这一讲覆盖什么

第三次测验覆盖第六章：特征值与特征向量、微分方程组、对称矩阵、正定矩阵、相似矩阵、SVD。第七章的线性变换不在此次范围，留到期末。

复习不是重讲定义，而是用旧题演示看到某种条件时应该先想到什么。下面按题型整理。

时间轴

题型一：微分方程组与反对称矩阵

题目结构

给定一个 $3\times3$ 矩阵 $A$，求 $\dfrac{du}{dt}=Au$ 的通解，并判断解的长期行为与周期性。

解题顺序

1. 先看矩阵是否奇异。若有线性相关的行或列，立即得到一个特征值 $0$。
2. 求其余特征值。解 $\det(A-\lambda I)=0$。
3. 写通解：

$$u(t)=c_1e^{\lambda_1t}x_1+c_2e^{\lambda_2t}x_2+c_3e^{\lambda_3t}x_3.$$

4. 用初值定常数。

例题

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

第一行与第三行互为相反数，行列式为零，所以 $\lambda_1=0$。

这个矩阵满足

$$A^{\mathsf T}=-A,$$

是反对称矩阵。反对称矩阵的特征值一定是纯虚数，因此另外两个特征值形如 $\pm i\omega$。

直接计算特征多项式：

$$\det(A-\lambda I) =-\lambda(\lambda^2+2).$$

所以

$$\lambda_1=0,\qquad \lambda_2=i\sqrt2,\qquad \lambda_3=-i\sqrt2.$$

通解中，$e^{0t}=1$ 不增不减；$e^{\pm i\sqrt2\,t}$ 的模长恒为 $1$，在单位圆上转动。因此解既不爆炸也不衰减，而是振荡。

周期性

若存在 $T>0$ 使所有 $e^{\lambda_iT}=1$，解就回到初值。对 $\lambda=i\sqrt2$：

$$e^{i\sqrt2\,T}=1 \iff \sqrt2\,T=2\pi \iff T=\pi\sqrt2.$$

这也是为什么反对称矩阵对应「能量守恒」的振荡系统：$A^{\mathsf T}=-A$ 时，$\lVert u(t)\rVert^2$ 对时间求导为零。

补充：若想得到实的解，可把 $e^{\pm i\omega t}$ 组合成 $\cos\omega t$ 与 $\sin\omega t$；这与纯虚特征值的结论一致。

矩阵指数

通解也可以写成

$$u(t)=e^{At}u(0).$$

若 $A$ 可对角化，$A=S\Lambda S^{-1}$，则

$$e^{At}=Se^{\Lambda t}S^{-1},$$

其中 $e^{\Lambda t}$ 只是对角元取 $e^{\lambda_it}$ 的对角矩阵。注意这个公式依赖「有足够多特征向量」；不可对角化时不能直接用。

题型二：由特征值与特征向量反推矩阵性质

题目结构

给出一个 $3\times3$ 矩阵的特征值与特征向量（其中一个特征值含未知参数 $c$），问：是否可对角化？何时对称？是否正定？是否 Markov？$\frac12A$ 是否投影矩阵？

核心判断链

特征向量是否足够，决定可对角化。 题目给出的三个特征向量正交，因而线性无关。所以无论 $c$ 取什么值，矩阵都可对角化。

特征向量正交 + 特征值全为实数，等价于对称。 题目已给正交特征向量，所以只要 $c$ 是实数，矩阵就是对称的。

正定性看特征值是否全为正。 这里有一个特征值为 $0$，所以永远不可能正定；若 $c\ge0$，最多是半正定。

Markov 矩阵的特征值满足 $|\lambda|\le1$，且有一个特征值等于 $1$。 题目中有一个特征值为 $2$，所以永远不是 Markov 矩阵。

投影矩阵的特征值只能是 $0$ 或 $1$，且 $P^2=P$、$P^{\mathsf T}=P$。 问 $\frac12A$ 是否为投影，就是问 $A$ 的特征值除以 $2$ 后是否只落在 $\{0,1\}$。因此 $c=0$ 或 $c=2$ 时成立。

可复用的判断表

题型三：SVD 的正用与反用

正用：给矩阵，求分解

1. 算 $A^{\mathsf T}A$；
2. 求它的特征值与特征向量，归一化得到 $V$ 与 $\sigma_i=\sqrt{\lambda_i}$；
3. 由 $u_i=\dfrac{Av_i}{\sigma_i}$ 求 $U$（这正是修正符号的正确做法）；
4. 也可以用 $AA^{\mathsf T}$ 的特征向量求 $U$，但符号必须用 $Av_i=\sigma_iu_i$ 核对。

反用：给分解，问矩阵

• 若 $\Sigma=\operatorname{diag}(3,2)$，两个奇异值都非零，矩阵可逆，秩为 $2$。
• 若 $\Sigma=\operatorname{diag}(3,0)$，矩阵奇异，秩为 $1$，零空间由 $v_2$ 张成，左零空间由 $u_2$ 张成。
• 若出现负的「奇异值」，那不是 SVD——奇异值按定义非负。

符号问题的根源

$A^{\mathsf T}A$ 只确定 $V$ 的每一列到一个符号。任取一个符号后，$U$ 必须由

$$Av_i=\sigma_iu_i$$

确定，而不是独立地对 $AA^{\mathsf T}$ 取特征向量。否则 $U\Sigma V^{\mathsf T}$ 可能整体差一个符号。

题型四：对称且正交的矩阵

题目结构

设 $A$ 既对称又正交，判断一系列命题，并证明 $\frac12(A+I)$ 是投影矩阵。

特征值的范围

• 对称 $\Rightarrow$ 特征值为实数；
• 正交 $\Rightarrow$ 特征值模长为 $1$。

两者结合：

$$\lambda\in\{1,-1\}.$$

逐条判断

证明 $\frac12(A+I)$ 是投影

投影矩阵需要两个性质：对称，以及平方等于自身。

对称性显然。平方：

$$\left(\frac{A+I}{2}\right)^2 =\frac{A^2+2A+I}{4}.$$

因为 $A$ 正交，$A^{\mathsf T}A=I$；又因为 $A$ 对称，$A^{\mathsf T}=A$，所以

$$A^2=A^{\mathsf T}A=I.$$

代入：

$$\frac{I+2A+I}{4}=\frac{A+I}{2}.$$

所以它确实是投影矩阵。

补充：用特征值看更直接。$A$ 的特征值是 $\pm1$，则 $A+I$ 的特征值是 $0$ 或 $2$，除以 $2$ 后是 $0$ 或 $1$——正是投影矩阵的特征值。

复习清单

考前可以按这个顺序自查：

1. 看到微分方程，先问矩阵是否奇异、是否对称/反对称。
2. 看到「特征向量正交」，立刻想到对称、投影、SVD 的可能。
3. 看到「是否正定」，先看有没有零或负特征值，再考虑主元与行列式。
4. 看到 SVD，先写 $A^{\mathsf T}A$，再用 $Av_i=\sigma_iu_i$ 定符号。
5. 看到「对称 + 正交」，特征值只能是 $\pm1$。

理解检验

1. 为什么反对称矩阵的特征值一定是纯虚数？解的模长为什么不变？

参考答案：设 $A^{\mathsf T}=-A$，$Ax=\lambda x$，$x\ne0$。左乘 $x^H$：

$$\lambda\,x^Hx=x^HAx .$$

对左边取共轭转置（它是标量，等于自身的共轭）：

$$\overline{x^HAx}=x^HA^Hx=x^HA^{\mathsf T}x=-x^HAx .$$

所以 $x^HAx$ 是纯虚数（或零），而 $x^Hx=\lVert x\rVert^2>0$，故 $\lambda$ 是纯虚数。模长不变看能量：$\frac{d}{dt}\lVert u\rVert^2=2u^{\mathsf T}u'=2u^{\mathsf T}Au$，而 $u^{\mathsf T}Au$ 是标量且 $(u^{\mathsf T}Au)^{\mathsf T}=u^{\mathsf T}A^{\mathsf T}u=-u^{\mathsf T}Au$，所以它恒为零。因此 $\lVert u(t)\rVert^2$ 是常数，解在球面上运动，既不爆炸也不衰减。

2. 一个矩阵有 $n$ 个正交特征向量，能否推出它对称？还需要什么条件？

参考答案：还不能，必须加上「全部特征值为实数」。若 $n$ 个特征向量两两正交（可再单位化），把它们作为列组成正交矩阵 $Q$，则 $A=Q\Lambda Q^{-1}=Q\Lambda Q^{\mathsf T}$。此时 $A^{\mathsf T}=Q\Lambda^{\mathsf T}Q^{\mathsf T}$；只有当 $\Lambda$ 是实对角矩阵时 $\Lambda^{\mathsf T}=\Lambda$，才有 $A^{\mathsf T}=A$。若特征值含复数，$\Lambda^{\mathsf T}=\bar\Lambda\ne\Lambda$，$A$ 一般不对称（实矩阵甚至不可能有实正交特征向量对应复特征值）。所以完整条件组合是：对称 $\iff$ 有 $n$ 个正交特征向量且特征值全为实（谱定理）。

3. 为什么 $\Sigma$ 中出现负数就说明分解写错了？

参考答案：奇异值定义为 $\sigma_i=\sqrt{\lambda_i(A^{\mathsf T}A)}$，$\lambda_i\ge0$，所以 $\sigma_i>0$ 或 $=0$，按定义不可能为负；标准形式还要求 $\sigma_1\ge\sigma_2\ge\cdots$。若计算中出现负「奇异值」，通常是把 $V$ 某列的符号处理错了：$A^{\mathsf T}A$ 只确定 $v_i$ 到相差一个符号，改变 $v_i$ 的符号时，应同时改变 $u_i$ 的符号以保持 $A=\sum\sigma_iu_iv_i^{\mathsf T}$，而不是把 $\sigma_i$ 变成负数。负号属于 $U$ 或 $V$，不属于 $\Sigma$。

4. 若 $A$ 对称且正交，$A+I$ 的特征值是什么？$\frac12(A+I)$ 的秩是多少？

参考答案：对称给出实特征值，正交给出模长为 $1$，所以 $A$ 的特征值是 $\pm1$。于是 $A+I$ 的特征值是 $0$（对应 $-1$）或 $2$（对应 $+1$），$\frac12(A+I)$ 的特征值是 $0$ 或 $1$。它是投影矩阵：对称，且

$$\left(\frac{A+I}{2}\right)^2=\frac{A^2+2A+I}{4}=\frac{I+2A+I}{4}=\frac{A+I}{2},$$

用了 $A$ 正交且对称导致的 $A^2=I$。投影矩阵的秩等于迹，也等于特征值 $1$ 的个数，所以 $\operatorname{rank}\frac12(A+I)$ 等于 $A$ 的特征值 $+1$ 的重数（若 $A=I$ 则为 $n$，若 $A=-I$ 则为 $0$）。

5. 已知 $A=U\Sigma V^{\mathsf T}$，如何直接从 $A$ 与 $V$ 读出 $U$，而不解 $AA^{\mathsf T}$？

参考答案：对每个 $\sigma_i>0$ 的列，直接用

$$u_i=\frac{Av_i}{\sigma_i}.$$

这是因为 $Av_i=\sigma_iu_i$，且 $\sigma_i>0$ 时除法有意义；这样得到的 $u_i$ 自动为单位向量（$\lVert Av_i\rVert=\sigma_i\lVert v_i\rVert=\sigma_i$），并且自动与 $v_i$ 的符号配套，不会出现符号不一致的问题。对 $\sigma_i=0$（$i>r$）的列，$Av_i=0$，无法由此确定 $u_i$，需要在左零空间中任取一组标准正交基补齐，使 $U$ 成为 $m\times m$ 正交矩阵。

与前后课程的关系

这一讲把第六章的题型集中演练。下一讲处理左逆、右逆与伪逆，把 SVD 用于非方阵的「求逆」；最后一讲做全课程总结，并把第七章的线性变换纳入期末范围。

What this review covers

The third quiz covers Chapter 6: eigenvalues and eigenvectors, systems of differential equations, symmetric matrices, positive definiteness, similar matrices, and SVD. Chapter 7 on linear transformations appears only on the final.

The review does not restate definitions. It works old exam problems to show what to look for first given a certain condition. Below, the problems are grouped by type.

Timeline

Type one: differential equations and skew-symmetric matrices

Problem shape

Given a $3\times3$ matrix $A$, find the general solution of $\dfrac{du}{dt}=Au$, then describe its long-term behavior and periodicity.

Solution order

1. Check whether $A$ is singular. Dependent rows or columns immediately give an eigenvalue $0$.
2. Find the remaining eigenvalues from $\det(A-\lambda I)=0$.
3. Write the general solution:

$$u(t)=c_1e^{\lambda_1t}x_1+c_2e^{\lambda_2t}x_2+c_3e^{\lambda_3t}x_3.$$

4. Match the initial condition.

The worked example

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

The first and third rows are negatives of each other, so the determinant is zero and $\lambda_1=0$.

The matrix satisfies

$$A^{\mathsf T}=-A,$$

so it is skew-symmetric. Eigenvalues of a skew-symmetric matrix are purely imaginary, so the other two have the form $\pm i\omega$.

Computing directly:

$$\det(A-\lambda I) =-\lambda(\lambda^2+2).$$

Hence

$$\lambda_1=0,\qquad \lambda_2=i\sqrt2,\qquad \lambda_3=-i\sqrt2.$$

In the solution, $e^{0t}=1$ neither grows nor decays; $e^{\pm i\sqrt2\,t}$ has magnitude one and moves around the unit circle. The solution neither blows up nor decays; it oscillates.

Periodicity

If some $T>0$ makes every $e^{\lambda_iT}=1$, the solution returns to its initial value. For $\lambda=i\sqrt2$:

$$e^{i\sqrt2\,T}=1 \iff \sqrt2\,T=2\pi \iff T=\pi\sqrt2.$$

This is why skew-symmetric matrices describe conservative oscillation: when $A^{\mathsf T}=-A$, the derivative of $\lVert u(t)\rVert^2$ is zero.

Background: real solutions are obtained by combining $e^{\pm i\omega t}$ into $\cos\omega t$ and $\sin\omega t$, consistent with purely imaginary eigenvalues.

The matrix exponential

The solution can also be written

$$u(t)=e^{At}u(0).$$

If $A$ is diagonalizable, $A=S\Lambda S^{-1}$, then

$$e^{At}=Se^{\Lambda t}S^{-1},$$

where $e^{\Lambda t}$ is diagonal with entries $e^{\lambda_it}$. The formula requires enough eigenvectors; it does not apply to non-diagonalizable matrices.

Type two: inferring matrix properties from spectrum

Problem shape

A $3\times3$ matrix is given by its eigenvalues (one containing a parameter $c$) and its eigenvectors. Ask: diagonalizable? symmetric? positive definite? Markov? Is $\frac12A$ a projection?

The decision chain

Enough eigenvectors determine diagonalizability. The three given eigenvectors are orthogonal, hence independent. So the matrix is diagonalizable for every $c$.

Orthogonal eigenvectors with all-real eigenvalues means symmetric. Since the eigenvectors are already orthogonal, any real $c$ makes the matrix symmetric.

Positive definiteness requires all eigenvalues positive. One eigenvalue is zero, so the matrix is never positive definite; if $c\ge0$, it is at most positive semidefinite.

A Markov matrix has $|\lambda|\le1$ and an eigenvalue equal to $1$. Here one eigenvalue is $2$, so it is never Markov.

A projection matrix has eigenvalues only $0$ or $1$, and satisfies $P^2=P$, $P^{\mathsf T}=P$. Asking whether $\frac12A$ is a projection means asking whether halving the eigenvalues leaves only $0$ and $1$. So $c=0$ or $c=2$.

Reusable table

Type three: SVD, forward and backward

Forward: matrix to factorization

1. Compute $A^{\mathsf T}A$.
2. Find its eigenvalues and eigenvectors; normalize to get $V$ and $\sigma_i=\sqrt{\lambda_i}$.
3. Get $U$ from $u_i=\dfrac{Av_i}{\sigma_i}$ — this is also the correct way to fix signs.
4. One may instead use eigenvectors of $AA^{\mathsf T}$, but the signs must be checked against $Av_i=\sigma_iu_i$.

Backward: factorization to matrix

• If $\Sigma=\operatorname{diag}(3,2)$, both singular values are nonzero: the matrix is invertible with rank $2$.
• If $\Sigma=\operatorname{diag}(3,0)$, the matrix is singular with rank $1$; the null space is spanned by $v_2$ and the left null space by $u_2$.
• A negative "singular value" is not an SVD at all — singular values are nonnegative by definition.

Where the sign problem comes from

$A^{\mathsf T}A$ determines each column of $V$ only up to sign. Once a sign is chosen, $U$ must be fixed by

$$Av_i=\sigma_iu_i,$$

not chosen independently as eigenvectors of $AA^{\mathsf T}$. Otherwise $U\Sigma V^{\mathsf T}$ may differ by an overall sign.

Type four: symmetric and orthogonal matrices

Problem shape

Let $A$ be both symmetric and orthogonal. Decide several statements and prove that $\frac12(A+I)$ is a projection.

The eigenvalue range

• Symmetric implies real eigenvalues.
• Orthogonal implies eigenvalues of magnitude $1$.

Together:

$$\lambda\in\{1,-1\}.$$

Statement by statement

Proving $\frac12(A+I)$ is a projection

A projection needs two properties: symmetry and idempotence.

Symmetry is immediate. For the square:

$$\left(\frac{A+I}{2}\right)^2 =\frac{A^2+2A+I}{4}.$$

Because $A$ is orthogonal, $A^{\mathsf T}A=I$; because it is symmetric, $A^{\mathsf T}=A$. Hence

$$A^2=A^{\mathsf T}A=I.$$

Substituting:

$$\frac{I+2A+I}{4}=\frac{A+I}{2}.$$

So it is indeed a projection.

Background: eigenvalues give a shorter view. The eigenvalues of $A$ are $\pm1$, so those of $A+I$ are $0$ or $2$, and halving gives $0$ or $1$ — exactly the spectrum of a projection.

A checklist for the quiz

1. For a differential equation, first ask whether the matrix is singular and whether it is symmetric or skew-symmetric.
2. On seeing "orthogonal eigenvectors", think of symmetric matrices, projections, and SVD.
3. On seeing "positive definite", check for zero or negative eigenvalues before pivots or determinants.
4. For SVD, compute $A^{\mathsf T}A$ first, then fix signs with $Av_i=\sigma_iu_i$.
5. For "symmetric and orthogonal", the eigenvalues can only be $\pm1$.

Check your understanding

1. Why are the eigenvalues of a skew-symmetric matrix purely imaginary, and why is the solution's norm constant?

Reference answer: Let $A^{\mathsf T}=-A$, $Ax=\lambda x$ with $x\ne0$. Left-multiply by $x^H$:

$$\lambda\,x^Hx=x^HAx .$$

Take the conjugate transpose of the scalar $x^HAx$:

$$\overline{x^HAx}=x^HA^Hx=x^HA^{\mathsf T}x=-x^HAx .$$

Hence $x^HAx$ is purely imaginary (or zero), while $x^Hx=\lVert x\rVert^2>0$, so $\lambda$ is purely imaginary. For the norm, $\frac{d}{dt}\lVert u\rVert^2=2u^{\mathsf T}u'=2u^{\mathsf T}Au$; and $u^{\mathsf T}Au$ is a scalar satisfying $(u^{\mathsf T}Au)^{\mathsf T}=u^{\mathsf T}A^{\mathsf T}u=-u^{\mathsf T}Au$, so it vanishes identically. Thus $\lVert u(t)\rVert^2$ is constant: the solution moves on a sphere, neither growing nor decaying.

2. If a matrix has $n$ orthogonal eigenvectors, does that alone imply symmetry? What else is needed?

Reference answer: Not alone — one also needs all eigenvalues to be real. If the $n$ eigenvectors are orthogonal (and can be normalized), use them as columns of an orthogonal matrix $Q$, so $A=Q\Lambda Q^{-1}=Q\Lambda Q^{\mathsf T}$. Then $A^{\mathsf T}=Q\Lambda^{\mathsf T}Q^{\mathsf T}$, and $\Lambda^{\mathsf T}=\Lambda$ only when $\Lambda$ is a real diagonal matrix, giving $A^{\mathsf T}=A$. With complex eigenvalues, $\Lambda^{\mathsf T}=\bar\Lambda\ne\Lambda$ and $A$ is generally not symmetric (a real matrix cannot even have real orthogonal eigenvectors for complex eigenvalues). The full statement of the spectral theorem is: symmetric $\iff$ $n$ orthogonal eigenvectors with real eigenvalues.

3. Why does a negative number in $\Sigma$ prove the factorization is wrong?

Reference answer: Singular values are defined as $\sigma_i=\sqrt{\lambda_i(A^{\mathsf T}A)}$ with $\lambda_i\ge0$, so by definition $\sigma_i>0$ or $=0$ and can never be negative; the standard form also orders $\sigma_1\ge\sigma_2\ge\cdots$. A negative "singular value" in a computation usually means the sign of a column of $V$ was mishandled: $A^{\mathsf T}A$ fixes each $v_i$ only up to sign, and when $v_i$ is negated one must negate $u_i$ as well to preserve $A=\sum_i\sigma_iu_iv_i^{\mathsf T}$, rather than turning $\sigma_i$ negative. The sign belongs to $U$ or $V$, never to $\Sigma$.

4. If $A$ is symmetric and orthogonal, what are the eigenvalues of $A+I$, and what is the rank of $\frac12(A+I)$?

Reference answer: Symmetry gives real eigenvalues and orthogonality gives modulus $1$, so the eigenvalues of $A$ are $\pm1$. Hence $A+I$ has eigenvalues $0$ (from $-1$) or $2$ (from $+1$), and $\frac12(A+I)$ has eigenvalues $0$ or $1$. It is a projection: it is symmetric, and

$$\left(\frac{A+I}{2}\right)^2=\frac{A^2+2A+I}{4}=\frac{I+2A+I}{4}=\frac{A+I}{2},$$

using $A^2=I$, which follows from orthogonality plus symmetry. The rank of a projection equals its trace, i.e. the number of eigenvalues equal to $1$; so $\operatorname{rank}\frac12(A+I)$ is the multiplicity of $+1$ as an eigenvalue of $A$ (namely $n$ if $A=I$, and $0$ if $A=-I$).

5. Given $A=U\Sigma V^{\mathsf T}$, how do you read $U$ directly from $A$ and $V$ without diagonalizing $AA^{\mathsf T}$?

Reference answer: For each column with $\sigma_i>0$, set

$$u_i=\frac{Av_i}{\sigma_i}.$$

This is valid because $Av_i=\sigma_iu_i$ and $\sigma_i>0$; the resulting $u_i$ is automatically a unit vector, since $\lVert Av_i\rVert=\sigma_i\lVert v_i\rVert=\sigma_i$, and it automatically carries the sign that matches $v_i$, so no sign inconsistency can arise. For the columns with $\sigma_i=0$ ($i>r$), $Av_i=0$ gives no information about $u_i$; complete $U$ by choosing any orthonormal basis of the left null space so that $U$ becomes an $m\times m$ orthogonal matrix.

How this connects to the course

This review drills the problem types of Chapter 6. The next lecture treats left inverses, right inverses, and the pseudoinverse, using SVD to "invert" non-square matrices; the final lecture synthesizes the course and brings Chapter 7's linear transformations into the final exam.