Consider the set of vectors \(S = \{A_1,A_2,A_3,A_4\}\) in \(\matrixring_2(\R)\text{,}\) where
\begin{align*}
A_1 \amp= \left[\begin{array}{rr} 0 \amp -1 \\ 2 \amp 1 \end{array}\right], \amp
A_3 \amp= \left[\begin{array}{rr} 0 \amp 1 \\ -2 \amp 0 \end{array}\right], \amp\\
A_2 \amp= \left[\begin{array}{rr} 1 \amp 2 \\ 4 \amp -1 \end{array}\right], \amp
A_4 \amp= \left[\begin{array}{rr} 0 \amp 0 \\ 1 \amp -2 \end{array}\right].
\end{align*}
Is this set a spanning set for all of \(\matrixring_2(\R)\text{?}\) That is, is \(\matrixring_2(\R) = \Span S\text{?}\) We already know a spanning set for \(\matrixring_2(\R)\) — the set of standard basis vectors \(\basisfont{B} = \{E_{11},E_{12},E_{21},E_{22}\}\text{,}\) where
\begin{align*}
E_{11} \amp= \begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}, \amp
E_{12} \amp= \begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix}, \amp\\
E_{21} \amp= \begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix}, \amp
E_{22} \amp= \begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix}.
\end{align*}
That is, we already know that
\(\matrixring_2(\R) = \Span \basisfont{B}\text{.}\) So we can turn our question into: is
\(\Span S = \Span \basisfont{B}\text{?}\) With this new version of our problem, we can use the same method as in the previous example. However, we don’t need to explicitly verify that each vector in
\(S\) can be expressed as a linear combination of the vectors in
\(\basisfont{B}\text{.}\) Besides being obvious, this fact is already implied by our assertion that
\(\matrixring_2(\R) = \Span \basisfont{B}\text{,}\) since clearly each vector in
\(S\) is a vector in
\(\matrixring_2(\R)\text{.}\) So it just remains to verify that each vector in
\(\basisfont{B}\) can be expressed as a linear combination of the vectors in
\(S\text{.}\) Let’s begin with vector
\(E_{11}\text{.}\) We use the same strategy as in the examples in
Subsection 16.4.3: express
\(E_{11}\) as a linear combination of the vectors in
\(S\) with unknown scalar coefficients, set up equations in those unknown scalars, and determine whether the resulting linear system is consistent.
\begin{align*}
\begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}
\amp= k_1 \left[\begin{array}{rr} 0 \amp -1 \\ 2 \amp 1 \end{array}\right]
+ k_2 \left[\begin{array}{rr} 1 \amp 2 \\ 4 \amp -1 \end{array}\right]
+ k_3 \left[\begin{array}{rr} 0 \amp 1 \\ -2 \amp 0 \end{array}\right]
+ k_4 \left[\begin{array}{rr} 0 \amp 0 \\ 1 \amp -2 \end{array}\right]\\
\amp= \begin{bmatrix} k_2 \amp -k_1+2k_2+k_3 \\ 2k_1+4k_2-2k_3+k_4 \amp k_1-k_2-2k_4 \end{bmatrix}
\end{align*}
Comparing entries on left and right sides leads to the system of equations
\begin{equation*}
\left\{\begin{array}{rcrcrcrcr}
\amp \amp k_2 \amp \amp \amp \amp \amp = \amp 1 \text{,} \\
-k_1 \amp + \amp 2k_2 \amp + \amp k_3 \amp \amp \amp = \amp 0 \text{,} \\
2k_1 \amp + \amp 4k_2 \amp - \amp 2k_3 \amp + \amp k_4 \amp = \amp 0 \text{,} \\
k_1 \amp - \amp k_2 \amp \amp \amp - \amp 2k_4 \amp = \amp 0 \text{,}
\end{array}\right.
\end{equation*}
which can be put in an augmented matrix and reduced.
\begin{equation*}
\left[\begin{array}{rrrr|r}
0 \amp 1 \amp 0 \amp 0 \amp 1 \\
-1 \amp 2 \amp 1 \amp 0 \amp 0 \\
2 \amp 4 \amp -2 \amp 1 \amp 0 \\
1 \amp -1 \amp 0 \amp -2 \amp 0
\end{array}\right]
\qquad\rowredarrow\qquad
\left[\begin{array}{rrrr|r}
1 \amp 0 \amp 0 \amp 0 \amp -15\\
0 \amp 1 \amp 0 \amp 0 \amp 1\\
0 \amp 0 \amp 1 \amp 0 \amp -17\\
0 \amp 0 \amp 0 \amp 1 \amp -8
\end{array}\right]
\end{equation*}
The reduced augmented matrix above tells us that
\begin{align*}
k_1 \amp= -15, \amp k_2 \amp= 1, \amp k_3 \amp= -17, \amp k_4 \amp= -8,
\end{align*}
and so
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \\ 0 \amp 0 \end{bmatrix}
=-15 \left[\begin{array}{rr} 0 \amp -1 \\ 2 \amp 1 \end{array}\right]
+ \left[\begin{array}{rr} 1 \amp 2 \\ 4 \amp -1 \end{array}\right]
- 17 \left[\begin{array}{rr} 0 \amp 1 \\ -2 \amp 0 \end{array}\right]
- 8 \left[\begin{array}{rr} 0 \amp 0 \\ 1 \amp -2 \end{array}\right],
\end{equation*}
though we only cared about the existence of a solution, not the actual solution itself.
In a similar manner, one can calculate that
\begin{align*}
\begin{bmatrix} 0 \amp 1 \\ 0 \amp 0 \end{bmatrix}
\amp= 4 \left[\begin{array}{rr} 0 \amp -1 \\ 2 \amp 1 \end{array}\right]
+ 5 \left[\begin{array}{rr} 0 \amp 1 \\ -2 \amp 0 \end{array}\right]
+ 2 \left[\begin{array}{rr} 0 \amp 0 \\ 1 \amp -2 \end{array}\right],\\
\\
\begin{bmatrix} 0 \amp 0 \\ 1 \amp 0 \end{bmatrix}
\amp= 2 \left[\begin{array}{rr} 0 \amp -1 \\ 2 \amp 1 \end{array}\right]
+ 2 \left[\begin{array}{rr} 0 \amp 1 \\ -2 \amp 0 \end{array}\right]
+ \left[\begin{array}{rr} 0 \amp 0 \\ 1 \amp -2 \end{array}\right],\\
\\
\begin{bmatrix} 0 \amp 0 \\ 0 \amp 1 \end{bmatrix}
\amp= \left[\begin{array}{rr} 0 \amp -1 \\ 2 \amp 1 \end{array}\right]
+ \left[\begin{array}{rr} 0 \amp 1 \\ -2 \amp 0 \end{array}\right].
\end{align*}
We have now verified that each vector in
\(\basisfont{B}\) can be expressed as a linear combination of the vectors in
\(S\text{.}\) As discussed above, we already knew that each vector in
\(S\) can be expressed as a linear combination of the vectors in
\(\basisfont{B}\text{.}\) Therefore,
Statement 2 of
Proposition 16.5.6 tells us that
\(\Span S = \Span \basisfont{B}\text{.}\) Since we already knew that
\(\Span\basisfont{B}\) is equal to the entire space
\(\matrixring_2(\R)\text{,}\) we must also have
\(\Span S\) equal to this entire space.
