Linear Equations & Linear Independent
Start from linear equations with n equations and n unknown variables.
$
\begin{eqnarray}
x_1-2x_2+x_3&=&0\\
2x_2-8x_3&=&8\\
-4x_1+5x_2+9x_3&=&-9
\end{eqnarray}
$
The matrix form:
$A=$
$
\begin{bmatrix}
1&-2&1\\
0&2&-8\\
-4&5&9
\end{bmatrix}
$
$X=$
$
\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}
$
$b=$
$
\begin{bmatrix}
0\\
8\\
-9
\end{bmatrix}
$
The problem is: finding $X$ that satisfies $AX=b$.
Intuitively, we can draw the row picture of matrix $A$ and find $X$ as the intersection point of the three equations. But it is not clear to sovle the problem in such 3-D space. And lets see from the column picture of matrix $A$. We will find it more clear how to find $X$ to satisfies $AX=b$.
Actually, vector $X$ defines a linear combinaton of columns in $A$. And we will get the solution if $b$ can be generated by any linear combination of columns in $A$.
Linear Independent
For vectors $A=\{v_1,…,v_p\}$ in $\mathbb{R}^n$ , if $AX=0$ has only trivial solution, then we called columns in $A$ is linear independent. Otherwise, linear dependent.
Inverse
Nonsingular matrix $A$: exists $A^{-1}$ satisfies $AA^{-1}=I$.
Singulatr matrix $A$: inverse does not exists.
A reversible:
a. & $(A^{-1})^{-1}=A$.
b. if $B$ reversible, $(AB)^{-1}=B^{-1}A^{-1}$.
c. $(A^{T})^{-1}=(A^{-1})^{T}$.
d. $A\Leftrightarrow I$.
e. $A$ has n main elements.
f. $AX=0$ has only trivial solution.
g. columns in $A$ linear independent.
h. linear transform $x\mapsto Ax$ is one-to-one.
i. for $b$ in $mathbb{R}^n$ equation $Ax=b$ has exactly one solution.
j. columns in $A$ generate $mathbb{R}^n$.
k. linear transform $x\mapsto Ax$ map $\mathbb{R}^n$ to $\mathbb{R}^n$.
l. exists $n\times n$ matrix $C$ and $D$, $CA=I$ and $AD=I$.
m. columns in $A$ is a basis of $\mathbb{R}^n$.
n. $Col\ A=\mathbb{R}^n$.
o. $dim\ Col\ A=n$.
p. $rank\ A=n$.
q. $Nul\ A={0}$.
r. $dim\ Nul\ A=0$.
s. 0 is not the feature value of $A$.
t. $det\ A\neq 0$.
u. $(Col\ A) zhengjiaobu=\{0\}$.
v. $(Nul\ A) zhengjiaobu=\mathbb{R}^n$.
w. $Row\ A=\mathbb{R}^n$.
x. $A$ has n non-zero singular value.
dim H is the number of columns in a basis of $H$.
rank A is the dim of $colA$.
for a $m\times n$ matrix, $rank\ A+dim\ Null\ A\ =\ n$.