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133 - FML 3 - 18.1W - e.g. 46_$\tkco{mock 1 ✓}$
Related Topics
Wize University Linear Algebra Textbook > Vector Spaces and Subspaces
Column Space and Null Space (Range and Kernel)
3 Activities
The matrix
M
‾
=
[
4
0
1
0
1
10
3
−
3
0
−
1
0
−
1
−
9
−
2
7
0
1
0
2
16
6
−
6
0
−
2
0
−
2
−
18
−
5
]
\M = \begin{bmatrix} 4 & 0 & 1 & 0 & 1 & 10 & 3 \\[5pt] -3 & 0 & -1 & 0 & -1 & -9 & -2 \\[5pt] 7 & 0 & 1 & 0 & 2 & 16 & 6 \\[5pt] -6 & 0 & -2 & 0 & -2 & -18 & -5 \end{bmatrix}
M
=
4
−
3
7
−
6
0
0
0
0
1
−
1
1
−
2
0
0
0
0
1
−
1
2
−
2
10
−
9
16
−
18
3
−
2
6
−
5
Has row-echelon form given by:
M
‾
R
E
F
=
[
1
0
1
4
0
1
4
5
2
3
4
0
0
1
0
1
6
−
1
0
0
0
0
1
3
0
0
0
0
0
0
0
1
]
\M_{_{REF}} = \begin{bmatrix} 1 & 0 & \frac14 & 0 & \frac14 & \frac52 & \frac34 \\[5pt] 0 & 0 & 1 & 0 & 1 & 6 & -1 \\[5pt] 0 & 0 & 0 & 0 & 1 & 3 & 0 \\[5pt] 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}
M
R
E
F
=
1
0
0
0
0
0
0
0
4
1
1
0
0
0
0
0
0
4
1
1
1
0
2
5
6
3
0
4
3
−
1
0
1
Part 1
Part 2
A basis for the row-space is given by (select all that apply):
The first, second, third and fourth rows of
M
‾
\M
M
The second, third and fourth rows of
M
‾
\M
M
The first, third and fourth rows of
M
‾
\M
M
The first, second, third and fourth rows of
M
‾
R
E
F
\M_{_{REF}}
M
R
E
F
The second, third and fourth rows of
M
‾
R
E
F
\M_{_{REF}}
M
R
E
F
The first, third and fourth rows of
M
‾
R
E
F
\M_{_{REF}}
M
R
E
F
None of the other options.
I don't know
Previous Part
Check Part 1
Next Part
Check Submission
More Column Space and Null Space (Range and Kernel) Questions:
Practice: Null Space and Range
Suppose that
T
:
R
3
→
R
3
T:\mathbb{R}^3\rightarrow\mathbb{R}^3
T
:
R
3
→
R
3
is defined by
T
(
x
⃗
)
:
=
A
x
⃗
T(\vec x):=A\vec x
T
(
x
)
:=
A
x
, where
A
=
[
a
⃗
1
a
⃗
2
a
⃗
3
]
=
[
1
1
2
2
2
4
−
1
3
6
]
A=\left[\begin{array}{rrr}\vec a_1&\vec a_2&\vec a_3\end{array}\right]=\left[\begin{array}{rrr}1&1&2\\2&2&4\\-1&3&6\end{array}\right]
A
=
[
a
1
a
2
a
3
]
=
1
2
−
1
1
2
3
2
4
6
Find
R
a
n
g
e
(
T
)
Range(T)
R
an
g
e
(
T
)
and
N
u
l
l
(
T
)
Null(T)
N
u
l
l
(
T
)
And write each as a span of as few vectors as possible.
Example: Rank-Nullity Theorem
Let
M
2
×
2
M_{2\times2}
M
2
×
2
be the vector space of
2
×
2
2\times 2
2
×
2
matrices with real number entries, and define the transformation
L
:
M
2
×
2
→
M
2
×
2
L:M_{2\times2}\to M_{2\times2}
L
:
M
2
×
2
→
M
2
×
2
by:
L
(
B
)
=
A
B
L(B)=AB
L
(
B
)
=
A
B
where
A
=
[
3
1
3
1
]
A=\left[\begin{array}{rr}3&1\\3&1\end{array}\right]
A
=
[
3
3
1
1
]
Practice: Rank-Nullity Theorem
Let
P
2
P_2
P
2
be the vector space of polynomials of degree at most 2, with the standard operations of polynomial addition and scalar multiplication:
P
2
(
x
)
=
{
p
(
x
)
=
a
x
2
+
b
x
+
c
∣
a
,
b
,
c
∈
R
}
P_2(x)=\{p(x)=ax^2+bx+c\;|\;a,b,c\in\mathbb{R}\}
P
2
(
x
)
=
{
p
(
x
)
=
a
x
2
+
b
x
+
c
∣
a
,
b
,
c
∈
R
}
Let
L
:
P
2
→
P
2
L:P_2\to P_2
L
:
P
2
→
P
2
be the transformation defined by:
133 - FML 3 - 18.1W - e.g. 46_$\tkcth{mock 1 ✓}$
The matrix
M
‾
=
[
4
0
1
0
1
10
3
−
3
0
−
1
0
−
1
−
9
−
2
7
0
1
0
2
16
6
−
6
0
−
2
0
−
2
−
18
−
5
]
\M = \begin{bmatrix} 4 & 0 & 1 & 0 & 1 & 10 & 3 \\[5pt] -3 & 0 & -1 & 0 & -1 & -9 & -2 \\[5pt] 7 & 0 & 1 & 0 & 2 & 16 & 6 \\[5pt] -6 & 0 & -2 & 0 & -2 & -18 & -5 \end{bmatrix}
M
=
4
−
3
7
−
6
0
0
0
0
1
−
1
1
−
2
0
0
0
0
1
−
1
2
−
2
10
−
9
16
−
18
3
−
2
6
−
5
Has reduced-row-echelon form given by:
19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{45}$_
Let
A
‾
=
[
a
i
1
⋯
a
i
n
]
\A = \big[ a_{i1} \; \cdots \; a_{in} \big]
A
=
[
a
i
1
⋯
a
in
]
, where
{
a
i
2
,
a
i
3
,
a
i
5
}
\{ a_{i2},\, a_{i3},\, a_{i5} \}
{
a
i
2
,
a
i
3
,
a
i
5
}
is a basis for the column space of
A
‾
\A
A
. The the subset
U
=
{
y
⃗
∈
R
n
:
A
‾
y
⃗
=
b
⃗
,
b
⃗
∈
span
{
a
i
2
,
a
i
3
,
a
i
5
}
}
U = \left\{ \vec{y} \, \in \, \mathbb{R}^{n} \, : \, \A \vec{y} = \vb,\quad \vb \, \in \, \text{span}\{ a_{i2},\, a_{i3},\, a_{i5}\} \right\}
U
=
{
y
∈
R
n
:
A
y
=
b
,
b
∈
span
{
a
i
2
,
a
i
3
,
a
i
5
}
}
is a subspace of
R
n
\mathbb{R}^{n}
R
n
.
19.4F_Mid_Builder_$\tkcth{8.4.17}\tkcf{p2}$__$\key{F.Quiz}$
A basis for
ker
M
‾
\ker\M
ker
M
(
i.e.
null
M
‾
\text{null}\,\M
null
M
) for the matrix
M
‾
=
[
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
]
\M = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
M
=
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
is given by the set of vectors (select all that apply):
19.4F_Mid_Builder_$\tkcth{8.4.17}\tkcf{p1}$_$\key{F.Quiz}$
A basis for
Im
M
‾
\text{Im}\, \M
Im
M
for the matrix
M
‾
=
[
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
]
\M = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
M
=
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
is given by the set of vectors (select all that apply):
19.4F_Mid_Builder_$\tkcth{8.4.}\tkcf{20}\tkcth{p2}$_$\tkco{Mock 3}$
A basis for the kernel of the matrix
M
‾
=
[
5
0
−
7
1
0
1
4
0
0
0
1
0
5
0
0
1
]
\M = \begin{bmatrix} 5 & 0 & -7 & 1 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 5 & 0 & 0 & 1 \\ \end{bmatrix}
M
=
5
0
0
5
0
1
0
0
−
7
4
1
0
1
0
0
1
is given by the set of vectors (select all that apply):
19.4F_Final_Builder_Ch_17.3_Least_Squares_$\tkco{eg3}$_$\key{Final}$_Builder_$\tkcth{17.3.}\tkcf{3}$_
The system given by:
{
x
−
2
y
=
−
4
2
x
−
4
y
=
7
\left\{ \begin{array}{rr} x - 2y \quad =& -4 \\ 2x - 4y \quad =& 7 \end{array} \right.
{
x
−
2
y
=
2
x
−
4
y
=
−
4
7
can be represented as a single matrix equation
A
‾
x
⃗
=
b
⃗
\A \, \vx = \vb
A
x
=
b
, where
A
‾
\A
A
is the matrix of coefficients and
b
⃗
\vb
b
is the constant vector
[
−
4
7
]
\colvec{-4}{7}
[
−
4
7
]
.
19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p2}$_$\tkct{redo solution: *not* obvious =D}$
A basis for the kernel of the matrix
M
‾
=
[
0
3
0
−
5
0
1
3
0
0
0
1
0
0
0
0
1
]
\M = \begin{bmatrix} 0 & 3 & 0 & -5 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}
M
=
0
0
0
0
3
1
0
0
0
3
1
0
−
5
0
0
1
is given by the set of vectors (select all that apply):
Dimension of Null Spaces
Prove that
d
i
m
(
n
u
l
l
(
A
B
)
)
≥
d
i
m
(
n
u
l
l
(
B
)
)
dim(null(AB))\geq dim(null(B))
d
im
(
n
u
l
l
(
A
B
))
≥
d
im
(
n
u
l
l
(
B
))
for any matrices
A
,
B
A,B
A
,
B
for which
A
B
AB
A
B
is defined.
19.4F_WML_6_$\tkco{eg6}$_$\key{Final}$_Builder_$\tkcth{17.1.}\tkct{6}$_
Find a basis for the kernel of the matrix of coefficients for the system:
[
1
0
2
2
1
4
1
1
2
]
[
x
1
x
2
x
3
]
=
[
0
0
0
]
\begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 4 \\ 1 & 1 & 2 \end{bmatrix} % \colvecth{x_1}{x_2}{x_3} % = % \colvecth{0}{0}{0}
1
2
1
0
1
1
2
4
2
x
1
x
2
x
3
=
0
0
0
Use your result to show that
any
vector in the kernel of the matrix is orthogonal to
any
vector in the space spanned by the rows of the matrix.
Practice Question: Bases of Column and Row Spaces
Practice Question: Bases of Column and Row Spaces
Find
bases
for each of
r
o
w
(
A
)
,
c
o
l
(
A
)
row(A) ,col(A)
r
o
w
(
A
)
,
co
l
(
A
)
when
A
=
[
0
1
−
2
2
5
1
1
2
2
0
1
1
2
−
3
1
]
A=\left[\begin{array}{rrrr}0&1&-2&2&5\\1&1&2&2&0\\1&1&2&-3&1\end{array}\right]
A
=
0
1
1
1
1
1
−
2
2
2
2
2
−
3
5
0
1
$\tkco{move to mock ✓}$ 133 - FML 3 - 18.1W - e.g. 50 $\tkcth{copy - not update for now}$ $\tkcf{mock2}$
Given the matrix
M
‾
=
[
−
2
−
4
4
−
2
1
2
−
4
−
4
]
\bcb{\ul{M} = \begin{bmatrix} -2 & -4 & 4 & -2 \\ 1 & 2 & -4 & -4 \end{bmatrix}}
M
=
[
−
2
1
−
4
2
4
−
4
−
2
−
4
]
$\tkco{move to mock ✓}$ $\tkct{needs vid / solution!}$ 133 - FML 3 - 18.1W - e.g. 50
Given the matrix
M
‾
=
[
−
2
−
4
4
−
2
1
2
−
4
−
4
]
\bcb{\ul{M} = \begin{bmatrix} -2 & -4 & 4 & -2 \\ 1 & 2 & -4 & -4 \end{bmatrix}}
M
=
[
−
2
1
−
4
2
4
−
4
−
2
−
4
]
, find a basis for the row-, column- and null-space of
M
‾
\bcb{\ul{M}}
M
.
Practice Question: Rank-Nullity
Write
(
−
2
,
2
,
0
,
−
5
)
(-2,2,0,-5)
(
−
2
,
2
,
0
,
−
5
)
as a sum
r
⃗
+
n
⃗
\vec r+\vec n
r
+
n
, where
r
⃗
∈
r
o
w
(
A
)
,
n
⃗
∈
n
u
l
l
(
A
)
\vec r\in row(A),\vec n\in null(A)
r
∈
r
o
w
(
A
)
,
n
∈
n
u
l
l
(
A
)
, and where
A
:
=
[
1
4
1
−
3
1
1
1
1
]
A:=\left[\begin{array}{rrrr}1&4&1&-3\\1&1&1&1\end{array}\right]
A
:=
[
1
1
4
1
1
1
−
3
1
]
.
19.4F_133_8.2_Mock_F1_$\tkco{eg8}$_$\key{Final}$_Builder_$\tkcth{8.2.}\tkcf{8}$_
If
dim
(
ker
(
A
‾
)
)
=
0
\textrm{dim}(\textrm{ker}(\A)) = 0
dim
(
ker
(
A
))
=
0
, for an
n
×
n
n \times n
n
×
n
matrix
A
‾
\A
A
, then
Im
(
A
‾
)
\text{Im}\, (\A)
Im
(
A
)
is
R
n
\mathbb{R}^{n}
R
n
, and every vector in
R
n
\mathbb{R}^{n}
R
n
is perpendicular to the space spanned by the rows of
A
‾
\A
A
.
Bases for Row and Column Spaces
Find bases for the row and column spaces of
A
=
[
1
2
3
4
5
6
7
8
9
10
11
12
]
A=\left[\begin{array}{rrr}1&2&3\\4&5&6\\7&8&9\\10&11&12\end{array}\right]
A
=
1
4
7
10
2
5
8
11
3
6
9
12
Let
A
=
(
1
2
−
2
0
1
4
−
3
−
6
6
2
4
−
4
)
A = \left ( \begin{array}{ccc} 1 & 2 & -2\\ 0 & 1 & 4\\ -3 & -6 & 6 \\ 2 & 4 & -4 \end{array} \right )
A
=
1
0
−
3
2
2
1
−
6
4
−
2
4
6
−
4
be a real
4
×
3
4 \times 3
4
×
3
matrix. Find a basis for the kernel of
A
A
A
, and for the image of
A
A
A
. Using the dimensions of the kernel and image of
A
A
A
, verify the rank-nullity theorem.
Bases of Row, Column, & Null Spaces
Suppose that
A
=
[
1
−
1
3
−
1
1
−
1
−
2
2
3
1
−
1
1
]
A=\left[\begin{array}{rrr}1&-1&3\\-1&1&-1\\-2&2&3\\1&-1&1\end{array}\right]
A
=
1
−
1
−
2
1
−
1
1
2
−
1
3
−
1
3
1
.
Find a basis for each of
r
o
w
(
A
)
,
c
o
l
(
A
)
row(A),col(A)
r
o
w
(
A
)
,
co
l
(
A
)
.
What are
r
a
n
k
(
A
)
&
d
i
m
(
n
u
l
l
(
A
)
)
rank(A)\text{ \& }dim(null(A))
r
ank
(
A
)
&
d
im
(
n
u
l
l
(
A
))
?
19.4F_Mid_Builder_$\tkcth{8.4.19}\tkcf{p1}$_$\tkct{redo solution}$
A basis for
Im
M
‾
\text{Im}\, \M
Im
M
for the matrix
M
‾
=
[
0
3
0
−
5
0
1
3
0
0
0
1
0
0
0
0
1
]
\M = \begin{bmatrix} 0 & 3 & 0 & -5 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}
M
=
0
0
0
0
3
1
0
0
0
3
1
0
−
5
0
0
1
is given by the set of vectors (select all that apply):
Let
L
:
R
2
⟶
R
3
L:\mathbb{R}^2\longrightarrow \mathbb{R}^3
L
:
R
2
⟶
R
3
be a linear transformation
Prove that a basis for
I
m
(
L
)
Im(L)
I
m
(
L
)
can not be a basis for
R
3
\mathbb{R}^3
R
3
Basis and Dimension
Practice: Finding a Basis
Let
A
=
[
1
−
1
1
]
A=\left[\begin{array}{rrr} 1&-1&1\\ \end{array}\right]
A
=
[
1
−
1
1
]
and consider the special subspace of
R
3
\mathbb{R}^3
R
3
called the
null space
:
N
=
{
x
⃗
∈
R
3
∣
A
x
⃗
=
0
⃗
}
N=\{\vec x\in\mathbb{R}^3\;|\;A\vec x=\vec{0}\}
N
=
{
x
∈
R
3
∣
A
x
=
0
}
133 - FML 3 - 18.1W - e.g. 37
Given
A
=
[
−
4
2
14
1
−
1
1
4
0
−
3
0
9
1
1
1
−
2
0
]
⟶
RREF
[
1
0
−
3
0
0
1
1
0
0
0
0
1
0
0
0
0
]
\bcb{A = \begin{bmatrix} -4 & 2 & 14 & 1 \\ -1 & 1 & 4 & 0 \\ -3 & 0 & 9 &1 \\ 1 & 1 & -2 & 0 \end{bmatrix} \overset{\text{\footnotesize{RREF}}}{\longrightarrow} \begin{bmatrix} 1 & 0 & -3 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 &1 \\ 0 & 0 & 0 & 0 \end{bmatrix} }
A
=
−
4
−
1
−
3
1
2
1
0
1
14
4
9
−
2
1
0
1
0
⟶
RREF
1
0
0
0
0
1
0
0
−
3
1
0
0
0
0
1
0
, determine the dimension of the row space, column space and null space of
A
\bcb{A}
A
, respectively, and provide a basis for each.
Practice: Matrix Invertibility
Suppose
A
A
A
is an
n
×
n
n\times n
n
×
n
matrix whose
rows
span
R
n
\mathbb{R}^n
R
n
Show that for all choices of vectors
x
⃗
1
,
x
⃗
2
,
.
.
.
,
x
⃗
k
∈
R
n
\vec x_1,\vec x_2,...,\vec x_k\in\mathbb{R}^n
x
1
,
x
2
,
...
,
x
k
∈
R
n
:
s
p
a
n
{
x
⃗
1
,
x
⃗
2
,
.
.
.
,
x
⃗
k
}
=
R
n
span\{\vec x_1,\vec x_2,...,\vec x_k\}=\mathbb{R}^n
s
p
an
{
x
1
,
x
2
,
...
,
x
k
}
=
R
n
if and only if
s
p
a
n
{
A
x
⃗
1
,
A
x
⃗
2
,
.
.
.
,
A
x
⃗
k
}
=
R
n
span\{A\vec x_1,A\vec x_2,...,A\vec x_k\}=\mathbb{R}^n
s
p
an
{
A
x
1
,
A
x
2
,
...
,
A
x
k
}
=
R
n
.
Let
L
:
R
3
⟶
R
2
L:\mathbb{R}^3\longrightarrow\mathbb{R}^2
L
:
R
3
⟶
R
2
be defined by:
L
(
(
x
,
y
,
z
)
)
=
(
x
+
2
y
+
3
z
,
−
x
+
2
y
+
5
z
)
L\big((x,y,z)\big)=(x+2y+3z,-x+2y+5z)
L
(
(
x
,
y
,
z
)
)
=
(
x
+
2
y
+
3
z
,
−
x
+
2
y
+
5
z
)
(a)
Prove that
L
L
L
is a linear transformation
133- 05.4F - Final - PartI Question#6
Let
A
\bcb{A}
A
be the matrix
A
=
[
1
−
2
1
4
4
−
1
2
1
2
3
2
−
4
0
2
1
]
\bcb{A=\begin{bmatrix} 1&-2&1&4&4\\ -1&2&1&2&3\\ 2&-4&0&2&1 \end{bmatrix}}
A
=
1
−
1
2
−
2
2
−
4
1
1
0
4
2
2
4
3
1
Let
r
\bcb{r}
r
be the rank of
A
\bcb{A}
A
and
q
\bcb{q}
q
the dimension of the null space of
A
\bcb{A}
A
(i.e. the nullity of
A
\bcb{A}
A
. What are the values of
r
\bcb{r}
r
and
q
\bcb{q}
q
?
Basis and Dimension
Practice: Finding a Basis
Let
A
=
[
1
−
1
1
]
A=\left[\begin{array}{rrr} 1&-1&1\\ \end{array}\right]
A
=
[
1
−
1
1
]
and consider the special subspace of
R
3
\mathbb{R}^3
R
3
called the
null space
:
N
=
{
x
⃗
∈
R
3
∣
A
x
⃗
=
0
⃗
}
N=\{\vec x\in\mathbb{R}^3\;|\;A\vec x=\vec{0}\}
N
=
{
x
∈
R
3
∣
A
x
=
0
}
Suppose that
A
A
A
is a
5
×
7
5\times7
5
×
7
matrix and
d
=
dim
(
n
u
l
l
(
A
)
)
d=\dim\left(null\left(A\right)\right)
d
=
dim
(
n
u
l
l
(
A
)
)
. Which of the following is always true?
Which of the following is a one-dimensional subspace of
R
3
\mathbb{R}^3
R
3
?