这一讲的位置

18.06 的前半程处理矩形矩阵：消元、秩、四个基本子空间。从这一讲开始转向方阵，先讲行列式，再讲特征值。行列式是给每个方阵配一个数，最重要的用途是判断可逆性：

$$A \text{ 可逆} \iff \det A \neq 0$$

但 Strang 刻意不先给公式。n 阶行列式的「大公式」有 $n!$ 项，直接看会淹没结构。他改为先列三条性质，并声明：这三条性质唯一地定义了行列式，其余全部可以推出来。

时间轴

三条定义性质

3A 与 3B 合起来就是：行列式对每一行分别是线性的，但其他行必须保持不变。这一点极易误解。行列式没有整体线性：

$$\det(A+B) \neq \det A + \det B$$

它只有「一次动一行」的线性。理解了这句话，后面的推导都不会走错。

由三条性质立刻得到的东西

• 置换矩阵：由单位矩阵做行交换得到，所以行列式只能是 $+1$ 或 $-1$，取决于交换次数的奇偶。
• 二阶情形：设 $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$。把第一行拆成 $(a,0)+(0,b)$，再拆第二行，得到四个矩阵；其中两个含零列、行列式为零，剩下两个分别给出 $ad$ 与 $-bc$，于是 $\det A = ad-bc$。

这条推导是整讲的模板：拆行 → 大部分项因零列消失 → 少数幸存项由置换矩阵定符号。

从三条性质推出的性质

性质 4：两行相同，行列式为零

若第 1 行等于第 3 行，交换这两行，矩阵不变，所以行列式不变；但性质 2 说符号必须翻转。一个数等于自己的相反数，只能是零。

这个证明只用了性质 2，没有依赖任何公式，因此对任意阶都成立。

性质 5：一行减去另一行的倍数，行列式不变

设把第 $k$ 行换成「第 $k$ 行减去 $\ell$ 倍第 $i$ 行」。由 3B 拆成两个行列式：原矩阵的行列式，加上一个「第 $k$ 行等于 $-\ell$ 倍第 $i$ 行」的行列式。后者由 3A 提出因子 $-\ell$，剩下的是「两行成比例」，再由性质 4（比例行可通过行交换与线性组合化为相同行）得零。

于是：

$$\det(\text{消元后的矩阵}) = \det A$$

这正是消元法能用来算行列式的原因。

性质 6：含零行，行列式为零

由 3A 取 $t=0$：把零行写成 $0$ 倍某行，行列式就是 $0$ 乘以原行列式。

性质 7：三角矩阵的行列式等于对角线乘积

设 $U$ 是上三角矩阵，对角元为 $d_1,\dots,d_n$。先用性质 5，把对角线上方的元素全部消掉（用下面的行减上面的行），得到一个对角矩阵，行列式不变。再由 3A 逐行提出 $d_1,\dots,d_n$，剩下单位矩阵，由性质 1 得

$$\det U = d_1 d_2 \cdots d_n$$

这一步也顺便说明了实际计算方式：消元成三角，再把主元乘起来。

性质 8：行列式等于主元乘积，符号由行交换次数决定

如果消元过程中做了奇数次行交换，整体要乘 $-1$：

$$\det A = (-1)^{\#\text{exchanges}} \, d_1 d_2 \cdots d_n$$

二阶验证：主元是 $a$ 与 $d-\frac{c}{a}b$，乘积为 $ad-bc$，与公式一致。若 $a=0$，必须先交换行；若交换也无法得到主元，说明矩阵奇异。

可逆性的完整证明

• 若 $A$ 可逆，消元得到 $n$ 个非零主元，$\det A = \prod d_i \neq 0$。
• 若 $A$ 奇异，消元会出现零行，由性质 6，$\det A = 0$。

两者合起来就是「可逆 $\iff$ 行列式非零」。

两个「乘法型」性质

性质 9：$\det(AB) = \det A \cdot \det B$

这是整讲最意外的性质。行列式对加法完全没有好行为，对乘法却是乘性的。几个直接推论：

$$\det(A^{-1}) = \frac{1}{\det A}, \qquad \det(A^2) = (\det A)^2, \qquad \det(2A) = 2^n \det A$$

最后一条常被记错。把整个矩阵乘 2，相当于每一行都乘 2，由 3A 每行提出一个 2，共提出 $n$ 个，所以是 $2^n$。几何上这也合理：把 $n$ 维盒子的每条边放大 2 倍，体积放大 $2^n$ 倍。

性质 10：$\det(A^T) = \det A$

二阶直接验证：$ad-bc$ 转置后仍是 $ad-bc$。

证明用 LU 分解（忽略需要行交换的情形）：$A = LU$，$A^T = U^T L^T$。由性质 9，

$$\det A = \det L \det U, \qquad \det A^T = \det U^T \det L^T$$

$L$ 是对角线全为 1 的下三角矩阵，$\det L = 1$；$L^T$ 同理为 1。$U$ 与 $U^T$ 都是三角矩阵，对角线相同，由性质 7 行列式相等。两边因此相等。

这条性质的实用价值：所有关于行的命题，对列也同样成立。列交换变号、零列使行列式为零、列线性性，都可以通过转置把列变成行再转回来。

关于「符号良定义」的一个细节

性质 2 只说交换一次变号。要保证定义自洽，必须确认：同一个置换无论用多少次交换实现，奇偶性都相同。否则做 7 次交换与做 10 次交换会给出相反的符号，行列式就没有定义。

Strang 在结尾点出这一点：每个置换非奇即偶，这是行列式由性质 1–3 唯一确定的前提。

核心回顾

• 三条性质定义行列式：$\det I=1$、交换变号、逐行线性。
• 两行相同、零行、成比例行都给出零。
• 消元不改变行列式；三角阵行列式等于对角线乘积。
• 实际算法：消元 → 乘主元 → 按行交换次数定符号。
• 行列式对乘法是乘性的，对加法不是。
• 转置不改变行列式，因此行列性质对列同样成立。

常见误区

1. 把逐行线性当成整体线性。$\det(A+B)$ 一般不等于 $\det A+\det B$。
2. 忘记行交换的符号。用消元算行列式时，做了奇数次交换必须取负。
3. 把 $\det(2A)$ 写成 $2\det A$。正确结果是 $2^n\det A$。
4. 以为 $\det A=0$ 只说明「某行是零」。零行只是奇异的一种表现，两行线性相关同样使行列式为零。

理解检验

1. 用三条性质证明：若两行成比例，行列式为零。

参考答案：先证两行相同的情形。设第 $i$ 行与第 $k$ 行相同，交换这两行，矩阵不变，所以行列式不变；但性质 2 要求变号，于是 $\det A=-\det A$，只能有 $\det A=0$。再设第 $k$ 行 $=c\cdot$ 第 $i$ 行。由 3A 把因子 $c$ 提出：

$$\det A=c\cdot\det A',$$

其中 $A'$ 的第 $k$ 行等于第 $i$ 行，即两行相同，所以 $\det A'=0$，从而 $\det A=0$。$c=0$ 时第 $k$ 行是零行，由 3A 取 $t=0$ 也直接得零。

2. 为什么消元不改变行列式？请写出用 3A、3B、性质 4 的完整推理。

参考答案：一次消元把第 $k$ 行换成 $R_k-\ell R_i$，其余行不动。把新矩阵的第 $k$ 行拆成 $R_k+(-\ell R_i)$，由 3B：

$$\det(\text{消元后})= \det(\text{原矩阵})+\det(\text{第 }k\text{ 行换成 }-\ell R_i\text{ 的矩阵}).$$

第二个行列式用 3A 提出因子 $-\ell$，剩下「第 $k$ 行等于第 $i$ 行」的矩阵，由性质 4 其行列式为零。于是

$$\det(\text{消元后})=\det A+(-\ell)\cdot 0=\det A .$$

关键点：3B 要求只有一行变化，消元恰好只改动一行，所以拆分合法。

3. 一个 $5\times5$ 矩阵的所有元素乘以 3，行列式变成原来的多少倍？

参考答案：$3^5=243$ 倍。把整个矩阵乘 3 等于每一行都乘 3；由 3A 逐行提出因子，$5$ 行共提出 $3^5$。注意不是 $3$ 倍：$\det(3A)=3^n\det A$。

4. 若 $\det A=2$，求 $\det(A^{-1})$ 与 $\det(A^2)$。

参考答案：由 $\det(AB)=\det A\det B$，取 $B=A^{-1}$ 得 $\det I=1=\det A\cdot\det(A^{-1})$，所以 $\det(A^{-1})=1/2$。同理 $\det(A^2)=(\det A)^2=4$。

5. 性质 10 为什么能说明「列交换也变号」？

参考答案：交换 $A$ 的两列，等价于交换 $A^T$ 的两行。由性质 2，$\det(A^T)$ 变号；再由性质 10，$\det A=\det(A^T)$，所以 $\det A$ 也变号。一般地，任何关于行的命题（零行、成比例行、逐行线性）都可以转置成关于列的命题。

与前后课程的关系

上一阶段用消元研究矩形矩阵的秩与子空间；这一讲把消元用于方阵，得到行列式的计算与可逆性判据。下一讲给出行列式的显式公式（大公式与代数余子式），再下一讲用行列式求逆矩阵与体积，最后进入特征值。

Where this lecture sits

The first half of 18.06 dealt with rectangular matrices: elimination, rank, the four subspaces. This lecture turns to square matrices, starting with determinants and heading toward eigenvalues. A determinant assigns one number to every square matrix, and its most important job is the invertibility test:

$$A \text{ is invertible} \iff \det A \neq 0$$

Strang deliberately refuses to start with the formula. The general $n \times n$ formula has $n!$ terms, which hides the structure. Instead he states three properties and claims: these three properties define the determinant, and everything else follows.

Timeline

The three defining properties

Together, 3A and 3B say the determinant is linear in each row separately, holding the other rows fixed. This is easy to misread. The determinant is not linear in the whole matrix:

$$\det(A+B) \neq \det A + \det B$$

It is linear only when a single row changes. Once that sentence is clear, the derivations stop going wrong.

Immediate consequences

• Permutation matrices. They come from the identity by row exchanges, so their determinant is $+1$ or $-1$ according to the parity of the number of exchanges.
• The $2 \times 2$ case. For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, split the first row into $(a,0)+(0,b)$ and then split the second row. Four matrices appear; two contain a zero column and vanish, while the other two give $ad$ and $-bc$. Hence $\det A = ad-bc$.

That derivation is the template for the whole lecture: split rows, let most terms die on zero columns, and let permutation parity fix the signs of the survivors.

Properties that follow

Property 4: Two equal rows give zero

If row 1 equals row 3, exchanging them leaves the matrix unchanged, so the determinant is unchanged. But property 2 demands a sign flip. A number equal to its own negative must be zero.

The proof uses only property 2, so it holds at every size.

Property 5: Subtracting a multiple of one row from another changes nothing

Replace row $k$ by row $k$ minus $\ell$ times row $i$. By 3B this splits into two determinants: the original, plus one whose row $k$ is $-\ell$ times row $i$. Property 3A pulls out $-\ell$, leaving a matrix with proportional rows, which is zero by property 4. Therefore

$$\det(\text{matrix after elimination}) = \det A$$

This is why elimination can be used to compute determinants.

Property 6: A zero row gives zero

By 3A with $t=0$: a zero row is $0$ times any row, so the determinant is $0$ times the original.

Property 7: A triangular matrix has determinant equal to its diagonal product

Let $U$ be upper triangular with diagonal $d_1,\dots,d_n$. Property 5 clears every entry above the diagonal using the rows below, leaving a diagonal matrix with the same determinant. Then 3A pulls out $d_1,\dots,d_n$ one row at a time, leaving the identity, whose determinant is 1. Hence

$$\det U = d_1 d_2 \cdots d_n$$

This also gives the practical algorithm: eliminate to triangular form, then multiply the pivots.

Property 8: The determinant is the pivot product, with a sign from row exchanges

If elimination needed an odd number of row exchanges, multiply by $-1$:

$$\det A = (-1)^{\#\text{exchanges}} \, d_1 d_2 \cdots d_n$$

Check on $2 \times 2$: the pivots are $a$ and $d-\frac{c}{a}b$, whose product is $ad-bc$. If $a=0$, an exchange is required; if no pivot can be found, the matrix is singular.

The complete invertibility proof

• If $A$ is invertible, elimination produces $n$ nonzero pivots, so $\det A = \prod d_i \neq 0$.
• If $A$ is singular, elimination produces a zero row, so $\det A = 0$ by property 6.

Together: invertible $\iff$ determinant nonzero.

Two multiplicative properties

Property 9: $\det(AB) = \det A \cdot \det B$

This is the most surprising property in the lecture. The determinant behaves terribly under addition but multiplicatively under products. Immediate consequences:

$$\det(A^{-1}) = \frac{1}{\det A}, \qquad \det(A^2) = (\det A)^2, \qquad \det(2A) = 2^n \det A$$

The last one is commonly misremembered. Doubling the whole matrix doubles each of the $n$ rows, and 3A extracts one factor of 2 per row, so the factor is $2^n$. Geometrically, doubling every edge of an $n$-dimensional box multiplies its volume by $2^n$.

Property 10: $\det(A^T) = \det A$

Direct check for $2 \times 2$: $ad-bc$ is unchanged by transposition.

The proof uses $LU$ (ignoring row exchanges): $A = LU$ and $A^T = U^T L^T$. By property 9,

$$\det A = \det L \det U, \qquad \det A^T = \det U^T \det L^T$$

$L$ is lower triangular with ones on the diagonal, so $\det L = 1$, and likewise for $L^T$. $U$ and $U^T$ are triangular with the same diagonal, so property 7 makes their determinants equal. The two sides match.

The practical value: every statement about rows is also true about columns. Column exchanges flip the sign, a zero column kills the determinant, and linearity holds column by column, because transposing turns columns into rows and back.

Why the sign is well defined

Property 2 says one exchange flips the sign. For the definition to be consistent, the same permutation must have the same parity no matter how it is built. Otherwise seven exchanges and ten exchanges would give opposite signs, and the determinant would not be well defined.

Strang closes by noting that every permutation is either odd or even. That fact is what makes the determinant uniquely determined by properties 1–3.

Core review

• Three properties define the determinant: $\det I=1$, exchange flips the sign, linear in each row.
• Equal rows, a zero row, or proportional rows all give zero.
• Elimination leaves the determinant unchanged; a triangular determinant is the diagonal product.
• The algorithm: eliminate, multiply pivots, apply the sign from row exchanges.
• The determinant is multiplicative, not additive.
• Transposition leaves it unchanged, so row facts become column facts.

Common misconceptions

1. Treating row-wise linearity as global linearity. $\det(A+B)$ is generally not $\det A+\det B$.
2. Forgetting the exchange sign. An odd number of row exchanges in elimination requires a minus sign.
3. Writing $\det(2A)=2\det A$. The correct value is $2^n\det A$.
4. Thinking $\det A=0$ means a row is zero. A zero row is only one way to be singular; linearly dependent rows also make the determinant zero.

Check your understanding

1. Use the three properties to prove that proportional rows give determinant zero.

Reference answer: First handle equal rows. If rows $i$ and $k$ coincide, swapping them leaves the matrix unchanged, so the determinant is unchanged; property 2 demands a sign flip, so $\det A=-\det A$ and hence $\det A=0$. Now suppose row $k$ equals $c$ times row $i$. By 3A the factor $c$ comes out:

$$\det A=c\cdot\det A',$$

where $A'$ has row $k$ equal to row $i$, so $\det A'=0$ and therefore $\det A=0$. When $c=0$ the row is zero, which 3A with $t=0$ also sends to zero.

2. Why does elimination leave the determinant unchanged? Give the full argument using 3A, 3B, and property 4.

Reference answer: One elimination step replaces row $k$ by $R_k-\ell R_i$ and leaves the other rows alone. Split the new $k$-th row as $R_k+(-\ell R_i)$; by 3B,

$$\det(\text{after elimination})= \det A+\det(\text{matrix with row }k\text{ equal to }-\ell R_i).$$

In the second determinant, 3A pulls out $-\ell$, leaving the matrix whose $k$-th row equals row $i$; by property 4 that determinant is zero. Hence

$$\det(\text{after elimination})=\det A+(-\ell)\cdot 0=\det A .$$

The key point is that 3B applies because only one row changes, which is exactly what elimination does.

3. If every entry of a $5\times5$ matrix is multiplied by 3, what happens to the determinant?

Reference answer: It is multiplied by $3^5=243$. Scaling the whole matrix by 3 scales each of the five rows; 3A pulls a factor 3 out of each row, giving $3^5$. It is not just 3 times: $\det(3A)=3^n\det A$.

4. If $\det A = 2$, find $\det(A^{-1})$ and $\det(A^2)$.

Reference answer: From $\det(AB)=\det A\det B$ with $B=A^{-1}$, $1=\det I=\det A\det(A^{-1})$, so $\det(A^{-1})=1/2$. Likewise $\det(A^2)=(\det A)^2=4$.

5. Why does property 10 imply that exchanging two columns also flips the sign?

Reference answer: Swapping two columns of $A$ is the same as swapping two rows of $A^T$. Property 2 says $\det(A^T)$ flips sign, and property 10 says $\det A=\det(A^T)$, so $\det A$ flips as well. More generally, any row statement (zero row, proportional rows, row linearity) transfers to columns by transposing.

How this connects

The previous stage used elimination to study rank and subspaces of rectangular matrices. This lecture applies elimination to square matrices and produces the determinant and the invertibility test. The next lecture gives explicit formulas (the big formula and cofactors), then Cramer's rule and volume, before the course moves to eigenvalues.