复习范围

这次测验覆盖三块内容：

1. 正交性：$Q^TQ=I$、投影到直线与子空间、最小二乘、Gram-Schmidt。
2. 行列式：三条定义性质、由它们推出的性质、大公式与代数余子式、逆矩阵公式。
3. 特征值与特征向量（6.1–6.2）：$Ax=\lambda x$、特征方程、对角化与矩阵的幂。

微分方程（6.3）不在这次范围内。

时间轴

题型一：投影到直线

题目：$a=(2,1,2)$，求投影到 $a$ 所在直线的矩阵 $P$。

单列矩阵的投影公式是

$$P=\frac{aa^T}{a^Ta}$$

先算分母：

$$a^Ta=4+1+4=9$$

分子是外积：

$$aa^T= \begin{bmatrix} 4&2&4\\ 2&1&2\\ 4&2&4 \end{bmatrix}$$

所以

$$P=\frac19 \begin{bmatrix} 4&2&4\\ 2&1&2\\ 4&2&4 \end{bmatrix}$$

追问：秩、特征值、特征向量

• 秩为 1，因为每一列都是 $a$ 的倍数。
• 列空间就是 $a$ 所在的直线。
• 秩为 1 说明零空间是二维的，所以 $\lambda=0$ 是二重特征值。
• 剩下的特征值由迹决定：$\operatorname{tr}P=1$，所以第三个特征值是 1。
• $\lambda=1$ 的特征向量就是 $a$ 本身：投影到直线上，$a$ 不动。

也可以用投影矩阵的通用性质：$P^2=P$，所以特征值只能是 0 或 1。

题型二：投影型差分方程

题目：$u_{k+1}=Pu_k$，$u_0=(9,9,0)$，求 $u_k$。

第一步：

$$u_1=Pu_0=a\frac{a^Tu_0}{a^Ta}$$

计算内积：

$$a^Tu_0=2\cdot9+1\cdot9+2\cdot0=27$$

所以

$$u_1=\frac{27}{9}a=3a=\begin{bmatrix}6\\3\\6\end{bmatrix}$$

关键观察：$u_1$ 已经在直线上。再投影一次不会改变它，因为 $P^2=P$。因此

$$u_k=\begin{bmatrix}6\\3\\6\end{bmatrix}\quad(k\ge1)$$

与一般矩阵的对比

若矩阵不是投影矩阵，就必须走完整流程：

1. 求特征值；
2. 求特征向量；
3. 把 $u_0$ 展开成 $c_1x_1+\cdots+c_nx_n$；
4. 得到 $u_k=\sum c_i\lambda_i^k x_i$。

投影矩阵的特殊之处在于：两个特征值为零，对应项从第一步起就消失，只剩 $\lambda=1$ 的那一项。

题型三：过原点的最小二乘

题目：用 $y=Dt$ 拟合点 $(1,4),(2,5),(3,8)$。

写成方程组：

$$\begin{bmatrix}1\\2\\3\end{bmatrix}D = \begin{bmatrix}4\\5\\8\end{bmatrix}$$

这个方程组无解（三个点不共线）。最小二乘解满足正规方程

$$A^TA\hat D=A^Tb$$

计算：

$$A^TA=1^2+2^2+3^2=14$$ $$A^Tb=1\cdot4+2\cdot5+3\cdot8=38$$

所以

$$\hat D=\frac{38}{14}=\frac{19}{7}$$

两张图不要混淆

• ty 平面图：三个数据点与拟合直线 $y=\frac{19}{7}t$。这里没有把点投影到直线上。
• 向量空间图：把 $b=(4,5,8)$ 投影到 $A$ 的列空间。列空间是由 $(1,2,3)$ 张成的一条直线。投影得到的向量是 $A\hat D$。

最小二乘的「投影」发生在第二张图里。

题型四：Gram-Schmidt

题目：$a_1=(1,2,3)$，$a_2=(1,1,1)$，求同一平面内的一组正交基。

保留 $a_1$，把 $a_2$ 中沿 $a_1$ 方向的分量减掉：

$$B=a_2-\frac{a_1^Ta_2}{a_1^Ta_1}a_1$$

计算：

$$a_1^Ta_2=1+2+3=6,\qquad a_1^Ta_1=14$$ $$B=(1,1,1)-\frac{6}{14}(1,2,3) = \left(\frac47,\frac17,-\frac27\right)$$

验证正交性：

$$1\cdot\frac47+2\cdot\frac17+3\cdot\left(-\frac27\right) = \frac47+\frac27-\frac67=0$$

题目只要正交，不必单位化；若要标准正交，再除以 $\|B\|$。

题型五：特征值的性质

设 $4\times4$ 矩阵 $A$ 的特征值为 $\lambda_1,\dots,\lambda_4$。

这些关系都来自两条恒等式：特征值之和等于迹，之积等于行列式。

题型六：三对角全 1 矩阵的递推

设 $A_n$ 是 $n$ 阶三对角矩阵，三条对角线全为 1。记 $D_n=\det A_n$。

递推公式

沿第一行展开。第一个 1 的余子式是 $D_{n-1}$。第二个 1 位于 $(1,2)$，符号为负；它的余子式再沿第一列展开，得到 $D_{n-2}$。因此

$$D_n=D_{n-1}-D_{n-2}$$

初值：

$$D_1=1,\qquad D_2=\begin{vmatrix}1&1\\1&1\end{vmatrix}=0$$

化成一阶系统

令向量 $\begin{bmatrix}D_n\\D_{n-1}\end{bmatrix}$，则

$$\begin{bmatrix}D_n\\D_{n-1}\end{bmatrix} = \begin{bmatrix}1&-1\\1&0\end{bmatrix} \begin{bmatrix}D_{n-1}\\D_{n-2}\end{bmatrix}$$

特征值

$$\det\begin{bmatrix}1-\lambda&-1\\1&-\lambda\end{bmatrix} = \lambda^2-\lambda+1$$

解得

$$\lambda=\frac{1\pm i\sqrt3}{2}=e^{\pm i\pi/3}$$

两个特征值的模都是 1，所以 $D_n$ 不收敛也不发散，而是周期变化。又因为 $\lambda^6=1$，矩阵的六次方是单位矩阵，序列周期为 6。

这与前面 Fibonacci 的例子形成对照：Fibonacci 的递推是 $D_n=D_{n-1}+D_{n-2}$，特征值一实一虚、模不为 1，所以指数增长；这里中间是减号，特征值落在单位圆上，所以周期。

题型七：矩阵族的可逆性

设 $A_3$ 为

$$A_3=\begin{bmatrix}0&1&0\\1&0&2\\0&2&0\end{bmatrix}$$

第三列是第一列的 2 倍，所以奇异。

特征多项式：

$$\det(A_3-\lambda I)=-\lambda^3+5\lambda=-\lambda(\lambda^2-5)$$

特征值为

$$0,\ \sqrt5,\ -\sqrt5$$

迹为 0，与特征值之和一致。

设 $A_4$ 为

$$A_4=\begin{bmatrix} 0&1&0&0\\ 1&0&2&0\\ 0&2&0&3\\ 0&0&3&0 \end{bmatrix}$$

用余子式展开算行列式，得到 $\det A_4=9\ne0$，所以可逆。可逆意味着列空间是整个 $\mathbb{R}^4$，投影到列空间的矩阵就是单位矩阵。

一般规律（课堂上的猜测）：奇数阶奇异，偶数阶可逆，行列式有规律可循。

核心回顾

• 投影到直线：$P=aa^T/a^Ta$，秩 1，特征值 1 与 0。
• 投影型差分方程：一次投影后就到达不动点。
• 最小二乘：正规方程 $A^TA\hat x=A^Tb$；投影发生在 $b$ 所在空间。
• Gram-Schmidt：减去在已有方向上的投影。
• 特征值性质：可逆看零、逆取倒数、加 $cI$ 平移。
• 三对角递推：$D_n=D_{n-1}-D_{n-2}$，特征值在单位圆上，周期 6。
• 矩阵族：先判断可逆，再谈列空间与投影。

常见误区

1. 在最小二乘里把数据点投影到拟合直线上。投影发生在向量空间，不在散点图。
2. Gram-Schmidt 忘记减去投影，只做正交化的一半。
3. 算完特征值不验证迹与行列式。这是最快的自检。
4. 把三对角递推的符号写反。中间的减号来自 $(1,2)$ 位置的负号。
5. 看到矩阵族就直接求投影矩阵。先判断可逆性，可逆时答案可能平凡。

理解检验

1. 为什么投影矩阵的零空间维数等于 $n-1$？

参考答案：$P=\dfrac{aa^T}{a^Ta}$，每一列都是 $a$ 的倍数，所以 $\operatorname{rank}P=1$（$a\ne0$）。由秩–零化度定理，

$$\dim N(P)=n-\operatorname{rank}P=n-1 .$$

几何上，$P$ 把向量投影到直线 $\operatorname{span}\{a\}$ 上，只有该方向在列空间里；垂直于 $a$ 的 $n-1$ 维超平面整体被压成零。对应地，$\lambda=0$ 的代数重数是 $n-1$，$\lambda=1$ 的重数是 1。

2. 若 $u_0$ 已经在列空间内，$u_1$ 与 $u_0$ 有什么关系？

参考答案：$u_0\in C(P)=\operatorname{span}\{a\}$，投影到列空间不动，所以 $u_1=Pu_0=u_0$。由 $P^2=P$，之后每一步都不再变化：$u_k=u_0$ 对所有 $k\ge0$ 成立。在特征向量分解中，$u_0$ 只含 $\lambda=1$ 的分量，两个 $\lambda=0$ 分量系数为零，所以没有任何衰减项。

3. 用正规方程求 $y=Dt$ 过原点的拟合，并解释 $\hat D$ 的单位。

参考答案：$A=(1,2,3)^T$，$b=(4,5,8)^T$。正规方程 $A^TA\hat D=A^Tb$：

$$A^TA=1+4+9=14,\qquad A^Tb=1\cdot4+2\cdot5+3\cdot8=38,$$

所以 $\hat D=\frac{38}{14}=\frac{19}{7}$。单位是「$y$ 的单位除以 $t$ 的单位」：若 $y$ 用米、$t$ 用秒，$\hat D\approx2.71\ \mathrm{m/s}$，正是直线的斜率（速度）。过原点的模型没有截距，只有一个参数。

4. 写出三对角矩阵序列的前六项行列式。

参考答案：递推 $D_n=D_{n-1}-D_{n-2}$，初值 $D_1=1$、$D_2=0$：

$$D_1=1,\quad D_2=0,\quad D_3=0-1=-1,\quad D_4=-1-0=-1,\quad D_5=-1-(-1)=0,\quad D_6=0-(-1)=1 .$$

序列是 $1,0,-1,-1,0,1,\dots$，周期 6；$D_6=D_1=1$ 与特征值落在单位圆上、满足 $\lambda^6=1$ 一致。

5. $A_4$ 可逆，为什么它到列空间的投影矩阵是 $I$？

参考答案：$A_4$ 是 $4\times4$ 且 $\det A_4=9\ne0$，所以它的四列线性无关，列空间是整个 $\mathbb{R}^4$。任何向量都已经在列空间里，投影不改变它。公式验证：

$$P=A(A^TA)^{-1}A^T=A\,A^{-1}(A^T)^{-1}A^T=I .$$

推广：$m\ge n$ 且列满秩时，$A(A^TA)^{-1}A^T$ 是投影到 $C(A)$ 的矩阵；只有 $A$ 为可逆方阵时它才恰好等于 $I$。

与前后课程的关系

复习课把正交、行列式、特征值三条线并置。接下来的课程进入对称矩阵与正定性：实特征值、正交特征向量、二次型的符号，都会在这里汇合。

Scope

The quiz covers three blocks:

1. Orthogonality: $Q^TQ=I$, projections onto lines and subspaces, least squares, Gram-Schmidt.
2. Determinants: the three defining properties, their consequences, the big formula and cofactors, the inverse formula.
3. Eigenvalues and eigenvectors (6.1–6.2): $Ax=\lambda x$, the characteristic equation, diagonalization and powers.

Differential equations (6.3) are not on this quiz.

Timeline

Problem type 1: projection onto a line

Problem: For $a=(2,1,2)$, find the matrix $P$ that projects onto the line through $a$.

For a single-column matrix the formula is

$$P=\frac{aa^T}{a^Ta}$$

The denominator:

$$a^Ta=4+1+4=9$$

The numerator is the outer product:

$$aa^T= \begin{bmatrix} 4&2&4\\ 2&1&2\\ 4&2&4 \end{bmatrix}$$

Hence

$$P=\frac19 \begin{bmatrix} 4&2&4\\ 2&1&2\\ 4&2&4 \end{bmatrix}$$

Rank, eigenvalues, eigenvectors

• The rank is 1, since every column is a multiple of $a$.
• The column space is the line through $a$.
• Rank 1 means the null space is two-dimensional, so $\lambda=0$ has multiplicity two.
• The remaining eigenvalue comes from the trace: $\operatorname{tr}P=1$, so the third eigenvalue is 1.
• The eigenvector for $\lambda=1$ is $a$ itself: a vector on the line does not move under projection.

Equivalently, use $P^2=P$, which forces every eigenvalue to be 0 or 1.

Problem type 2: a projection difference equation

Problem: Solve $u_{k+1}=Pu_k$ with $u_0=(9,9,0)$.

First step:

$$u_1=Pu_0=a\frac{a^Tu_0}{a^Ta}$$

The inner product:

$$a^Tu_0=2\cdot9+1\cdot9+2\cdot0=27$$

so

$$u_1=\frac{27}{9}a=3a=\begin{bmatrix}6\\3\\6\end{bmatrix}$$

The key observation: $u_1$ already lies on the line, and $P^2=P$, so projecting again changes nothing. Therefore

$$u_k=\begin{bmatrix}6\\3\\6\end{bmatrix}\quad(k\ge1)$$

Contrast with a general matrix

For a general matrix, the full procedure is:

1. find the eigenvalues;
2. find the eigenvectors;
3. expand $u_0=c_1x_1+\cdots+c_nx_n$;
4. write $u_k=\sum c_i\lambda_i^k x_i$.

For a projection, two eigenvalues are zero, so those terms vanish immediately and only the $\lambda=1$ term remains.

Problem type 3: least squares through the origin

Problem: Fit $y=Dt$ to $(1,4),(2,5),(3,8)$.

As a system:

$$\begin{bmatrix}1\\2\\3\end{bmatrix}D = \begin{bmatrix}4\\5\\8\end{bmatrix}$$

which has no solution, since the three points are not collinear. The least-squares solution satisfies the normal equation

$$A^TA\hat D=A^Tb$$

Compute:

$$A^TA=1^2+2^2+3^2=14$$ $$A^Tb=1\cdot4+2\cdot5+3\cdot8=38$$

so

$$\hat D=\frac{38}{14}=\frac{19}{7}$$

Two pictures, not one

• The $ty$-plane picture: the three data points and the fitted line $y=\frac{19}{7}t$. Nothing is projected onto the line here.
• The vector-space picture: $b=(4,5,8)$ is projected onto the column space of $A$, which is the line spanned by $(1,2,3)$ in $\mathbb{R}^3$. The projection is $A\hat D$.

The word "projection" in least squares belongs to the second picture.

Problem type 4: Gram-Schmidt

Problem: For $a_1=(1,2,3)$ and $a_2=(1,1,1)$, find an orthogonal basis of their plane.

Keep $a_1$ and subtract from $a_2$ its component along $a_1$:

$$B=a_2-\frac{a_1^Ta_2}{a_1^Ta_1}a_1$$

Compute:

$$a_1^Ta_2=1+2+3=6,\qquad a_1^Ta_1=14$$ $$B=(1,1,1)-\frac{6}{14}(1,2,3) = \left(\frac47,\frac17,-\frac27\right)$$

Check orthogonality:

$$1\cdot\frac47+2\cdot\frac17+3\cdot\left(-\frac27\right) = \frac47+\frac27-\frac67=0$$

Only orthogonality was requested; normalize by dividing by $\|B\|$ if an orthonormal basis is needed.

Problem type 5: eigenvalue identities

Let a $4\times4$ matrix $A$ have eigenvalues $\lambda_1,\dots,\lambda_4$.

All three follow from the two identities: eigenvalues sum to the trace and multiply to the determinant.

Problem type 6: the tridiagonal recurrence

Let $A_n$ be the $n\times n$ tridiagonal matrix with ones on all three diagonals, and let $D_n=\det A_n$.

The recurrence

Expand along the first row. The first 1 has cofactor $D_{n-1}$. The second 1 sits at $(1,2)$ with a minus sign; expanding its cofactor along the first column leaves $D_{n-2}$. Therefore

$$D_n=D_{n-1}-D_{n-2}$$

Initial values:

$$D_1=1,\qquad D_2=\begin{vmatrix}1&1\\1&1\end{vmatrix}=0$$

A first-order system

With the vector $\begin{bmatrix}D_n\\D_{n-1}\end{bmatrix}$,

$$\begin{bmatrix}D_n\\D_{n-1}\end{bmatrix} = \begin{bmatrix}1&-1\\1&0\end{bmatrix} \begin{bmatrix}D_{n-1}\\D_{n-2}\end{bmatrix}$$

Eigenvalues

$$\det\begin{bmatrix}1-\lambda&-1\\1&-\lambda\end{bmatrix} = \lambda^2-\lambda+1$$

so

$$\lambda=\frac{1\pm i\sqrt3}{2}=e^{\pm i\pi/3}$$

Both eigenvalues have modulus 1, so $D_n$ neither converges nor blows up; it is periodic. Since $\lambda^6=1$, the sixth power of the matrix is the identity and the sequence has period six.

Contrast with Fibonacci: there the recurrence is $D_n=D_{n-1}+D_{n-2}$, the eigenvalues are real and one has modulus above 1, so the sequence grows exponentially. Here the minus sign places the eigenvalues on the unit circle, producing periodicity.

Problem type 7: a family of matrices

Let

$$A_3=\begin{bmatrix}0&1&0\\1&0&2\\0&2&0\end{bmatrix}$$

The third column is twice the first, so the matrix is singular.

The characteristic polynomial:

$$\det(A_3-\lambda I)=-\lambda^3+5\lambda=-\lambda(\lambda^2-5)$$

so the eigenvalues are

$$0,\ \sqrt5,\ -\sqrt5$$

Their sum is 0, matching the trace.

Now let

$$A_4=\begin{bmatrix} 0&1&0&0\\ 1&0&2&0\\ 0&2&0&3\\ 0&0&3&0 \end{bmatrix}$$

Cofactor expansion gives $\det A_4=9\ne0$, so $A_4$ is invertible. Its column space is all of $\mathbb{R}^4$, and the projection onto it is the identity matrix.

The lecture's conjecture: odd sizes are singular, even sizes are invertible, and the determinants follow a pattern.

Core review

• Projection onto a line: $P=aa^T/a^Ta$, rank 1, eigenvalues 1 and 0.
• A projection difference equation reaches its fixed point after one step.
• Least squares: solve $A^TA\hat x=A^Tb$; the projection happens in the space where $b$ lives.
• Gram-Schmidt subtracts the component along directions already chosen.
• Eigenvalue identities: invertibility means no zero; inverses invert; adding $cI$ shifts.
• Tridiagonal recurrence: $D_n=D_{n-1}-D_{n-2}$, eigenvalues on the unit circle, period six.
• For a matrix family, decide invertibility first; the projection may then be trivial.

Common misconceptions

1. Projecting the data points onto the fitted line in least squares. The projection is in vector space, not in the scatter plot.
2. Skipping the subtraction in Gram-Schmidt.
3. Not checking eigenvalues against trace and determinant. It is the fastest sanity check.
4. Flipping the sign in the tridiagonal recurrence. The minus comes from the $(1,2)$ position.
5. Computing a projection matrix for a matrix family before checking invertibility. If it is invertible, the answer is often trivial.

Check your understanding

1. Why is the null space of a rank-one projection $(n-1)$-dimensional?

Reference answer: $P=\dfrac{aa^T}{a^Ta}$ has every column a multiple of $a$, so $\operatorname{rank}P=1$ (for $a\ne0$). By rank–nullity,

$$\dim N(P)=n-\operatorname{rank}P=n-1 .$$

Geometrically, $P$ projects onto the line $\operatorname{span}\{a\}$; only that direction survives, while the $(n-1)$-dimensional hyperplane perpendicular to $a$ is collapsed to zero. Correspondingly $\lambda=0$ has algebraic multiplicity $n-1$ and $\lambda=1$ has multiplicity 1.

2. If $u_0$ already lies in the column space, how do $u_1$ and $u_0$ compare?

Reference answer: Since $u_0\in C(P)=\operatorname{span}\{a\}$, projecting onto the column space leaves it unchanged, so $u_1=Pu_0=u_0$. Because $P^2=P$, nothing changes afterwards: $u_k=u_0$ for all $k\ge0$. In the eigenvector expansion, $u_0$ contains only the $\lambda=1$ component; the coefficients of the two $\lambda=0$ components are zero, so no decaying term remains.

3. Solve the through-origin fit by the normal equation and interpret the units of $\hat D$.

Reference answer: $A=(1,2,3)^T$, $b=(4,5,8)^T$. The normal equation $A^TA\hat D=A^Tb$ gives

$$A^TA=1+4+9=14,\qquad A^Tb=1\cdot4+2\cdot5+3\cdot8=38,$$

so $\hat D=\frac{38}{14}=\frac{19}{7}$. Its unit is "$y$ unit per $t$ unit": if $y$ is in metres and $t$ in seconds, $\hat D\approx2.71\ \mathrm{m/s}$, the slope (velocity) of the line. The through-origin model has no intercept and only this one parameter.

4. Write out the first six determinants of the tridiagonal sequence.

Reference answer: With $D_n=D_{n-1}-D_{n-2}$, $D_1=1$ and $D_2=0$:

$$D_1=1,\quad D_2=0,\quad D_3=0-1=-1,\quad D_4=-1-0=-1,\quad D_5=-1-(-1)=0,\quad D_6=0-(-1)=1 .$$

The sequence $1,0,-1,-1,0,1,\dots$ has period 6; $D_6=D_1=1$ agrees with the eigenvalues lying on the unit circle and satisfying $\lambda^6=1$.

5. Why is the projection onto the column space of $A_4$ the identity?

Reference answer: $A_4$ is $4\times4$ with $\det A_4=9\ne0$, so its four columns are linearly independent and its column space is all of $\mathbb{R}^4$. Every vector already lies in the column space, so projection changes nothing. In formula form,

$$P=A(A^TA)^{-1}A^T=A\,A^{-1}(A^T)^{-1}A^T=I .$$

More generally, for full column rank with $m\ge n$, $A(A^TA)^{-1}A^T$ projects onto $C(A)$; it equals $I$ exactly when $A$ is an invertible square matrix.

How this connects

The review places orthogonality, determinants, and eigenvalues side by side. The next lectures turn to symmetric matrices and positive definiteness, where real eigenvalues, orthogonal eigenvectors, and quadratic forms all meet.