Four Fundamental Spaces finder

Fundamental Spaces Visualizer

Matrix Input

Type integers. Updates apply on calculate.

RREF Process (A)

2
4
-2
1
2
-1
E1(1/2)
1
2
-1
1
2
-1
E21(-1)
1
2
-1
0
0
0
RREF
1
2
-1
0
0
0

RREF Process (Aᵗ)

2
1
4
2
-2
-1
E1(1/2)
1
1/2
4
2
-2
-1
E21(-4)
1
1/2
0
0
-2
-1
E31(2)
1
1/2
0
0
0
0
RREF AT
1
1/2
0
0
0
0

Column Space C(A)

The RREF has a pivot in column 1, so column 1 of the original matrix is a basis for the column space C(A).

21

Row Space R(A)

The RREF of Aᵗ has a pivot in column 1, so row 1 of the original matrix is a basis vector for the row space R(A).

AT =
2
1
4
2
-2
-1
RREF
1
1/2
0
0
0
0
R(A) = C(AT)
= span
24-2

Null Space N(A)

Columns 2, 3 are free variables, so the null space N(A) basis vectors are found by setting one free variable to 1 (others 0) and solving for the pivot variables.

RREF =
x1
x2
x3
1
2
-1
0
0
0
-2x2 + x3
x2
x3
=
x2
-210
+
x3
101
N(A) = span
-210
101

Left Null Space N(Aᵗ)

Column 2 of Aᵗ is a free variable, so the left null space N(Aᵗ) has one basis vector found by setting that free variable to 1 and the others to 0.

RREF (AT)
x1
x2
1
1/2
0
0
0
0
-1/2x2
x2
N(AT) = span
-1/21

Geometric Visualization

Ambient space: ℝ2Basis vectors: 1
x
y