线性变换和反转矩阵
Section outline
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Linear Transformations and Matrix Inversion
::线性转换和矩阵转换From a while back we have the system A ⋅ → x = → b and know that we find the solution set by solving the augmented matrix [ A ∣ → b ] and plugging back in the entries x 1 , x 2 , . . . , x n to the vector → x .
::从前一阵子,我们有了Axb系统,知道我们通过解决增强的矩阵(Ab)和在条目x1x2xxxxxxxxxxlxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxmmmmmmmmets[Ab],找到确定的解决办法。We know that this is equivalent to taking the inverse of A and multiplying by → b getting → x = A − 1 → b . However, finding A − 1 is tedious and requires lots of boring arithmetic. But what if there was a different and easier way to think about this? There is! And we can use what we just learned about Linear Transformations.
::我们知道这相当于对A的反比和乘法的乘法。但是,对A-1的发现是乏味的,需要大量无聊的算术。但如果有不同和更容易的方法来思考这个问题呢?确实如此。我们可以使用我们刚刚学到的线性变换学。When we write a linear transformation T ( → x ) we are multiplying our vector → x by some standard matrix A , in other words, T ( → x ) = A ⋅ → x . We can write our standard matrix A in terms of the coordinates where the basis vectors are mapped. In our case, we are just going to use the standard basis vectors e 1 = [ 1 0 ⋅ ⋅ ⋅ 0 ] e 2 = [ 0 1 ⋅ ⋅ ⋅ 0 ]
::当我们写入直线变换 T( x) 时, 我们正在将矢量 x 乘以一些标准矩阵A, 换句话说, T( x) = A x。 我们可以用绘制基矢量的坐标来写标准矩阵A。 以我们的情况来说, 我们只是要使用标准基矢量 e1 = [100] e2 = [010] 。We get
::我们得到T ( → x ) = [ T ( e 1 ) T ( e 2 ) ⋅ ⋅ ⋅ T ( e n ) ] ⋅ → x
::T(x) = [T(e1) T(e2) @T(en)] xSo to find A − 1 we can solve the systems A → x 1 = e 1 and A → x 2 = e 2 to get [ → x 1 → x 2 ] which will be our standard matrix as x 1 and x 2 are T ( e 1 ) and T ( e 2 ) because solving the systems gives us the coordinates where the basis vectors are mapped to.
::因此,要找到 A-1we 可以解析 Ax1=e1 和 Ax2=e2 系统, 以获得 [x1=x2], 这将是我们的标准矩阵, 如 x1 和 x2are T(e1) 和 T(e2), 因为解析系统给我们提供了基准矢量映射到的坐标 。Summing up, A − 1 = [ A − 1 e 1 A − 1 e 2 ]
::总结,A-1=[A-1e1A-1e2]Let’s try a couple of examples:
::让我们举几个例子:Calculate the inverse of A where
::计算 A 的反对A = [ − 2 1.5 1 − 0.5 ]
::A=[-21.51-0.5]First, we put augment this matrix to find the coordinates of → x 1 which is [ A ∣ e 1 ] and then we will do the same for → x 2 which is [ A ∣ e 2 ] .
::首先,我们增加这个矩阵,以找到x1的坐标,也就是[Ae1],然后我们将对x2,也就是[Ae2]也这样做。Let us begin:
::让我们开始:[ A ∣ e 1 ] = [ − 2 1.5 1 1 − 0.5 0 ]
::[Ae1]=[-21.511-0.50]Now we try to go to reduced row echelon form using row reduction
::现在,我们尝试用减排方式去减排 echelon 格式[ − 2 1.5 1 1 − 0.5 0 ]
[ − 2 1.5 1 0 0.5 1 ]
[ − 2 0 − 2 0 0.5 1 ]
[ 1 0 1 0 0.5 1 ]
[ 1 0 1 0 1 2 ]
Now we have that A − 1 e 1 = → x 1 = [ 1 2 ]
::现在我们有了A - 1e1\\\\\\\ x1=[12]Now let’s do the same to find A − 1 e 2 .
::现在让我们也这样做来找到A-1e2。We get our augmented matrix [ A ∣ e 2 ] as
::我们的扩大矩阵表[Ae2]改为:[ − 2 1.5 0 1 − 0.5 1 ]
[ − 2 1.5 0 0 0.5 2 ]
[ − 2 0 − 6 0 0.5 2 ]
[ 1 0 3 0 0.5 2 ]
[ 1 0 3 0 1 4 ]
Now we get that → x 2 = A − 1 e 2 = [ 3 4 ] .
::现在我们得到"x2=A -1e2=[34]。Putting this all together, (putting x 1 and x 2 ) as the columns of our matrix A − 1 we get
::把这些全部组合在一起,(要求x1和x2)作为我们A-1矩阵的列A − 1 = [ 1 3 2 4 ]
::A-1=[1324]Let’s try another example with a 3 × 3 matrix. Before I would just like to reiterate the importance of showing your row reduction steps. It is a pain, but if you don’t and you mess up you are in big trouble. Also, can you now see why we can only take the inverse of square matrices?
::让我们用一个 3x3 矩阵来尝试另一个例子。 在我重申显示您减排步骤的重要性之前,我只想重申这一点。 这是一个痛苦,但如果你不这样做,你就会陷入大麻烦。 另外,你现在能不能明白为什么我们只能接受平方矩阵的反面?The matrix has to have the same number of pivot columns as the matrix has rows and columns. In other words, it has to be row equivalent to the identity, meaning that row operations yield the identity matrix. If row operations cannot lead to I then we cannot solve [ A ∣ e 1 ] or [ A ∣ e 2 ] because we would get no or infinite solutions and we could not express that as a standard matrix of a linear transformation.
::矩阵必须拥有与矩阵有行和列相同的主轴列数。 换句话说, 它必须是与身份相等的行, 意思是行操作产生身份矩阵。 如果行操作不能导致我, 那么我们就无法解决 [Ae1] 或[Ae2] , 因为我们得不到任何或无限的解决方案, 我们无法将此表达为线性转换的标准矩阵。Now let’s try our 3 × 3 example.
::现在让我们试一下我们的3x3的例子。Let A = [ 1 0 5 2 4 1 0 0 1 ]
::A=[105241001]Calculate A − 1 .
::计算 A - 1 。For a 3 × 3 matrix, we are going to have to solve [ A ∣ e i ] , but we will have to do this 3 times as there are 3 standard basis vectors where again, e 1 = [ 1 0 0 ] ,
::对于 3x3 矩阵, 我们需要解决 [Aei] , 但是我们不得不做三次, 因为有 3 个标准基矢量, 并且 e1 = [100],e 2 = [ 0 1 0 ] and e 3 = [ 0 0 1 ]
::e2=[010]和e3=[001]Now, we solve to get column one which we will write as → c 1
::现在,我们解决了 获得第1列 我们将写为"c1"[ 1 0 5 1 2 4 1 0 0 0 1 0 ]
We first row reduce:
::第一排减速:[ 1 0 5 1 2 4 1 0 0 0 1 0 ]
[ 1 0 5 1 0 4 − 9 − 2 0 0 1 0 ]
[ 1 0 5 1 0 4 0 − 2 0 0 1 0 ]
[ 1 0 0 1 0 4 0 − 2 0 0 1 0 ]
[ 1 0 0 1 0 1 0 − 1 2 0 0 1 0 ]
So we get → c 1 = [ 1 − 1 2 0 ]
::因此,我们得到"c1"=[1 - 120]Now, we take the reduced row echelon form of [ A ∣ e 2 ] to get the second column, → c 2 :
::现在,我们采取减排的梯层形式[Ae2], 来获得第二列, c2:[ 1 0 5 0 2 4 1 1 0 0 1 0 ]
And we begin:
::我们开始:[ 1 0 5 0 2 4 1 1 0 0 1 0 ]
[ 1 0 5 0 0 4 − 9 1 0 0 1 0 ]
[ 1 0 5 0 0 4 0 1 0 0 1 0 ]
[ 1 0 0 0 0 4 0 1 0 0 1 0 ]
[ 1 0 0 0 0 2 0 1 4 0 0 1 0 ]
Now we get our → c 2 written as → c 2 = [ 0 1 4 0 ]
::现在我们得到我们的"c2" 以"c2="[0140]写成"c2="0140"Now, we take the reduced row echelon form of [ A ∣ e 3 ] to get the third column, → c 3 :
::现在,我们采取减排的梯队形式[Ae3] 来获得第三列, c3:[ 1 0 5 0 2 4 1 0 0 0 1 1 ]
And we begin:
::我们开始:[ 1 0 5 0 2 4 1 0 0 0 1 1 ]
[ 1 0 5 0 0 4 − 9 0 0 0 1 1 ]
[ 1 0 0 − 5 0 4 0 9 0 0 1 1 ]
[ 1 0 5 0 0 4 0 9 0 0 1 1 ]
[ 1 0 0 − 5 0 1 0 9 4 0 0 1 1 ]
We now get → c 3 = [ − 5 9 4 1 ] and can combine our three → c i vectors into the columns of our inverse of matrix A to get:
::我们现在得到“c3”=[-5941],可以把我们的三个“ci”矢量结合到我们矩阵A的反面列中,以便:A − 1 = [ 1 0 − 5 − 1 2 1 4 9 4 0 0 1 ]
::A-1=[10-5-121.494001]For further exploration:Because each matrix is the same other than the augment we just did the same row reduction to each matrix. Remember how row reduction can be expressed as matrix multiplication? Why can't we just express the row reductions we made as matrix multiplication and multiply that matrix by the basis vectors and then make that each column? Well, we can!Let's take our matrix A , row reduce it to I then do the same for each basis vector, plug it into the matrix and bam, we have A − 1 . For now I will leave doing this as an exercise for the reader as it is not as connected to linear transformations, our topic for this chapter.I just want you all to pause and think about how interconnected linear algebra is. Ponder what this all means geometrically. We find out what gets us to the basis vectors if we treat A as a standard matrix and then we transform the space we are in so that we can get back to our original points.