Section outline

  • General

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  • 列空间

    In earlier lectures we have talked a lot about spanning sets and taking the span of a set of vectors. Now, when we have a matrix we can look at the columns as vectors and finding the span of those columns is known as the column space of the matrix.
    ::在先前的讲座中,我们谈论了很多关于横跨一组矢量和跨越一组矢量的问题。现在,当我们有一个矩阵时,我们可以将列看成矢量,发现这些列的宽度被称为矩阵的列空间。

    Let  A = [ 1 2 5 − 1 3 4 0 1 − 2 ]  , calculate the column space of  A , denoted by col ( A )
    ::Let A=[125-13401-2],计算 A 的列空间,以col(A) 表示

    col ( A ) = c 1 [ 1 − 1 0 ] + c 2 [ 2 3 1 ] + c 3 [ 5 4 − 2 ] , c 1 , c 2 , c 3 ∈ R  
    :sadA) = c1[1- 10] + c2[231] + c3[54-2], c1, c2, c3R

    This is obvious, so how do we simplify this answer?
    ::这是显而易见的,我们如何简化这一答案?

    To make this more useful, let's first put the columns in reduced row echelon form.  Doing this we get 
    ::为了让这更有用,让我们首先把柱子放在减排的梯层表上。这样我们得到的是

    rref ( [ 1 2 5 − 1 3 4 0 1 − 2 ] ) = I 3 × 3 = [ 1 0 0 0 1 0 0 0 1 ]
    ::rref([125-13401-2])=I3x3=[100000001]

    Now, we know that the column space of the matrix is just  c 1 [ 1 0 0 ] + c 2 [ 0 1 0 ] + c 3 [ 0 0 1 ] = [ c 1 c 2 c 3 ] ∀ c 1 , c 2 , c 3 ∈ R = R 3
    ::现在,我们知道矩阵的柱体空间只有c1[100]+c2[010]+c3[001]=[c1c2c3]c1,c2,c3,c3R=R3

    So now, the column space spans the three dimensional reals
    ::现在,柱体空间横跨三维真实

     

     

    Now, let's look at what happens when the columns are linearly independent and linearly dependent.
    ::现在,让我们看看当柱子线上独立和线上依赖时会发生什么。

    We know that when the columns are linearly independent then, by the invertible matrix theorem, we can apply row reduction to yield the identity matrix. This then means that the column space of that matrix is equal to the column space of the identity matrix which is then the n dimensional real numbers where n is the dimension of the matrix.
    ::我们知道,当列线线上独立时,通过倒置矩阵定理,我们可以应用减排来生成身份矩阵。这意味着该矩阵的列空间等于身份矩阵的列空间,而身份矩阵的列空间则等于n维实际数字,n是矩阵的维度。

     

    Now, going on to linear dependence, it depends on the number of linearly independent columns of the matrix. For example say you take
    ::现在,继续线直线依赖性,它取决于矩阵中线上独立的列数。例如,您使用

    [ 1 1 2 − 1 3 2 4 − 7 − 3 ]

    We see that the columns are linearly dependent because the sum of the first two equals the third. This then means that the third column is in the span of the first two, so the column space is just the span of the first two columns. 
    ::我们看到列的线性依赖, 因为前两个列的总和等于第三列。 这意味着第三列在前两个列的间距内, 因此列的间距只是前两个列的间距 。

    Hence, the span is just a plane. spanned by the vectors in the first two columns.
    ::因此,光圈只是一个平面,在前两列的矢量上横过。

    Now, on your own try to see if these matrices have linearly independent or dependent columns and similarly, calculate the column space.
    ::现在,你自己试试看这些矩阵是否具有线性独立或依赖性列,同样,计算列空间。

    1. [ 1 3 1 2 4 0 2 17 13 ] 2. [ 1 2 5 2 7 9 10 11 14 ] 3. [ 1 2 3 5 ] 4. [ 1 − 1 2 − 2 ] 5. [ 1 4 5 2 7 9 3 10 14 ]

     

    Let's now look at a new definition called the rank of the matrix which comes up a lot in advanced mathematics. There word rank is used quite often in higher mathematics, but in this lesson we will only learn about the linear algebraic sense of rank.
    ::现在让我们来看看一个新的定义, 叫做矩阵的等级, 它在高级数学中产生了很多东西。 高数学中常常使用单词等级, 但在这个教训中, 我们只学习线性代数等级感 。

    The dimension of the column space of a matrix is known as the rank of a matrix. 
    ::矩阵列空间的尺寸称为矩阵的等级。

    Let's look at some properties of the rank:
    ::让我们看看军衔的一些属性:

    For an n by m matrix,
    ::对于 n x m 矩阵,

    1. If n is less than m then the maximum rank could be n and no more
    ::1. 如果n小于m,最高军衔可以是n,但不超过n

    2. If n is greater than m then the maximum rank is m
    ::2. n大于m的,最高军衔为m

    To find the rank, reduce all of the columns of the matrix into row-echelon form.Then once you find all of the pivot columns, calculate the total number of pivot columns and then you have the rank.
    ::要找到排位, 请将矩阵的所有列排减为行- 梯层形式。 一旦您找到全部的支流列, 请计算轴列的总数, 然后您就可以获得排位 。

    This video may help you understand it better:
    ::这段影片也许能帮助您更好地理解:

     

    Now on your own, try some of the problems in the section above, and now calculate the rank of those matrices. Here they are again:
      1. [ 1 3 1 2 4 0 2 17 13 ] 2. [ 1 2 5 2 7 9 10 11 14 ] 3. [ 1 2 3 5 ] 4. [ 1 − 1 2 − 2 ] 5. [ 1 4 5 2 7 9 3 10 14 ]