Discover Linear Algebra

Example 20.4.1. From the discovery guide.

First, let's carry out some of the examples from Discovery 20.2.

1. In Discovery 20.2.c, we considered a certain subspace of \(\R^3\text{.}\)

   An arbitrary vector in \(\R^3\) requires three parameters to describe its three components: \(\uvec{x} = (a,b,c)\text{.}\) If we restrict to just those vectors whose first and third components are equal, we can replace \(c\) by \(a\text{,}\) to get

   \begin{equation*} \uvec{x} = (a,b,a) = (a,0,a) + (0,b,0) = a(1,0,1) + b(0,1,0). \end{equation*}

   So a basis for this subspace of \(\R^3\) is \(\basisfont{B} = \{(1,0,1),(0,1,0)\}\text{,}\) and the dimension is \(2\text{.}\)

2. In Discovery 20.2.g, we considered a certain subspace of \(\matrixring_2(\R)\text{.}\)

   An arbitrary matrix in \(\matrixring_2(\R)\) requires four parameters to describe its four entries:

   \begin{equation*} A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}. \end{equation*}

   If we restrict to those matrices whose entries sum to zero, so that \(a+b+c+d=0\text{,}\) then we can isolate \(d=-a-b-c\) and substitute that into the matrix:

   \begin{align*} A \amp= \begin{bmatrix} a \amp b \\ c \amp -a-b-c \end{bmatrix} \\ \amp = \left[\begin{array}{rr} a \amp 0 \\ 0 \amp -a \end{array}\right] + \left[\begin{array}{rr} 0 \amp b \\ 0 \amp -b \end{array}\right] + \left[\begin{array}{rr} 0 \amp 0 \\ c \amp -c \end{array}\right]\\ \amp = a\left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp -1 \end{array}\right] + b\left[\begin{array}{rr} 0 \amp 1 \\ 0 \amp -1 \end{array}\right] + c\left[\begin{array}{rr} 0 \amp 0 \\ 1 \amp -1 \end{array}\right]. \end{align*}

   So this subspace of \(\matrixring_2(\R)\) has dimension \(3\text{,}\) with basis

   \begin{equation*} \basisfont{B} = \left\{ \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp -1 \end{array}\right], \left[\begin{array}{rr} 0 \amp 1 \\ 0 \amp -1 \end{array}\right], \left[\begin{array}{rr} 0 \amp 0 \\ 1 \amp -1 \end{array}\right] \right\}. \end{equation*}

3. In Discovery 20.2.j, we considered a certain subspace of \(\poly_5(\R)\text{.}\)

   An arbitrary polynomial in \(\poly_5(\R)\) requires six parameters, one for each power of \(x\text{,}\) along with a parameter for the constant term:

   \begin{equation*} p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5. \end{equation*}

   If we restrict to only odd polynomials, we need to eliminate the constant term and the even powers of \(x\text{:}\)

   \begin{equation*} p(x) = a_1 x + a_3 x^3 + a_5 x^5. \end{equation*}

   (Equivalently, we have applied the homogeneous conditions \(a_0 = 0\text{,}\) \(a_2 = 0\text{,}\) and \(a_4 = 0\text{.}\)) So this subspace of \(\poly_5(\R)\) has dimension \(3\text{,}\) with basis \(\basisfont{B} = \{x,x^3,x^5\}\text{.}\)

Example 20.4.2. Dimensions of familiar spaces via parameters.

Now let's examine how the dimensions of our favourite example spaces relate to our parametric point of view. We considered specific examples of these in parts of Discovery 20.2, but here we'll work more generally.

1. An arbitrary vector in \(\R^n\) requires \(n\) parameters, one for each component:
   \begin{equation*} \uvec{x} = (x_1,x_2,\dotsc,x_n). \end{equation*}
   If we expanded this into a linear combination, each parameter would be attached to a standard basis vector \(\uvec{e}_j\text{.}\) Since we've got \(n\) parameters and a corresponding \(n\) standard basis vectors, we have
   \begin{equation*} \dim \R^n = n. \end{equation*}
2. An arbitrary \(m\times n\) matrix in \(\matrixring_{m \times n}(\R)\) requires \(mn\) parameters, one for each entry:
   \begin{equation*} A = [a_{ij}], \qquad 1 \le i \le m, \;\; 1 \le j \le n. \end{equation*}
   If we expanded this into a linear combination, each parameter would be attached to a standard basis matrix \(E_{ij}\text{,}\) with zeros in all entries except for a single \(1\) in the \(\nth[(i,j)]\) entry. Since we've got \(mn\) parameters and a corresponding \(mn\) standard basis matrices, we have
   \begin{equation*} \dim \matrixring_{m \times n}(\R) = mn. \end{equation*}
3. An arbitrary polynomial in \(\poly_n(\R)\text{,}\) the space of polynomials of degree \(n\) or less, requires \(n+1\) parameters, one for each power of \(x\) plus an extra one for the constant term:
   \begin{equation*} p(x) = a_0 + a_1x + a_2 x^2 + \dotsb + a_n x^n. \end{equation*}
   This is already naturally expressed as a linear combination, and each parameter is attached to a polynomial from the standard basis \(\basisfont{B} = \{1,x,x^2,\dotsc,x^n\}\text{.}\) Since we've got \(n+1\) parameters and a corresponding \(n+1\) standard basis polynomials, we have
   \begin{equation*} \dim \poly_n(\R) = n+1. \end{equation*}

Example 20.4.3. The solution space of a homogeneous system.

In Remark 20.3.2, we noted how assigning parameters after row reducing a homogeneous system corresponded directly to a parameter-based procedure for determine the basis for a space. Let's illustrate this correspondence with an example.

Consider the homogeneous system in Discovery 2.4, which we solved in Example 2.4.4. In Example 17.4.8, we used the Subspace Test to verify that the solution set of a homogeneous system with an \(m\times n\) coefficient matrix is a subspace of \(\R^n\text{.}\) The system from Discovery 2.4 has a \(4 \times 4\) coefficient matrix that we reduced:

Assigning parameters to free variables \(x_2,x_4\text{,}\) we obtained the general solution in parametric form:

We can use these expressions as components in a general solution vector, and expand it out to a linear combination, just as in the previous examples in this subsection:

Since two parameters are needed to describe the solution vectors for this system, the solution space has dimension \(2\text{,}\) and a basis for this subspace is