces follow from it.

Check your understanding

1. How do "one right side unsolvable, another uniquely solvable" imply $r=n<m$?

Reference answer: The right side is a three-dimensional vector, so $m=3$. The existence of an unsolvable right side $Ax=(1,0,0)^{\mathsf T}$ means $A$ cannot be solved for every $\mathbf b\in\mathbb R^3$; the column space is not all of $\mathbb R^3$, hence $r<m$, i.e. $r<3$. On the other hand $Ax=(0,1,0)^{\mathsf T}$ has exactly one solution, so the null space is trivial (otherwise adding any null-space vector would produce more solutions) and the columns are independent, $r=n$. Combining gives $r=n<3$. Here $n$ may be $1$ or $2$, but it must equal $r$.

2. Why is $A^{\mathsf T}A$ invertible while $AA^{\mathsf T}$ is not positive definite?

Reference answer: $A$ is $3\times n$ with $r=n$, so its columns are independent and $A^{\mathsf T}A$ is $n\times n$ of full rank, hence invertible; it is even positive definite, since $x^{\mathsf T}A^{\mathsf T}Ax=\lVert Ax\rVert^2>0$ for $x\ne0$. But $AA^{\mathsf T}$ is $3\times3$ with rank $r=n<3$, so it is singular, has a zero eigenvalue, and is at most positive semidefinite — never positive definite. The two products differ in both size and rank and cannot be treated interchangeably; they are both invertible only when $A$ is square and invertible.

3. If $v_1-v_2+v_3=0$, is the solution of $Ax=v_1$ unique? Why?

Reference answer: No. The condition says $Ac=v_1-v_2+v_3=0$ for $c=(1,-1,1)^{\mathsf T}$, so $c\in N(A)$ and $c\ne0$ — the null space is nontrivial. The system $Ax=v_1$ has the obvious solution $x=e_1=(1,0,0)^{\mathsf T}$ (the coefficient on the first column); then for every $k$,

$$A(e_1+kc)=v_1+k\cdot0=v_1,$$

giving infinitely many solutions. Uniqueness is equivalent to a trivial null space, which fails here.

4. Why do eigenvectors of a Markov matrix for eigenvalues other than $1$ have zero component sum, and how does that determine the steady-state coefficient?

Reference answer: A Markov matrix has columns summing to $1$, i.e. $\mathbf 1^{\mathsf T}A=\mathbf 1^{\mathsf T}$. For $Ax_i=\lambda_ix_i$, left-multiply by $\mathbf 1^{\mathsf T}$:

$$\lambda_i\mathbf 1^{\mathsf T}x_i=\mathbf 1^{\mathsf T}Ax_i=\mathbf 1^{\mathsf T}x_i,$$

so $(\lambda_i-1)\mathbf 1^{\mathsf T}x_i=0$. If $\lambda_i\ne1$, then $\mathbf 1^{\mathsf T}x_i=0$: the components sum to zero. To get the steady-state coefficient, sum the components on both sides of $u_0=\sum c_ix_i$: every non-$1$ term contributes zero, leaving $c_2\mathbf 1^{\mathsf T}x_2=\mathbf 1^{\mathsf T}u_0$, so $c_2=\mathbf 1^{\mathsf T}u_0/\mathbf 1^{\mathsf T}x_2=10/10=1$. This avoids solving for the coefficients directly.

5. What is the exact condition for the least-squares solution to be zero, and how do you construct such a $b$?

Reference answer: The least-squares solution satisfies the normal equations $A^{\mathsf T}A\hat x=A^{\mathsf T}b$, so $\hat x=0$ exactly when $A^{\mathsf T}b=0$, i.e. $b$ is orthogonal to every column of $A$ — equivalently $b\in N(A^{\mathsf T})$ (the left null space), or $b\perp C(A)$. To construct one, take a nonzero left-null-space vector: here the columns of $A$ are $(1,1,1)^{\mathsf T}$ and $(0,1,2)^{\mathsf T}$, so solve $b_1+b_2+b_3=0$ and $b_2+2b_3=0$; setting $b_3=1$ gives $b_2=-2$, $b_1=1$, so $b'=(1,-2,1)^{\mathsf T}$. Then the projection $p=A\hat x=0$ and the least-squares solution is $\hat x=0$.

Closing

The power of linear algebra is that different problems share the same structure. Rank, subspaces, eigenvalues, and factorizations — master those four, and the computations are just their unfolding.