线性代数的几何表示
1 例一:两个方程两个未知数
{2x−y=0−x+2y=3 \begin{cases} 2x-y=0\\ -x+2y=3 \end{cases} {2x−y=0−x+2y=3
[2−1−12]×[xy]=[03] \begin{gathered} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \end{gathered}\times \begin{gathered} \begin{bmatrix} x \\ y \end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix} 0 \\ 3 \end{bmatrix} \end{gathered} [2−1−12]×[xy]=[03]
Ax=b Ax=b Ax=b
1.1 行图像
{2x−y=0−x+2y=3 \begin{cases} 2x-y=0\\ -x+2y=3 \end{cases} {2x−y=0−x+2y=3
x=1y=2 x=1\\ y=2 x=1y=2
1.2 列图像
x×[2−1]+y×[−12]=[03] x\times \begin{gathered} \begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{gathered}+ y\times \begin{gathered} \begin{bmatrix} -1\\ 2\end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix} 0\\ 3\end{bmatrix} \end{gathered} x×[2−1]+y×[−12]=[03]
[2−1][−12][03] \begin{gathered} \begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{gathered} \begin{gathered} \begin{bmatrix} -1\\ 2 \end{bmatrix} \end{gathered} \begin{gathered} \begin{bmatrix} 0 \\ 3 \end{bmatrix} \end{gathered} [2−1][−12][03]
1×[2−1]+2×[−12]=[03] \begin{gathered} 1\times\begin{bmatrix} 2 \\ -1 \end{bmatrix}+ 2\times\begin{bmatrix} -1\\ 2 \end{bmatrix}= \begin{bmatrix} 0 \\ 3 \end{bmatrix} \end{gathered} 1×[2−1]+2×[−12]=[03]
x×[2−1]+y×[−12]=? x\times \begin{gathered} \begin{bmatrix} 2 \\ -1 \end{bmatrix} \end{gathered}+ y\times \begin{gathered} \begin{bmatrix} -1\\ 2\end{bmatrix} \end{gathered}=? x×[2−1]+y×[−12]=?
2 例二:三个方程三个未知数
{2x−y=0−x+2y−z=−1−3y+4z=4 \begin{cases} 2x-y=0\\ -x+2y-z=-1\\ -3y+4z=4 \end{cases} ⎩⎪⎨⎪⎧2x−y=0−x+2y−z=−1−3y+4z=4
[2−10−12−1−340]×[xyz]=[0−14] \begin{gathered} \begin{bmatrix} 2 & -1 &0 \\ -1 & 2 & -1 \\ -3 & 4 & 0 \end{bmatrix} \end{gathered}\times \begin{gathered} \begin{bmatrix} x \\ y \\z \end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix} 0 \\ -1 \\ 4 \end{bmatrix} \end{gathered} ⎣⎡2−1−3−1240−10⎦⎤×⎣⎡xyz⎦⎤=⎣⎡0−14⎦⎤
Ax=b Ax=b Ax=b
2.1 行图像
{2x−y=0−x+2y−z=−1−3y+4z=4 \begin{cases} 2x-y=0\\ -x+2y-z=-1\\ -3y+4z=4 \end{cases} ⎩⎪⎨⎪⎧2x−y=0−x+2y−z=−1−3y+4z=4
2.2 列图像
x×[2−1−3]+y×[−124]+z×[0−10]=[0−14] x\times \begin{gathered} \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} \end{gathered}+ y\times \begin{gathered} \begin{bmatrix} -1\\ 2\\4 \end{bmatrix} \end{gathered}+ z\times \begin{gathered} \begin{bmatrix} 0\\ -1\\0 \end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix} 0\\ -1\\ 4 \end{bmatrix} \end{gathered} x×⎣⎡2−1−3⎦⎤+y×⎣⎡−124⎦⎤+z×⎣⎡0−10⎦⎤=⎣⎡0−14⎦⎤
0×[2−1−3]+0×[−124]+1×[0−14]=[0−14] \begin{gathered} 0\times\begin{bmatrix} 2 \\ -1 \\-3 \end{bmatrix}+ 0\times\begin{bmatrix} -1\\ 2 \\ 4 \end{bmatrix}+ 1\times\begin{bmatrix} 0\\ -1 \\ 4 \end{bmatrix}= \begin{bmatrix} 0 \\ -1 \\4 \end{bmatrix} \end{gathered} 0×⎣⎡2−1−3⎦⎤+0×⎣⎡−124⎦⎤+1×⎣⎡0−14⎦⎤=⎣⎡0−14⎦⎤
x×[2−1−3]+y×[−124]+z×[0−10]=? x\times \begin{gathered} \begin{bmatrix} 2 \\ -1 \\ -3 \end{bmatrix} \end{gathered}+ y\times \begin{gathered} \begin{bmatrix} -1\\ 2\\4 \end{bmatrix} \end{gathered}+ z\times \begin{gathered} \begin{bmatrix} 0\\ -1\\0 \end{bmatrix} \end{gathered}=? x×⎣⎡2−1−3⎦⎤+y×⎣⎡−124⎦⎤+z×⎣⎡0−10⎦⎤=?
3 思考:对于更高维度的Ax=b,其全部线性组合结果会是什么
4 矩阵与向量的乘法
A=[2513] A = \begin{gathered} \begin{bmatrix} 2 & 5 \\ 1 &3 \end{bmatrix} \end{gathered} A=[2153]
x=[12] x = \begin{gathered} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{gathered} x=[12]
4.1 法一:列图像的思路
Ax=[2513]×[12]=1×[21]+2×[53]=[127] Ax = \begin{gathered} \begin{bmatrix} 2 & 5 \\ 1 &3 \end{bmatrix} \end{gathered}\times \begin{gathered} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{gathered}= 1\times \begin{gathered} \begin{bmatrix} 2 \\ 1 \end{bmatrix} \end{gathered}+ 2\times \begin{gathered} \begin{bmatrix} 5 \\ 3 \end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix} 12 \\ 7 \end{bmatrix} \end{gathered} Ax=[2153]×[12]=1×[21]+2×[53]=[127]
4.2 法二:点乘法
Ax=[2513]×[12]=[2×1+5×21×1+3×2]=[127] Ax = \begin{gathered} \begin{bmatrix} 2 & 5 \\ 1 &3 \end{bmatrix} \end{gathered}\times \begin{gathered} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix} 2\times1+5\times2\\1\times1+3\times2 \end{bmatrix} \end{gathered}= \begin{gathered} \begin{bmatrix} 12 \\ 7 \end{bmatrix} \end{gathered} Ax=[2153]×[12]=[2×1+5×21×1+3×2]=[127]