Construction of Subspaces

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The principle of superposition shows that the set of all solutions to a homogeneous system of linear equations is closed under addition and scalar multiplication and is a subspace. Indeed, there are two ways to describe subspaces: first as solutions to linear systems, and second as the span of a set of vectors. We shall see that solving a homogeneous linear system of equations just means writing the solution set as the span of a finite set of vectors.

Solutions to Homogeneous Systems Form Subspaces

Let $A$ be an $m\times n$ matrix. The null space of $A$ is the set of solutions to the homogeneous system of linear equations $\begin{equation} \label{Ax=0} Ax=0. \end{equation}$

Let $A$ be an $m\times n$ matrix. Then the null space of $A$ is a subspace of $\R ^n$ .

Proof: Suppose that $x$ and $y$ are solutions to (??). Then $A(x+y) = Ax+Ay = 0+0 = 0;$ so $x+y$ is a solution of (??). Similarly, for $r\in \R$ $A(rx) = rAx = r0 = 0;$ so $rx$ is a solution of (??). Thus, $x+y$ and $rx$ are in the null space of $A$ , and the null space is closed under addition and scalar multiplication. So Theorem ?? implies that the null space is a subspace of the vector space $\R ^n$ . $\blacksquare$

Solutions to Linear Systems of Differential Equations Form Subspaces

Let $C$ be an $n\times n$ matrix and let $W$ be the set of solutions to the linear system of ordinary differential equations

$\begin{equation} \label{Cx(t)} \frac{dx}{dt}(t) = Cx(t). \end{equation}$ We will see later that a solution to (??) has coordinate functions $x_j(t)$ in $\CCone$ . The principle of superposition then shows that $W$ is a subspace of $(\CCone )^n$ . Suppose $x(t)$ and $y(t)$ are solutions of (??). Then $\frac {d}{dt}(x(t)+y(t)) = \frac {dx}{dt}(t) + \frac {dy}{dt}(t) = Cx(t) + Cy(t) = C(x(t)+y(t));$ so $x(t)+y(t)$ is a solution of (??) and in $W$ . A similar calculation shows that $rx(t)$ is also in $W$ and that $W\subset (\CCone )^n$ is a subspace.

Writing Solution Subspaces as a Span

The way we solve homogeneous systems of equations gives a second method for defining subspaces. For example, consider the system $Ax=0,$ where $A=\left (\begin {array}{rccc} 2 & 1 & 4 & 0 \\ -1 & 0 & 2 & 1 \end {array}\right ).$ The matrix $A$ is row equivalent to the reduced echelon form matrix $E=\left (\begin {array}{ccrr} 1 & 0 & -2 & -1 \\ 0 & 1 & 8 & 2 \end {array}\right ).$ Therefore $x=(x_1,x_2,x_3,x_4)$ is a solution of $Ex=0$ if and only if $x_1 = 2x_3+x_4$ and $x_2 = -8x_3 - 2x_4$ . It follows that every solution of $Ex=0$ can be written as: $x = x_3\left (\begin {array}{r} 2 \\ -8\\ 1\\ 0 \end {array}\right ) +x_4\left (\begin {array}{r} 1 \\ -2\\ 0\\ 1 \end {array}\right ).$ Since row operations do not change the set of solutions, it follows that every solution of $Ax=0$ has this form. We have also shown that every solution is generated by two vectors by use of vector addition and scalar multiplication. We say that this subspace is spanned by the two vectors $\left (\begin {array}{r} 2 \\ -8\\ 1\\ 0 \end {array}\right ) \AND \left (\begin {array}{r} 1 \\ -2\\ 0\\ 1 \end {array}\right ).$ For example, a calculation verifies that the vector $\left (\begin {array}{r} -1 \\ -2\\ 1\\ -3 \end {array}\right )$ is also a solution of $Ax=0$ . Indeed, we may write it as

$\begin{equation} \label{eq:SpanSol} \left(\begin{array}{r} -1 \\ -2\\ 1\\ -3 \end{array}\right) = \left(\begin{array}{r} 2 \\ -8\\ 1\\ 0 \end{array}\right) - 3 \left(\begin{array}{r} 1 \\ -2\\ 0\\ 1 \end{array}\right). \end{equation}$

Spans

Let $v_1,\ldots ,v_k$ be a set of vectors in a vector space $V$ . A vector $v\in V$ is a linear combination of $v_1,\ldots ,v_k$ if $v = r_1v_1 + \cdots + r_kv_k$ for some scalars $r_1,\ldots ,r_k$ .

The set of all linear combinations of the vectors $v_1,\ldots ,v_k$ in a vector space $V$ is the span of $v_1,\ldots ,v_k$ and is denoted by $\Span \{v_1,\ldots ,v_k\}$ .

For example, the vector on the left hand side in (??) is a linear combination of the two vectors on the right hand side.

The simplest example of a span is $\R ^n$ itself. Let $v_j=e_j$ where $e_j\in \R ^n$ is the vector with a $1$ in the $j^{th}$ coordinate and $0$ in all other coordinates. Then every vector $x=(x_1,\ldots ,x_n)\in \R ^n$ can be written as $x = x_1e_1 + \cdots + x_ne_n.$ It follows that $\R ^n=\Span \{e_1,\ldots ,e_n\}.$ Similarly, the set $\Span \{e_1,e_3\}\subset \R ^3$ is just the $x_1x_3$ -plane, since vectors in this span are $x_1e_1+x_3e_3 = x_1(1,0,0) + x_3(0,0,1) = (x_1,0,x_3).$

Let $V$ be a vector space and let $w_1,\ldots ,w_k\in V$ . Then $W=\Span \{w_1,\ldots ,w_k\} \subset V$ is a subspace.

Proof: Suppose $x,y\in W$ . Then
$\begin{eqnarray*} x & = & r_1w_1 + \cdots + r_kw_k \\ y & = & s_1w_1 + \cdots + s_kw_k \end{eqnarray*}$
for some scalars $r_1,\ldots ,r_k$ and $s_1,\ldots ,s_k$ . It follows that $x+y = (r_1+s_1)w_1 + \cdots + (r_k+s_k)w_k$ and $rx = (rr_1)w_1 + \cdots + (rr_k)w_k$ are both in $\Span \{w_1,\ldots ,w_k\}$ . Hence $W\subset V$ is closed under addition and scalar multiplication, and is a subspace by Theorem ??. $\blacksquare$

For example, let

$\begin{equation} \label{e:vandw} v=(2,1,0) \AND w=(1,1,1) \end{equation}$ be vectors in $\R ^3$ . Then linear combinations of the vectors $v$ and $w$ have the form $\alpha v + \beta w = (2\alpha +\beta , \alpha +\beta , \beta )$ for real numbers $\alpha$ and $\beta$ . Note that every one of these vectors is a solution to the linear equation $\begin{equation} \label{ex1} x_1 - 2x_2 + x_3 = 0, \end{equation}$ that is, the $1^{st}$ coordinate minus twice the $2^{nd}$ coordinate plus the $3^{rd}$ coordinate equals zero. Moreover, you may verify that every solution of (??) is a linear combination of the vectors $v$ and $w$ in (??). Thus, the set of solutions to the homogeneous linear equation (??) is a subspace, and that subspace can be written as the span of all linear combinations of the vectors $v$ and $w$ .

In this language we see that the process of solving a homogeneous system of linear equations is just the process of finding a set of vectors that span the subspace of all solutions. Indeed, we can now restate Theorem ?? of Chapter ??. Recall that a matrix $A$ has rank $\ell$ if it is row equivalent to a matrix in echelon form with $\ell$ nonzero rows.

Let $A$ be an $m\times n$ matrix with rank $\ell$ . Then the null space of $A$ is the span of $n-\ell$ vectors.

We have now seen that there are two ways to describe subspaces — as solutions of homogeneous systems of linear equations and as a span of a set of vectors, the spanning set. Much of linear algebra is concerned with determining how one goes from one description of a subspace to the other.

Exercises

In Exercises ?? – ?? a single equation in three variables is given. For each equation write the subspace of solutions in $\R ^3$ as the span of two vectors in $\R ^3$ .

$4x - 2y + z = 0$ .

$x - y + 3z = 0$ .

$x + y + z = 0$ .

$y=z$ .

In Exercises ?? – ?? each of the given matrices is in reduced echelon form. Write solutions of the corresponding homogeneous system of linear equations as a span of vectors.

$A = \left (\begin {array}{rrrrr} 1 & 2 & 0 & 1 & 0 \\ 0 & 0 & 1 & 4 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right )$ .

$B = \left (\begin {array}{rrrr} 1 & 3 & 0 & 5 \\ 0 & 0 & 1 & 2 \end {array}\right )$ .

$A = \left (\begin {array}{rrr} 1 & 0 & 2 \\ 0 & 1 & 1\end {array}\right )$ .

$B = \left (\begin {array}{rrrrrr} 1 & -1 & 0 & 5 & 0 & 0\\ 0 & 0 & 1 & 2 & 0 & 2\\ 0 & 0 & 0 & 0 & 1 & 2\end {array}\right )$ .

Write a system of two linear equations of the form $Ax=0$ where $A$ is a $2\times 4$ matrix whose subspace of solutions in $\R ^4$ is the span of the two vectors $v_1 = \left (\begin {array}{r} 1 \\ -1 \\ 0 \\ 0 \end {array}\right ) \AND v_2 = \left (\begin {array}{r} 0 \\ 0 \\ 1 \\ -1 \end {array}\right ).$

Write the matrix $A=\mattwo {2}{2}{-3}{0}$ as a linear combination of the matrices $B=\mattwo {1}{1}{0}{0} \AND C=\mattwo {0}{0}{1}{0}.$

Is $(2,20,0)$ in the span of $w_1=(1,1,3)$ and $w_2=(1,4,2)$ ? Answer this question by setting up a system of linear equations and solving that system by row reducing the associated augmented matrix.

In Exercises ?? – ?? let $W\subset \CCone$ be the subspace spanned by the two polynomials $x_1(t) = 1$ and $x_2(t)=t^2$ . For the given function $y(t)$ decide whether or not $y(t)$ is an element of $W$ . Furthermore, if $y(t)\in W$ , determine whether the set $\{y(t),x_2(t)\}$ is a spanning set for $W$ .

$y(t) = 1-t^2$ ,

$y(t) = t^4$ ,

$y(t) = \sin t$ ,

$y(t) = 0.5 t^2$

Let $W\subset \R ^4$ be the subspace that is spanned by the vectors $w_1=(-1,2,1,5)\AND w_2=(2,1,3,0).$ Find a linear system of two equations such that $W=\Span \{w_1,w_2\}$ is the set of solutions of this system.

Let $V$ be a vector space and let $v\in V$ be a nonzero vector. Show that $\Span \{v,v\}=\Span \{v\}.$

Let $V$ be a vector space and let $v,w\in V$ be vectors. Show that $\Span \{v,w\}=\Span \{v,w,v+3w\}.$

Let $W=\Span \{w_1,\ldots ,w_k\}$ be a subspace of the vector space $V$ and let $w_{k+1}\in W$ be another vector. Prove that $W=\Span \{w_1,\ldots ,w_{k+1}\}$ .

Let $Ax=b$ be a system of $m$ linear equations in $n$ unknowns, and let $r=\rank (A)$ and $s=\rank (A|b)$ . Suppose that this system has a unique solution. What can you say about the relative magnitudes of $m,n,r,s$ ?