Skip to main content \(\newcommand{\avec}{{\mathbf a}}
\newcommand{\bvec}{{\mathbf b}}
\newcommand{\cvec}{{\mathbf c}}
\newcommand{\dvec}{{\mathbf d}}
\newcommand{\dtil}{\widetilde{\mathbf d}}
\newcommand{\evec}{{\mathbf e}}
\newcommand{\fvec}{{\mathbf f}}
\newcommand{\nvec}{{\mathbf n}}
\newcommand{\pvec}{{\mathbf p}}
\newcommand{\qvec}{{\mathbf q}}
\newcommand{\svec}{{\mathbf s}}
\newcommand{\tvec}{{\mathbf t}}
\newcommand{\uvec}{{\mathbf u}}
\newcommand{\vvec}{{\mathbf v}}
\newcommand{\wvec}{{\mathbf w}}
\newcommand{\xvec}{{\mathbf x}}
\newcommand{\yvec}{{\mathbf y}}
\newcommand{\zvec}{{\mathbf z}}
\newcommand{\rvec}{{\mathbf r}}
\newcommand{\mvec}{{\mathbf m}}
\newcommand{\zerovec}{{\mathbf 0}}
\newcommand{\onevec}{{\mathbf 1}}
\newcommand{\real}{{\mathbb R}}
\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2
\end{array}\right]}
\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2
\end{array}\right]}
\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3
\end{array}\right]}
\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3
\end{array}\right]}
\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4
\end{array}\right]}
\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4
\end{array}\right]}
\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\
#4 \\ #5 \\ \end{array}\right]}
\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\
#4 \\ #5 \\ \end{array}\right]}
\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}
\newcommand{\laspan}[1]{\text{Span}\{#1\}}
\newcommand{\bcal}{{\cal B}}
\newcommand{\ccal}{{\cal C}}
\newcommand{\scal}{{\cal S}}
\newcommand{\wcal}{{\cal W}}
\newcommand{\ecal}{{\cal E}}
\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}
\newcommand{\gray}[1]{\color{gray}{#1}}
\newcommand{\lgray}[1]{\color{lightgray}{#1}}
\newcommand{\rank}{\text{rank}}
\newcommand{\row}{\text{Row}}
\newcommand{\col}{\text{Col}}
\renewcommand{\row}{\text{Row}}
\newcommand{\nul}{\text{Nul}}
\newcommand{\var}{\text{Var}}
\newcommand{\corr}{\text{corr}}
\newcommand{\len}[1]{\left|#1\right|}
\newcommand{\bbar}{\overline{\bvec}}
\newcommand{\bhat}{\widehat{\bvec}}
\newcommand{\bperp}{\bvec^\perp}
\newcommand{\xhat}{\widehat{\xvec}}
\newcommand{\vhat}{\widehat{\vvec}}
\newcommand{\uhat}{\widehat{\uvec}}
\newcommand{\what}{\widehat{\wvec}}
\newcommand{\Sighat}{\widehat{\Sigma}}
\newcommand{\circledNumber}[1]{\boxed{#1}}
\newcommand{\IR}{\mathbb{R}}
\newcommand{\IC}{\mathbb{C}}
\renewcommand{\P}{\mathcal{P}}
\renewcommand{\Im}{\operatorname{Im}}
\newcommand{\RREF}{\operatorname{RREF}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\setList}[1]{\left\{#1\right\}}
\newcommand{\setBuilder}[2]{\left\{#1\,\middle|\,#2\right\}}
\newcommand{\unknown}{\,{\color{gray}?}\,}
\newcommand{\drawtruss}[2][1]{
\begin{tikzpicture}[scale=#1, every node/.style={scale=#1}]
\draw (0,0) node[left,magenta]{C} --
(1,1.71) node[left,magenta]{A} --
(2,0) node[above,magenta]{D} -- cycle;
\draw (2,0) --
(3,1.71) node[right,magenta]{B} --
(1,1.71) -- cycle;
\draw (3,1.71) -- (4,0) node[right,magenta]{E} -- (2,0) -- cycle;
\draw[blue] (0,0) -- (0.25,-0.425) -- (-0.25,-0.425) -- cycle;
\draw[blue] (4,0) -- (4.25,-0.425) -- (3.75,-0.425) -- cycle;
\draw[thick,red,->] (2,0) -- (2,-0.75);
#2
\end{tikzpicture}
}
\newcommand{\trussNormalForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, blue,->] (4,0) -- (3.5,0.5);
}
\newcommand{\trussCompletion}{
\trussNormalForces
\draw [thick, magenta,<->] (0.4,0.684) -- (0.6,1.026);
\draw [thick, magenta,<->] (3.4,1.026) -- (3.6,0.684);
\draw [thick, magenta,<->] (1.8,1.71) -- (2.2,1.71);
\draw [thick, magenta,->] (1.6,0.684) -- (1.5,0.855);
\draw [thick, magenta,<-] (1.5,0.855) -- (1.4,1.026);
\draw [thick, magenta,->] (2.4,0.684) -- (2.5,0.855);
\draw [thick, magenta,<-] (2.5,0.855) -- (2.6,1.026);
}
\newcommand{\trussCForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, magenta,->] (0,0) -- (0.4,0.684);
\draw [thick, magenta,->] (0,0) -- (0.5,0);
}
\newcommand{\trussStrutVariables}{
\node[above] at (2,1.71) {\(x_1\)};
\node[left] at (0.5,0.866) {\(x_2\)};
\node[left] at (1.5,0.866) {\(x_3\)};
\node[right] at (2.5,0.866) {\(x_4\)};
\node[right] at (3.5,0.866) {\(x_5\)};
\node[below] at (1,0) {\(x_6\)};
\node[below] at (3,0) {\(x_7\)};
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section 3.5 Subspaces of \(\real^p\)
Activity 3.5.1 .
Letβs consider the following matrix \(A\) and its reduced row echelon form.
\begin{equation*}
A = \left[\begin{array}{rrrr}
2 \amp -1 \amp 2 \amp 3 \\
1 \amp 0 \amp 0 \amp 2 \\
-2 \amp 2 \amp -4 \amp -2 \\
\end{array}\right]
\sim
\left[\begin{array}{rrrr}
1 \amp 0 \amp 0 \amp 2 \\
0 \amp 1 \amp -2 \amp 1 \\
0 \amp 0 \amp 0 \amp 0 \\
\end{array}\right]\text{.}
\end{equation*}
(a)
Are the columns of
\(A\) linearly independent? Do they span
\(\real^3\text{?}\)
(b)
Give a parametric description of the solution space to the homogeneous equation
\(A\xvec = \zerovec\text{.}\)
(c)
Explain how this parametric description produces two vectors
\(\wvec_1\) and
\(\wvec_2\) whose span is the solution space to the equation
\(A\xvec = \zerovec\text{.}\)
(d)
What can you say about the linear independence of the set of vectors
\(\wvec_1\) and
\(\wvec_2\text{?}\)
(e)
Letβs denote the columns of
\(A\) as
\(\vvec_1\text{,}\) \(\vvec_2\text{,}\) \(\vvec_3\text{,}\) and
\(\vvec_4\text{.}\) Explain why
\(\vvec_3\) and
\(\vvec_4\) can be written as linear combinations of
\(\vvec_1\) and
\(\vvec_2\text{.}\)
(f)
Explain why
\(\vvec_1\) and
\(\vvec_2\) are linearly independent and
\(\laspan{\vvec_1,\vvec_2} =
\laspan{\vvec_1, \vvec_2, \vvec_3, \vvec_4}\text{.}\)
Definition 3.5.1 .
A
subspace of
\(\real^p\) is a nonempty subset of
\(\real^p\) such that any linear combination of vectors in that set is also in the set.
Activity 3.5.2 .
We will look at some subspaces of
\(\real^2\text{.}\)
(a)
Explain why a line that does not pass through the origin is not a subspace of
\(\real^2\text{.}\)
(b)
Explain why any subspace of
\(\real^2\) must contain the zero vector
\(\zerovec\text{.}\)
(c)
Explain why the subset
\(S\) of
\(\real^2\) that consists of only the zero vector
\(\zerovec\) is a subspace of
\(\real^2\text{.}\)
(d)
Explain why the subspace
\(S=\real^2\) is itself a subspace of
\(\real^2\text{.}\)
(e)
If
\(\vvec\) and
\(\wvec\) are two vectors in a subspace
\(S\text{,}\) explain why
\(\laspan{\vvec,\wvec}\) is contained in the subspace
\(S\) as well.
Definition 3.5.2 .
A
basis for a subspace
\(S\) of
\(\real^p\) is a set of vectors in
\(S\) that are linearly independent and span
\(S\text{.}\) It can be seen that any two bases have the same number of vectors. Therefore, we say that the
dimension of the subspace
\(S\text{,}\) denoted
\(\dim
S\text{,}\) is the number of vectors in any basis.
Definition 3.5.3 .
If
\(A\) is an
\(m\times n\) matrix, we call the subset of vectors
\(\xvec\) in
\(\real^n\) satisfying
\(A\xvec = \zerovec\) the
null space of
\(A\text{.}\) We denote it as
\(\nul(A)\text{.}\)
We will explore some null spaces in the next couple of activities.
Activity 3.5.3 .
Consider the matrix
\begin{equation*}
A=\left[\begin{array}{rrr}
1 \amp 3 \amp -1 \\
-2 \amp 0 \amp -4 \\
1 \amp 2 \amp 0 \\
\end{array}\right]
\end{equation*}
(a)
Give a parametric description of the null space
\(\nul(A)\text{.}\)
(b)
Give a basis for and state the dimension of
\(\nul(A)\text{.}\)
(c)
The null space
\(\nul(A)\) is a subspace of
\(\real^p\) for which
\(p\text{?}\)
Activity 3.5.4 .
Consider the matrix \(A\) whose reduced row echelon form is given:
\begin{equation*}
A \sim
\left[\begin{array}{rrrr}
1 \amp 2 \amp 0 \amp -3 \\
0 \amp 0 \amp 1 \amp 2 \\
\end{array}\right]\text{.}
\end{equation*}
(a)
Give a parametric description of
\(\nul(A)\text{.}\)
(b)
Notice that the parametric description gives a set of vectors that span
\(\nul(A)\text{.}\) Explain why this set of vectors is linearly independent and hence forms a basis. What is the dimension of
\(\nul(A)\text{?}\)
(c)
For this matrix,
\(\nul(A)\) is a subspace of
\(\real^p\) for what
\(p\text{?}\)
Activity 3.5.5 .
(a)
What is the relationship between the dimensions of the matrix
\(A\text{,}\) the number of pivot positions of
\(A\) and the dimension of
\(\nul(A)\text{?}\)
(b)
Suppose that the columns of a matrix
\(A\) are linearly independent. What can you say about
\(\nul(A)\text{?}\)
(c)
If
\(A\) is an invertible
\(n\times n\) matrix, what can you say about
\(\nul(A)\text{?}\)
Activity 3.5.6 .
Suppose that
\(A\) is a
\(5\times 10\) matrix and that
\(\nul(A) = \real^{10}\text{.}\) What can you say about the matrix
\(A\text{?}\)
Definition 3.5.4 .
The
rank of a matrix
\(A\text{,}\) denoted
\(\rank(A)\text{,}\) is the number of pivot positions of
\(A\text{.}\)
Definition 3.5.5 .
If
\(A\) is an
\(m\times n\) matrix, we call the span of its columns the
column space of
\(A\) and denote it as
\(\col(A)\text{.}\)
We will explore some column spaces in the next couple of activities.
Activity 3.5.7 .
Consider the matrix
\begin{equation*}
A= \left[\begin{array}{rrr}
\vvec_1 \amp \vvec_2 \amp \vvec_3
\end{array}\right]
=
\left[\begin{array}{rrr}
1 \amp 3 \amp -1 \\
-2 \amp 0 \amp -4 \\
1 \amp 2 \amp 0 \\
\end{array}\right]\text{.}
\end{equation*}
(a)
Since
\(\col(A)\) is the span of the columns, the vectors
\(\vvec_1\text{,}\) \(\vvec_2\text{,}\) and
\(\vvec_3\) naturally span
\(\col(A)\text{.}\) Are these vectors linearly independent?
(b)
Show that
\(\vvec_3\) can be written as a linear combination of
\(\vvec_1\) and
\(\vvec_2\) by giving the linear combination. Then explain why
\(\col(A)=\laspan{\vvec_1,\vvec_2}\text{.}\)
(c)
Explain why the vectors
\(\vvec_1\) and
\(\vvec_2\) form a basis for
\(\col(A)\text{.}\) This shows that
\(\col(A)\) is a 2-dimensional subspace of
\(\real^3\) and is therefore a plane in
\(\real^3\text{.}\)
Activity 3.5.8 .
Consider the matrix \(A\) and its reduced row echelon form:
\begin{equation*}
A = \left[\begin{array}{rrrr}
-2 \amp -4 \amp 0 \amp 6 \\
1 \amp 2 \amp 0 \amp -3 \\
\end{array}\right]
\sim
\left[\begin{array}{rrrr}
1 \amp 2 \amp 0 \amp -3 \\
0 \amp 0 \amp 0 \amp 0 \\
\end{array}\right]\text{.}
\end{equation*}
We will call the columns \(\vvec_1\text{,}\) \(\vvec_2\text{,}\) \(\vvec_3\text{,}\) and \(\vvec_4\text{.}\)
(a)
Explain why
\(\vvec_2\text{,}\) \(\vvec_3\text{,}\) and
\(\vvec_4\) can be written as a linear combination of
\(\vvec_1\text{.}\)
(b)
Explain why
\(\col(A)\) is a 1-dimensional subspace of
\(\real^2\) and is therefore a line.
(c)
What is the relationship between the dimension
\(\dim~\col(A)\) and the rank
\(\rank(A)\text{?}\)
(d)
What is the relationship between the dimension of the column space
\(\col(A)\) and the null space
\(\nul(A)\text{?}\)
Activity 3.5.9 .
If
\(A\) is an invertible
\(9\times9\) matrix, what can you say about the column space
\(\col(A)\text{?}\)
Activity 3.5.10 .
If
\(\col(A)=\{\zerovec\}\text{,}\) what can you say about the matrix
\(A\text{?}\)
Activity 3.5.11 .
Consider a matrix
\(A\text{.}\) Which of the following is equal to the dimension of the null space of
\(A\text{?}\)
The number of pivot columns
The number of non-pivot columns
The number of non-pivot rows
Activity 3.5.12 .
Consider a matrix
\(A\text{.}\) Which of the following is equal to the dimension of the column space of
\(A\text{?}\)
The number of pivot columns
The number of non-pivot columns
The number of non-pivot rows
Activity 3.5.13 .
Consider the matrix
\begin{equation*}
A
=
\left[\begin{array}{ccc} 1 & -3 & 2\\ 2 & -6 & 0 \\ 0 & 0 & 1 \\ -1 & 3 & 1 \end{array}\right]
.
\end{equation*}
Verify that if \(A\) is an \(m\times n\) matrix, then
\begin{equation*}
\dim~\nul(A) + \dim~\col(A) = n
\end{equation*}
holds for \(A\text{.}\) This equation is also known as the Rank-Nullity Theorem.