这一讲的主线

线性代数的根本问题是解线性方程组。Strang 用同一个 $2\times 2$ 例子，从三个角度反复看它：

1. 行图像：一次看一个方程，每个方程是一条直线，解是交点。
2. 列图像：一次看一列，解是「用多少倍各列相加能得到右端向量」。
3. 矩阵形式：把整个系统写成 $A\mathbf{x}=\mathbf{b}$。

三者是同一件事的三种语言。行图像最直观，列图像最能推广，矩阵形式最适合计算。

时间轴

例子与矩阵

取方程组

$$\begin{aligned} 2x - y &= 0,\\ -x + 2y &= 3. \end{aligned}$$

系数矩阵、未知向量和右端向量分别是

$$A=\begin{bmatrix}2 & -1\\ -1 & 2\end{bmatrix},\qquad \mathbf{x}=\begin{bmatrix}x\\ y\end{bmatrix},\qquad \mathbf{b}=\begin{bmatrix}0\\ 3\end{bmatrix}.$$

于是方程组就是 $A\mathbf{x}=\mathbf{b}$。

行图像：两条直线的交点

第一个方程 $2x-y=0$ 过原点：取 $x=0$ 得 $y=0$，取 $x=1$ 得 $y=2$，两点确定直线。

第二个方程 $-x+2y=3$ 不过原点：取 $y=0$ 得 $x=-3$，取 $x=-1$ 得 $y=1$。

两条直线交于 $(1,2)$。代回检验：

$$2(1)-2=0,\qquad -(1)+2(2)=3.$$

行图像在二维里非常清楚。但到了三维，每个方程是一个平面，三个平面交于一点；再高维就连画都画不出来。行图像好懂，但不好推广。

列图像：列的线性组合

把方程组按列拆开：

$$x\begin{bmatrix}2\\ -1\end{bmatrix}+y\begin{bmatrix}-1\\ 2\end{bmatrix} =\begin{bmatrix}0\\ 3\end{bmatrix}.$$

问题变成：用多少倍的列 1 加上多少倍的列 2，能得到 $\mathbf{b}$？

在这个例子里，答案已经知道了：$x=1,\ y=2$。验证：

$$1\begin{bmatrix}2\\ -1\end{bmatrix}+2\begin{bmatrix}-1\\ 2\end{bmatrix} =\begin{bmatrix}2-2\\ -1+4\end{bmatrix} =\begin{bmatrix}0\\ 3\end{bmatrix}.$$

几何上，从原点出发沿 $[2,-1]$ 走一份，再沿 $[-1,2]$ 走两份，终点正好是 $(0,3)$。这就是线性组合：把若干向量分别乘上系数再相加。Strang 说这是整门课最基本的运算。

「所有组合能到达哪里」

如果让 $x,y$ 取遍所有实数，这两个列向量的所有线性组合会填满整个平面。也就是说，对这个矩阵而言，任何右端向量 $\mathbf{b}$ 都能被表示出来。这个问题会在后面反复出现，它的名字叫「列空间是否等于整个空间」。

三维例子：平面与三列

$$\begin{aligned} 2x-y &= 0,\\ -x+2y-z &= -1,\\ -3y+4z &= 4. \end{aligned}$$

矩阵与右端向量为

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

行图像

每一行是一个平面：

• 第一行 $2x-y=0$ 与 $z$ 无关，是一个竖直的平面；
• 第二行 $-x+2y-z=-1$ 是另一个平面；
• 第三行 $-3y+4z=4$ 也是一个平面。

两个平面交于一条直线，第三个平面再与这条直线交于一点。三个平面「位置正常」时，交点唯一，就是方程组的解。

Strang 没有真的画出交点，因为三维图已经很难看清。他要说明的是：行图像在三维里开始失效。

列图像

把方程组写成三列的组合：

$$x\begin{bmatrix}2\\ -1\\ 0\end{bmatrix} +y\begin{bmatrix}-1\\ 2\\ -3\end{bmatrix} +z\begin{bmatrix}0\\ -1\\ 4\end{bmatrix} =\begin{bmatrix}0\\ -1\\ 4\end{bmatrix}.$$

注意到右端向量恰好等于第三列。所以取 $x=0,\ y=0,\ z=1$ 即可。代回：

$$0\cdot\begin{bmatrix}2\\ -1\\ 0\end{bmatrix} +0\cdot\begin{bmatrix}-1\\ 2\\ -3\end{bmatrix} +1\cdot\begin{bmatrix}0\\ -1\\ 4\end{bmatrix} =\begin{bmatrix}0\\ -1\\ 4\end{bmatrix}.$$

解是 $(0,0,1)$。这个例子是特意构造的，好让答案一眼可见；真实问题要靠下一讲的消元法。

换一个右端向量

保持 $A$ 不变，把 $\mathbf{b}$ 换成前两列之和：

$$\mathbf{b}'=\begin{bmatrix}2\\ -1\\ 0\end{bmatrix}+\begin{bmatrix}-1\\ 2\\ -3\end{bmatrix} =\begin{bmatrix}1\\ 1\\ -3\end{bmatrix}.$$

此时显然取 $x=1,\ y=1,\ z=0$。行图像上是三个新的平面交于新的一点；列图像上是同样的三列，换了一组系数。

同一个矩阵，不同的 $\mathbf{b}$，就是不同的组合问题。

核心问题：是否对每个 $\mathbf{b}$ 都有解

Strang 把这一讲的重心落在一个问题上：

对任意右端向量 $\mathbf{b}$，$A\mathbf{x}=\mathbf{b}$ 都有解吗？

用列图像翻译，就是：

这些列向量的所有线性组合，能否填满整个空间？

这两个问法完全等价，因为 $A\mathbf{x}$ 按定义就是「以 $\mathbf{x}$ 的分量为系数，对 $A$ 的列做线性组合」。

对这个 $3\times 3$ 例子，答案是肯定的，因为三列不共面，它们的组合能到达三维空间中的每一点。这样的矩阵叫非奇异或可逆矩阵。

什么时候会失败

如果三列都落在同一个平面里，它们的任何线性组合也都在那个平面里。此时：

• 落在平面内的 $\mathbf{b}$ 有解；
• 平面外的 $\mathbf{b}$ 无解。

最典型的退化情形是第三列等于前两列之和：它没有带来任何新方向。这样的矩阵是奇异的，不可逆。

同样的直觉可以推到九维：九个向量张成的组合可能填满九维空间，也可能只填满其中一个八维「平面」。随机选一个 $9\times 9$ 矩阵，几乎必然是满秩的；但如果第九列恰好等于第八列，就有一个方向永远到不了。

矩阵乘向量：两种算法，一个含义

最后 Strang 回到 $A\mathbf{x}$ 本身，用例子说明两种乘法：

$$\begin{bmatrix}2 & 5\\ 1 & 3\end{bmatrix} \begin{bmatrix}1\\ 2\end{bmatrix} =?$$

按列做：线性组合

$$1\begin{bmatrix}2\\ 1\end{bmatrix}+2\begin{bmatrix}5\\ 3\end{bmatrix} =\begin{bmatrix}2\\ 1\end{bmatrix}+\begin{bmatrix}10\\ 6\end{bmatrix} =\begin{bmatrix}12\\ 7\end{bmatrix}.$$

这正是列图像：$A\mathbf{x}$ 是 $A$ 的各列以 $\mathbf{x}$ 的分量为权重的线性组合。

按行做：点积

第一行与 $\mathbf{x}$ 做点积：

$$2\cdot 1+5\cdot 2=12.$$

第二行与 $\mathbf{x}$ 做点积：

$$1\cdot 1+3\cdot 2=7.$$

两种算法结果相同。Strang 的建议是：计算时按行，理解时按列。后面讲列空间、秩、四个基本子空间时，列的观点会成为主线。

本讲结论

• 解方程组有三种等价视角：行（直线/平面相交）、列（列的线性组合）、矩阵（$A\mathbf{x}=\mathbf{b}$）。
• 列图像是能推广到高维的那个视角。
• $A\mathbf{x}$ 的定义就是「$A$ 的列的线性组合」。
• 「对每个 $\mathbf{b}$ 都有解」等价于「列的组合填满整个空间」。
• 列向量共面（线性相关）时矩阵奇异，存在无法到达的 $\mathbf{b}$。
• 具体求解要靠下一讲的消元法，而不是靠观察。

理解检验

1. 对上面的 $2\times 2$ 例子，分别用行图像和列图像解释解 $(1,2)$ 的含义。

参考答案：行图像：$(x,y)=(1,2)$ 是两条直线的交点。代回第一个方程 $2(1)-2=0$，代回第二个方程 $-(1)+2(2)=3$，两个方程同时成立，所以这个点同时在两条直线上。列图像：$(1,2)$ 是一组组合系数，

$$1\begin{bmatrix}2\\-1\end{bmatrix}+2\begin{bmatrix}-1\\2\end{bmatrix}=\begin{bmatrix}0\\3\end{bmatrix},$$

即沿第一列走 1 份、沿第二列走 2 份，终点正好落在右端向量 $\mathbf{b}=(0,3)$ 上。两种说法描述的是同一组数，只是前者按方程看、后者按列看。

2. 为什么「三列共面」会导致有 $\mathbf{b}$ 无解？请用线性组合的语言回答。

参考答案：三列的任意线性组合 $x\mathbf{c}_1+y\mathbf{c}_2+z\mathbf{c}_3$ 仍然落在这三列张成的集合里。如果三列共面，它们张成的是过原点的一个平面（二维子空间），不可能填满三维空间。于是平面内的 $\mathbf{b}$ 能写成组合、有解；平面外的 $\mathbf{b}$ 无论取什么 $x,y,z$ 都写不出来，方程无解。关键在于线性组合只能留在张成空间里，不能凭空产生新的方向——第三列若等于前两列之和，它就没有带来任何新方向。

3. 计算 $\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix}\begin{bmatrix}2\\ 1\end{bmatrix}$，并分别用按行、按列两种方式写出过程。

参考答案：结果是 $\begin{bmatrix}4\\10\end{bmatrix}$。按列（线性组合）：

$$2\begin{bmatrix}1\\3\end{bmatrix}+1\begin{bmatrix}2\\4\end{bmatrix} =\begin{bmatrix}2\\6\end{bmatrix}+\begin{bmatrix}2\\4\end{bmatrix} =\begin{bmatrix}4\\10\end{bmatrix}.$$

按行（点积）：第一行 $(1,2)\cdot(2,1)=1\cdot2+2\cdot1=4$；第二行 $(3,4)\cdot(2,1)=3\cdot2+4\cdot1=10$。两种算法给出同一个向量，这正是「计算按行、理解按列」的例子。

4. 随机 $9\times 9$ 矩阵几乎总是可逆的，这句话和「第九列等于第八列时不可逆」矛盾吗？

参考答案：不矛盾，两者说的是不同层面的事。「几乎总是」是概率陈述：在所有 $9\times 9$ 矩阵组成的空间里，奇异矩阵构成一个低维集合（行列式为零的曲面），随机取点落在这个集合上的概率为零。而「第九列等于第八列」是一个精确的特殊条件，满足它的矩阵确实奇异——但它只是那个零测集里的一个特例。概率为零不等于不可能发生；反过来，存在反例也不推翻「随机取到反例的概率为零」。更准确的说法是：一般的矩阵可逆，退化是特殊的巧合。

与前后课的关系

这一讲建立了语言：行、列、线性组合、奇异。下一讲用消元法给出系统求解的算法，并说明消元在什么情况下会失败——那正是矩阵奇异的计算表现。

The thread of the lecture

The fundamental problem of linear algebra is solving a system of linear equations. Strang looks at a single $2\times 2$ example from three angles:

1. Row picture. One equation at a time; each is a line, and the solution is where they meet.
2. Column picture. One column at a time; the solution says how much of each column to combine to reach the right-hand side.
3. Matrix form. The whole system written as $A\mathbf{x}=\mathbf{b}$.

They are three languages for one object. The row picture is the most visual, the column picture generalizes best, and the matrix form is the one computation uses.

Timeline

The example and its matrix

$$\begin{aligned} 2x - y &= 0,\\ -x + 2y &= 3. \end{aligned}$$

The coefficient matrix, unknown vector, and right-hand side are

$$A=\begin{bmatrix}2 & -1\\ -1 & 2\end{bmatrix},\qquad \mathbf{x}=\begin{bmatrix}x\\ y\end{bmatrix},\qquad \mathbf{b}=\begin{bmatrix}0\\ 3\end{bmatrix}.$$

So the system is $A\mathbf{x}=\mathbf{b}$.

Row picture: two lines meeting

The first equation $2x-y=0$ passes through the origin: $x=0$ gives $y=0$, and $x=1$ gives $y=2$. Two points determine the line.

The second equation $-x+2y=3$ does not pass through the origin: $y=0$ gives $x=-3$, and $x=-1$ gives $y=1$.

The lines meet at $(1,2)$. Check it:

$$2(1)-2=0,\qquad -(1)+2(2)=3.$$

In two dimensions the row picture is completely clear. In three dimensions each equation is a plane; in higher dimensions it cannot be drawn at all. The row picture is easy to see and hard to generalize.

Column picture: a combination of columns

Split the system by columns:

$$x\begin{bmatrix}2\\ -1\end{bmatrix}+y\begin{bmatrix}-1\\ 2\end{bmatrix} =\begin{bmatrix}0\\ 3\end{bmatrix}.$$

The question becomes: how many copies of column 1 plus how many of column 2 produce $\mathbf{b}$?

Here the answer is already known, $x=1,\ y=2$. Check:

$$1\begin{bmatrix}2\\ -1\end{bmatrix}+2\begin{bmatrix}-1\\ 2\end{bmatrix} =\begin{bmatrix}2-2\\ -1+4\end{bmatrix} =\begin{bmatrix}0\\ 3\end{bmatrix}.$$

Geometrically: walk one copy of $[2,-1]$ from the origin, then two copies of $[-1,2]$, and the endpoint is $(0,3)$. This is a linear combination — multiply vectors by coefficients and add. Strang calls it the most fundamental operation in the course.

Where all combinations can reach

If $x$ and $y$ range over all real numbers, the combinations of these two columns fill the whole plane. For this matrix, every right-hand side $\mathbf{b}$ can be represented. That question — does the set of all combinations fill the space? — returns throughout the course under the name of the column space.

A three-dimensional example: planes and three columns

$$\begin{aligned} 2x-y &= 0,\\ -x+2y-z &= -1,\\ -3y+4z &= 4. \end{aligned}$$ $$A=\begin{bmatrix} 2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -3 & 4 \end{bmatrix},\qquad \mathbf{b}=\begin{bmatrix}0\\ -1\\ 4\end{bmatrix}.$$

Row picture

Each row is a plane:

• $2x-y=0$ does not involve $z$, so it is a vertical plane;
• $-x+2y-z=-1$ is another plane;
• $-3y+4z=4$ is a third.

Two planes meet in a line; the third plane cuts that line at a point. When the planes are in general position, the intersection is a single point — the solution.

Strang does not actually locate the point. His point is that the row picture already strains at three dimensions.

Column picture

$$x\begin{bmatrix}2\\ -1\\ 0\end{bmatrix} +y\begin{bmatrix}-1\\ 2\\ -3\end{bmatrix} +z\begin{bmatrix}0\\ -1\\ 4\end{bmatrix} =\begin{bmatrix}0\\ -1\\ 4\end{bmatrix}.$$

The right-hand side is exactly the third column, so $x=0,\ y=0,\ z=1$ works:

$$0\cdot\begin{bmatrix}2\\ -1\\ 0\end{bmatrix} +0\cdot\begin{bmatrix}-1\\ 2\\ -3\end{bmatrix} +1\cdot\begin{bmatrix}0\\ -1\\ 4\end{bmatrix} =\begin{bmatrix}0\\ -1\\ 4\end{bmatrix}.$$

The solution is $(0,0,1)$. The example was chosen so the answer is visible; a real problem needs elimination, which is the next lecture.

A different right-hand side

Keep $A$ fixed and replace $\mathbf{b}$ with the sum of the first two columns:

$$\mathbf{b}'=\begin{bmatrix}2\\ -1\\ 0\end{bmatrix}+\begin{bmatrix}-1\\ 2\\ -3\end{bmatrix} =\begin{bmatrix}1\\ 1\\ -3\end{bmatrix}.$$

Now $x=1,\ y=1,\ z=0$ is obvious. In the row picture, three new planes meet at a new point; in the column picture, the same three columns are combined with new coefficients.

Same matrix, different $\mathbf{b}$, different combination problem.

The central question: is there a solution for every $\mathbf{b}$?

Strang turns the lecture toward one question:

For every right-hand side $\mathbf{b}$, does $A\mathbf{x}=\mathbf{b}$ have a solution?

In the column picture:

Do all linear combinations of the columns fill the whole space?

The two are the same statement, because $A\mathbf{x}$ is by definition the combination of the columns of $A$ with the entries of $\mathbf{x}$ as weights.

For this $3\times 3$ example the answer is yes: the three columns are not coplanar, so their combinations reach every point of three-dimensional space. Such a matrix is nonsingular, or invertible.

When it fails

If all three columns lie in one plane, every combination lies in that plane:

• a $\mathbf{b}$ inside the plane has a solution;
• a $\mathbf{b}$ outside it has none.

The clearest failure is when the third column equals the sum of the first two: it adds no new direction. The matrix is singular and not invertible.

The same intuition scales. Nine vectors in nine dimensions may span the whole space, or may only span an eight-dimensional plane inside it. A random $9\times 9$ matrix is almost certainly nonsingular; make the ninth column a copy of the eighth and some $\mathbf{b}$ become unreachable.

Matrix times vector: two algorithms, one meaning

Strang closes by asking what $A\mathbf{x}$ actually computes:

$$\begin{bmatrix}2 & 5\\ 1 & 3\end{bmatrix} \begin{bmatrix}1\\ 2\end{bmatrix} =?$$

By columns: a linear combination

$$1\begin{bmatrix}2\\ 1\end{bmatrix}+2\begin{bmatrix}5\\ 3\end{bmatrix} =\begin{bmatrix}2\\ 1\end{bmatrix}+\begin{bmatrix}10\\ 6\end{bmatrix} =\begin{bmatrix}12\\ 7\end{bmatrix}.$$

This is the column picture: $A\mathbf{x}$ combines the columns of $A$ using the entries of $\mathbf{x}$ as weights.

By rows: dot products

First row with $\mathbf{x}$:

$$2\cdot 1+5\cdot 2=12.$$

Second row with $\mathbf{x}$:

$$1\cdot 1+3\cdot 2=7.$$

Same answer. Strang's advice: compute by rows, understand by columns. The column view becomes the main line when the course reaches column spaces, rank, and the four fundamental subspaces.

Takeaways

• Solving a system has three equivalent views: rows (intersecting lines or planes), columns (linear combinations), matrix ($A\mathbf{x}=\mathbf{b}$).
• The column picture is the one that survives in high dimensions.
• $A\mathbf{x}$ means "a linear combination of the columns of $A$".
• "A solution for every $\mathbf{b}$" is the same as "the columns span the whole space".
• Coplanar columns make the matrix singular, and some $\mathbf{b}$ become unreachable.
• Actually finding the solution requires elimination, not inspection.

Check your understanding

1. Explain the solution $(1,2)$ of the $2\times 2$ example in both the row and column pictures.

Reference answer: Row picture: $(x,y)=(1,2)$ is the intersection of the two lines — substituting gives $2(1)-2=0$ and $-(1)+2(2)=3$, so the point lies on both. Column picture: $(1,2)$ is the pair of weights

$$1\begin{bmatrix}2\\-1\end{bmatrix}+2\begin{bmatrix}-1\\2\end{bmatrix}=\begin{bmatrix}0\\3\end{bmatrix},$$

i.e. one copy of column 1 plus two copies of column 2 lands exactly on $\mathbf{b}=(0,3)$. The same two numbers, read per equation or per column.

2. Why does three coplanar columns imply that some right-hand sides have no solution? Answer in the language of linear combinations.

Reference answer: Every combination $x\mathbf{c}_1+y\mathbf{c}_2+z\mathbf{c}_3$ stays inside the span of the three columns. If they are coplanar, that span is a plane through the origin, not all of three-dimensional space. A $\mathbf{b}$ in the plane can be written as a combination and is solvable; a $\mathbf{b}$ off the plane cannot, for any choice of $x,y,z$. Combinations can only reach what the columns already span — if column three equals column one plus column two, it adds no new direction.

3. Compute $\begin{bmatrix}1 & 2\\ 3 & 4\end{bmatrix}\begin{bmatrix}2\\ 1\end{bmatrix}$ twice: once by rows, once by columns.

Reference answer: The result is $\begin{bmatrix}4\\10\end{bmatrix}$. By columns (a linear combination):

$$2\begin{bmatrix}1\\3\end{bmatrix}+1\begin{bmatrix}2\\4\end{bmatrix} =\begin{bmatrix}2\\6\end{bmatrix}+\begin{bmatrix}2\\4\end{bmatrix} =\begin{bmatrix}4\\10\end{bmatrix}.$$

By rows (dot products): $(1,2)\cdot(2,1)=2+2=4$ and $(3,4)\cdot(2,1)=6+4=10$. Both routes give the same vector — compute by rows, understand by columns.

4. Is "a random $9\times 9$ matrix is almost surely invertible" in tension with "it is singular if column nine equals column eight"?

Reference answer: No, they concern different things. "Almost surely" is probabilistic: singular matrices form a lower-dimensional set (the surface where the determinant vanishes), so a random draw hits it with probability zero. "Column nine equals column eight" is an exact special condition, and matrices satisfying it are indeed singular — but they are a measure-zero special case. Probability zero does not mean impossible, and the existence of counterexamples does not contradict a probability-zero statement. The accurate reading is that invertibility is generic and degeneracy is a coincidence.

How it connects

This lecture builds the vocabulary: rows, columns, linear combinations, singularity. The next lecture supplies elimination — the systematic algorithm — and shows what failure looks like computationally.