Introduction to the null space of a matrix

$Ax=0$Ax=0

To prove the set of the solutions of $x$x can be a subspace.

1. When $x = 0$x=0, it works.
2. Fine two solutions of $x$x, $v_1$v1​ and $v_2$v2​, they satisfy
   $Av_1 = 0$Av1​=0
   $Av_2 = 0$Av2​=0
   Then:
   $A(v_1+v_2)=0$A(v1​+v2​)=0 and $A(cv_1) = 0, A(cv_2) = 0$A(cv1​)=0,A(cv2​)=0

So the null space is a valid subspace.

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Calculating the null space of a matrix

$[A|0]→[rref(A)|0]$[A∣0]→[rref(A)∣0]

$N(A)=N(rref(A))$N(A)=N(rref(A))

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Relation to linear independence

The column vectors of matrix $A$A are linearly independent if and only if the null space of matrix $A$A only contains 0.

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Column space of a matrix

$Ax = b$Ax=b has no solution, then $b$b is not in the column space of $A$A.
If $Ax=b$Ax=b has at least one solution, then $b$b is in the column space of $A$A.

Null space and column space basis

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Visualizing a column space as a plane in R3

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Proof: Any subspace basis has same number of elements

If set $A$A, the basis of $V$V has $n$n entries, then we suppose a set of $m$m vectors $B(m<n)$B(m<n), we can replace at least one entry in $B$B with an entry in $A$A one time, and do ti $m$m times, $B$B becomes a set of m entries in $A$A, and basis in $A$A are independent, so $m>n$m>n.
So, if $C$C(m elements) and $D$D(n elements) are both basis of a same subspace, then $m>=n$m>=n, also $n >= m$n>=m.

Dimension of a subspace = # of elements in a basis for the subspace.

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Dimension of the null space or nullity

nullity: Dimension of a null pace = # of free variables.
rank: Dimension of the column space = linearly independent column vectors.

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Showing relation between basis cols and pivot cols

If the pivot columns of rref of $A$A are linearly independent, then the null space of rref$A$A only contains 0.
We know that $N(A)=N(rref(A))$N(A)=N(rref(A)) since row elementary operation didn’t change the solutions of matrix equation.
So the null space of $A$A only contains 0.
So the pivot columns of $A$A are linearly independent.

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Showing that the candidate basis does span C(A)

Since $N(A) = N(rrefA)$N(A)=N(rrefA)
$A\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\\x_5\end{matrix}\right] = 0 \ \ \ \ \ \ \ \ \ \ rref(A)\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\\x_5\end{matrix}\right] = 0$A⎣⎢⎢⎢⎢⎡​x1​x2​x3​x4​x5​​⎦⎥⎥⎥⎥⎤​=0          rref(A)⎣⎢⎢⎢⎢⎡​x1​x2​x3​x4​x5​​⎦⎥⎥⎥⎥⎤​=0$x_1\vec{a_1} + x_2\vec{a_2}+x_3\vec{a_3}+x_4\vec{a_4}+x_5\vec{a_5}=0\tag1$x1​a1​​+x2​a2​​+x3​a3​​+x4​a4​​+x5​a5​​=0(1)

$x_1\vec{r_1} + x_2\vec{r_2}+x_3\vec{r_3}+x_4\vec{ra_4}+x_5\vec{r_5}=0 \tag 2$x1​r1​​+x2​r2​​+x3​r3​​+x4​ra4​​+x5​r5​​=0(2)

Since $x_3$x3​ and $x_5$x5​ are free variables in equation $(2)$(2), $x_3$x3​ and $x_5$x5​ are also free variables in equation $(1)$(1), so column 3 and column 5 in $A$A can also be represented by other columns in $A$A.
So column 1, column 2 and column 4 in $A$A can span the column space of $A$A.
Besides, we have proved that column 1, column 2 and column 4 in $A$A are independent.
So the corresponding columns of the basis columns in $rref(A)$rref(A) are the basis columns of $A$A.