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1 - The Column Space of a Matrix

0001-01-01 08:05 LMT
2020-05-24 22:11 CST
CC BY-NC 4.0

A matrix is a square / rectangle of numbers.

Column / Row Space

$Ax = A_1x_1 + A_2x_2 + \dots + A_nx_n$ is the linear combination of columns of $A$.

The column space of $A$, $\textsf{C}(A)$ is all vectors $A_x$, all linear combination of columns of $A$.

$\textsf{C}(A)$ is a plane.

Basic for the Column / Row Space

$A = \begin{bmatrix} 1 & 4 & 5 \\ 3 & 2 & 5 \\ 2 & 1 & 3 \end{bmatrix}$

Column 3 is not independent, $A_3 = A_1 + A_2$ so it is dependent.

Let $A = CR = \begin{bmatrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$,

The rows of $R$ are a basis for the row space.

$A = CR$ shows that column rank of $A$ = row rank of $A$.

The $r$ columns of $C$ are a basis for the column space of $A$, the $r$ rows of $R$ are a basis for the row space of $A$: dimension $r$.

Counting Theorem: an $n$-matrix of rank $r$ has $n-r$ independent solutions to $Ax = 0$.

Matrix $A$ with Rank $1$

If all columns of $A$ are multiples of column $1$, then all rows of $A$ are multiples of one row.

Proof using $A = CR$: one column $v$ in $C$ $\rightarrow$ one row $w$ in $R$. All rows are multiple of $w$.

($C$只有一列,所以$R$只有一行,因此矩阵的所有行都是$w$的倍数。)

Equation $A = CR$

Pros:

  • $C$ has columns directly from $A$
  • $R$ turns out to be the row reduced echelon form of $A$
  • Row rank = Column rank is clear ($C$ = column basis, $R$ = row basis.)

Cons:

  • $C$ and $R$ could be hard to deal with
  • If $A$ is invertible then $C = A$ and $R = I$: $A = AI$