Span Properties
If b → , c → ∈ S p a n { v 1 → , . . . , v n → } Then: b → + c → ∈ S p a n − b → ∈ S p a n 0 → ∈ S p a n (always true) λ ∈ R , b → ∈ S p a n → λ ∗ b → ∈ S p a n \text{If }
\overrightarrow{b}, \overrightarrow{c} \in Span\{\overrightarrow{v_1}, ..., \overrightarrow{v_n}\} \\
\text{Then: } \\
\overrightarrow{b} + \overrightarrow{c} \in Span \\
-\overrightarrow{b} \in Span \\
\overrightarrow{0} \in Span \text{ (always true)} \\
\lambda \in \mathbb{R}, \overrightarrow{b} \in Span \rightarrow \lambda * \overrightarrow{b} \in Span If b , c ∈ Sp an { v 1 , ... , v n } Then: b + c ∈ Sp an − b ∈ Sp an 0 ∈ Sp an (always true) λ ∈ R , b ∈ Sp an → λ ∗ b ∈ Sp an
Spans can either have no points (only the zero vector) or an infinite amount of points, but the magnitude of that infinity can vary.
S p a n { v 1 → , . . . , v n → } = R m Span\{\overrightarrow{v_1}, ..., \overrightarrow{v_n}\} = \mathbb{R}^m Sp an { v 1 , ... , v n } = R m Matrix-Vector Multiplication
A = ( a 11 . . . a 1 n ⋮ ⋱ ⋮ a m 1 . . . a m n ) V → = ( v 1 ⋮ v n ) A ∗ V → = ( a 11 . . . a 1 n ⋮ ⋱ ⋮ a m 1 . . . a m n ) ( v 1 ⋮ v n ) = ( a 11 v 1 + a 12 v 2 + . . . + a 1 n v n a 21 v 1 + a 22 v 2 + . . . + a 2 n v n ⋮ a m 1 v 1 + a m 2 v 2 + . . . + a m n v n ) A ∗ V → = ( a 1 → … a n → ) ( v 1 ⋮ v n ) = v 1 a 1 → + v 2 a 2 → + . . . + v n a n → A =
\begin{pmatrix}
a_{11} & ... & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & ... & a_{mn}
\end{pmatrix} \\
\overrightarrow{V} = \begin{pmatrix}v_1 \\ \vdots \\ v_n\end{pmatrix}\\
A * \overrightarrow{V} =
\begin{pmatrix}
a_{11} & ... & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & ... & a_{mn}
\end{pmatrix}
\begin{pmatrix}v_1 \\ \vdots \\ v_n\end{pmatrix} =
\begin{pmatrix}
a_{11}v_1 + a_{12}v_2 + ... + a_{1n}v_n \\
a_{21}v_1 + a_{22}v_2 + ... + a_{2n}v_n \\
\vdots \\
a_{m1}v_1 + a_{m2}v_2 + ... + a_{mn}v_n
\end{pmatrix} \\
A * \overrightarrow{V} = (\overrightarrow{a_1} \ldots \overrightarrow{a_n})\begin{pmatrix}v_1 \\ \vdots \\ v_n\end{pmatrix} =
v_1\overrightarrow{a_1} + v_2\overrightarrow{a_2} + ... + v_n\overrightarrow{a_n} A = ⎝ ⎛ a 11 ⋮ a m 1 ... ⋱ ... a 1 n ⋮ a mn ⎠ ⎞ V = ⎝ ⎛ v 1 ⋮ v n ⎠ ⎞ A ∗ V = ⎝ ⎛ a 11 ⋮ a m 1 ... ⋱ ... a 1 n ⋮ a mn ⎠ ⎞ ⎝ ⎛ v 1 ⋮ v n ⎠ ⎞ = ⎝ ⎛ a 11 v 1 + a 12 v 2 + ... + a 1 n v n a 21 v 1 + a 22 v 2 + ... + a 2 n v n ⋮ a m 1 v 1 + a m 2 v 2 + ... + a mn v n ⎠ ⎞ A ∗ V = ( a 1 … a n ) ⎝ ⎛ v 1 ⋮ v n ⎠ ⎞ = v 1 a 1 + v 2 a 2 + ... + v n a n Examples of Matrix-Vector Multiplication
A = ( 1 2 3 4 5 6 ) V → = ( 1 0 − 1 ) A ∗ V → = ( − 2 − 2 ) = − 2 ( 1 1 ) A = \begin{pmatrix}
1 & 2 & 3 \\
4 & 5 & 6
\end{pmatrix} \\
\overrightarrow{V} = \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix} \\
A * \overrightarrow{V} = \begin{pmatrix}-2 \\ -2\end{pmatrix} = -2\begin{pmatrix}1 \\ 1\end{pmatrix} A = ( 1 4 2 5 3 6 ) V = ⎝ ⎛ 1 0 − 1 ⎠ ⎞ A ∗ V = ( − 2 − 2 ) = − 2 ( 1 1 ) A = ( a b c d ) V → = ( x y ) A ∗ V → = ( a x + b y c x + d y ) A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \\
\overrightarrow{V} = \begin{pmatrix}x \\ y\end{pmatrix} \\
A * \overrightarrow{V} = \begin{pmatrix}ax + by \\ cx + dy\end{pmatrix} A = ( a c b d ) V = ( x y ) A ∗ V = ( a x + b y c x + d y ) 5 u → − 7 v → + 3 w → = ( u → v → w → ) ( 5 − 7 3 ) 5\overrightarrow{u} - 7\overrightarrow{v} + 3\overrightarrow{w} = \begin{pmatrix}\overrightarrow{u} & \overrightarrow{v} & \overrightarrow{w}\end{pmatrix} \begin{pmatrix}5 \\ -7 \\ 3\end{pmatrix} 5 u − 7 v + 3 w = ( u v w ) ⎝ ⎛ 5 − 7 3 ⎠ ⎞ SLEs as Matrix Equations
A ∗ x → = b → where x → , b → ∈ R m A = ( a 11 . . . a 1 n ⋮ ⋱ ⋮ a m 1 . . . a m n ) x → = ( x 1 ⋮ x n ) b → = ( b 1 ⋮ b m ) (where b are all constants) A ∗ x → = x 1 a 1 → + … + x n a n → To find a solution to this, we can create an augmented matrix. ( A b → ) = ( a 1 → … a n → b → ) A * \overrightarrow{x} = \overrightarrow{b} \text{ where } \overrightarrow{x}, \overrightarrow{b} \in \mathbb{R}^m \\
A =
\begin{pmatrix}
a_{11} & ... & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & ... & a_{mn}
\end{pmatrix} \\
\overrightarrow{x} = \begin{pmatrix}x_1 \\ \vdots \\ x_n\end{pmatrix} \\
\overrightarrow{b} = \begin{pmatrix}b_1 \\ \vdots \\ b_m\end{pmatrix} \text{ (where b are all constants)} \\
A * \overrightarrow{x} = x_1\overrightarrow{a_1} + \ldots + x_n\overrightarrow{a_n} \\
\text{To find a solution to this, we can create an augmented matrix.} \\
\left(
\begin{array}{c|c}
A & \overrightarrow{b}
\end{array}
\right) =
\left(
\begin{array}{ccc|c}
\overrightarrow{a_1} & \ldots & \overrightarrow{a_n} & \overrightarrow{b}
\end{array}
\right) A ∗ x = b where x , b ∈ R m A = ⎝ ⎛ a 11 ⋮ a m 1 ... ⋱ ... a 1 n ⋮ a mn ⎠ ⎞ x = ⎝ ⎛ x 1 ⋮ x n ⎠ ⎞ b = ⎝ ⎛ b 1 ⋮ b m ⎠ ⎞ (where b are all constants) A ∗ x = x 1 a 1 + … + x n a n To find a solution to this, we can create an augmented matrix. ( A b ) = ( a 1 … a n b ) Example
A = ( 3 0 1 1 − 1 0 1 3 4 ) , b → = ( 1 1 2 ) ( A b → ) = ( 3 0 1 1 1 − 1 0 1 1 3 4 2 ) = ( 1 0 0 − 1 8 0 1 0 − 9 8 0 0 1 11 8 ) Thus, A ( − 1 8 − 9 8 11 8 ) = b → A = \begin{pmatrix}
3 & 0 & 1 \\
1 & -1 & 0 \\
1 & 3 & 4
\end{pmatrix}, \overrightarrow{b} = \begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix} \\
\left(
\begin{array}{c|c}
A & \overrightarrow{b}
\end{array}
\right) =
\left(
\begin{array}{ccc|c}
3 & 0 & 1 & 1 \\
1 & -1 & 0 & 1 \\
1 & 3 & 4 & 2 \\
\end{array}
\right) =
\left(
\begin{array}{ccc|c}
1 & 0 & 0 & -\frac{1}{8} \\
0 & 1 & 0 & -\frac{9}{8} \\
0 & 0 & 1 & \frac{11}{8} \\
\end{array}
\right) \\
\text{Thus, } A \begin{pmatrix} -\frac{1}{8} \\ -\frac{9}{8} \\ \frac{11}{8} \end{pmatrix} = \overrightarrow{b} A = ⎝ ⎛ 3 1 1 0 − 1 3 1 0 4 ⎠ ⎞ , b = ⎝ ⎛ 1 1 2 ⎠ ⎞ ( A b ) = ⎝ ⎛ 3 1 1 0 − 1 3 1 0 4 1 1 2 ⎠ ⎞ = ⎝ ⎛ 1 0 0 0 1 0 0 0 1 − 8 1 − 8 9 8 11 ⎠ ⎞ Thus, A ⎝ ⎛ − 8 1 − 8 9 8 11 ⎠ ⎞ = b Properties of Matrix-Vector Multiplication
A ∗ ( v → + w → ) = A ∗ v → + A ∗ w → A ∗ ( c ∗ v → ) = c ∗ ( A ∗ v → ) A * (\overrightarrow{v} + \overrightarrow{w}) = A * \overrightarrow{v} + A * \overrightarrow{w} \\
A * (c * \overrightarrow{v}) = c * (A * \overrightarrow{v}) A ∗ ( v + w ) = A ∗ v + A ∗ w A ∗ ( c ∗ v ) = c ∗ ( A ∗ v ) Row Theorem
Suppose we have a 1 → , … , a n → ∈ R m Letting A = ( a 1 → , … , a n → ) , the following statements are equivalent. (They’re all true or all false) 1. S p a n { a 1 → , … , a n → } = R m 2. A x → = b → is consistent for every b → 3. The RREF of A has a pivot in every row \text{Suppose we have }\overrightarrow{a_1}, \ldots, \overrightarrow{a_n} \in \mathbb{R}^m \\
\text{Letting } A = (\overrightarrow{a_1}, \ldots, \overrightarrow{a_n}) \text{, the following statements are equivalent. (They're all true or all false)}\\
\text{1. } Span\{\overrightarrow{a_1}, \ldots, \overrightarrow{a_n}\} = \mathbb{R}^m \\
\text{2. } A\overrightarrow{x} = \overrightarrow{b} \text{ is consistent for every } \overrightarrow{b} \\
\text{3. The RREF of A has a pivot in every row} Suppose we have a 1 , … , a n ∈ R m Letting A = ( a 1 , … , a n ) , the following statements are equivalent. (They’re all true or all false) 1. Sp an { a 1 , … , a n } = R m 2. A x = b is consistent for every b 3. The RREF of A has a pivot in every row Example
u → , v → , w → ∈ R 4 Can S p a n { u → , v → , w → } = R 4 ? A = ( u 1 v 1 w 1 ⋮ ⋮ ⋮ u 4 v 4 w 4 ) Because there is at least one row that does not have a pivot, all of the parts of the theorem are not true. Therefore, it does not span R4. \overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w} \in \mathbb{R}^4 \\
\text{Can } Span\{\overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w}\} = \mathbb{R}^4\text{?}\\
A = \begin{pmatrix}
u_1 & v_1 & w_1 \\
\vdots & \vdots & \vdots \\
u_4 & v_4 & w_4
\end{pmatrix} \\
\text{Because there is at least one row that does not have a pivot, all of the parts of the theorem are not true. Therefore, it does not span R4.} u , v , w ∈ R 4 Can Sp an { u , v , w } = R 4 ? A = ⎝ ⎛ u 1 ⋮ u 4 v 1 ⋮ v 4 w 1 ⋮ w 4 ⎠ ⎞ Because there is at least one row that does not have a pivot, all of the parts of the theorem are not true. Therefore, it does not span R4.