Properties
Consider a linear function T: R n → R m T is onto ‾ if every w → ∈ R m can be written as T ( v → ) = w → T is one-to-one ‾ if T ( v → ) = T ( v → ) ⟹ v → = w → T is a bijection ‾ if it is both onto and one-to-one \text{Consider a linear function T: } \mathbb{R}^n \rightarrow \mathbb{R}^m \\
\text{T is \underline{onto} if every } \overrightarrow{w} \in \mathbb{R}^m \text{ can be written as } T(\overrightarrow{v}) = \overrightarrow{w} \\
\text{T is \underline{one-to-one} if } T(\overrightarrow{v}) = T(\overrightarrow{v}) \implies \overrightarrow{v} = \overrightarrow{w} \\
\text{T is a \underline{bijection} if it is both onto and one-to-one} Consider a linear function T: R n → R m T is onto if every w ∈ R m can be written as T ( v ) = w T is one-to-one if T ( v ) = T ( v ) ⟹ v = w T is a bijection if it is both onto and one-to-one Other ways to define properties
T is onto if and only if A T x → = b → is consistent for every b → ∈ R m T is one-to-one if and only if the columns of A T are linearly independent \text{T is onto if and only if } A_T \overrightarrow{x} = \overrightarrow{b} \text{ is consistent for every } \overrightarrow{b} \in \mathbb{R}^m \\
\text{T is one-to-one if and only if the columns of } A_T \text{ are linearly independent} T is onto if and only if A T x = b is consistent for every b ∈ R m T is one-to-one if and only if the columns of A T are linearly independent Examples
The function defined by this matrix multiplication is onto: ( 1 0 1 0 1 0 ) ( a b 0 ) = ( a b ) The function defined by this matrix multiplication is one-to-one: ( 1 0 0 1 0 0 ) ( a b ) = ( a b 0 ) The function defined by this matrix multiplication is a bijection: ( c o s θ − s i n θ s i n θ c o s θ ) \text{The function defined by this matrix multiplication is onto:}\\
\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix} a \\ b \\ 0 \end{pmatrix} =
\begin{pmatrix} a \\ b \end{pmatrix} \\
\text{The function defined by this matrix multiplication is one-to-one:} \\
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
0 & 0
\end{pmatrix}
\begin{pmatrix} a \\ b \end{pmatrix} =
\begin{pmatrix} a \\ b \\ 0\end{pmatrix} \\
\text{The function defined by this matrix multiplication is a bijection:} \\
\begin{pmatrix}
cos\theta & -sin\theta \\
sin\theta & cos\theta
\end{pmatrix} The function defined by this matrix multiplication is onto: ( 1 0 0 1 1 0 ) ⎝ ⎛ a b 0 ⎠ ⎞ = ( a b ) The function defined by this matrix multiplication is one-to-one: ⎝ ⎛ 1 0 0 0 1 0 ⎠ ⎞ ( a b ) = ⎝ ⎛ a b 0 ⎠ ⎞ The function defined by this matrix multiplication is a bijection: ( cos θ s in θ − s in θ cos θ ) Invertible Matrices
Consider a n x n matrix A. A is invertible if there exists A − 1 such that A − 1 ∗ A = A ∗ A − 1 = I n The matrix A also has to represent a bijective linear transformation. \text{Consider a n x n matrix A. A is invertible if there exists } A^{-1} \text{ such that } A^{-1} * A = A * A^{-1} = I_n \\
\text{The matrix A also has to represent a bijective linear transformation.} Consider a n x n matrix A. A is invertible if there exists A − 1 such that A − 1 ∗ A = A ∗ A − 1 = I n The matrix A also has to represent a bijective linear transformation. Theorem: An n x n matrix A is invertible if and only if the RREF of A is I n . In this case, row operations that take A to I n transform I n into A − 1 ( A ∣ I n ) → ( I n ∣ A − 1 ) \text{Theorem: An n x n matrix A is invertible if and only if the RREF of A is } I_n \text{.}\\
\text{In this case, row operations that take A to } I_n \text{ transform } I_n \text{ into } A^{-1} \\
(A | I_n) \rightarrow (I_n | A^{-1}) Theorem: An n x n matrix A is invertible if and only if the RREF of A is I n . In this case, row operations that take A to I n transform I n into A − 1 ( A ∣ I n ) → ( I n ∣ A − 1 ) Theorem: For an n x n matrix A, the following are equivalent: A is invertible A has a pivot in every row A has a pivot in every column \text{Theorem: For an n x n matrix A, the following are equivalent:} \\
\text{A is invertible} \\
\text{A has a pivot in every row} \\
\text{A has a pivot in every column} Theorem: For an n x n matrix A, the following are equivalent: A is invertible A has a pivot in every row A has a pivot in every column Examples
( 0 0 0 1 ) is not invertible because you cannot get a 1 in the top left cell. Proof: ( 0 0 0 1 ) ( a b c d ) = ( 0 0 c d ) \begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}
\text{ is not invertible because you cannot get a 1 in the top left cell. Proof: }\\
\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix} =
\begin{pmatrix}
0 & 0 \\
c & d
\end{pmatrix} ( 0 0 0 1 ) is not invertible because you cannot get a 1 in the top left cell. Proof: ( 0 0 0 1 ) ( a c b d ) = ( 0 c 0 d )
Invert ( 1 2 3 0 1 4 5 6 0 ) ( 1 2 3 1 0 0 0 1 4 0 1 0 5 6 0 0 0 1 ) → ( 1 2 3 1 0 0 0 1 4 0 1 0 0 − 4 − 15 − 5 0 1 ) → ( 1 2 3 1 0 0 0 1 4 0 1 0 0 0 1 − 5 4 1 ) ( 1 2 0 16 − 12 − 3 0 1 0 20 − 15 − 4 0 0 1 − 5 4 1 ) → ( 1 0 0 − 24 18 5 0 1 0 20 − 15 − 4 0 0 1 − 5 4 1 ) Therefore, the inverse of ( 1 2 3 0 1 4 5 6 0 ) is ( − 24 18 5 20 − 15 − 4 − 5 4 1 ) \text{Invert }
\begin{pmatrix}
1 & 2 & 3 \\
0 & 1 & 4 \\
5 & 6 & 0
\end{pmatrix} \\
\left(
\begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & 4 & 0 & 1 & 0\\
5 & 6 & 0 & 0 & 0 & 1
\end{array}
\right) \rightarrow
\left(
\begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & 4 & 0 & 1 & 0\\
0 & -4 & -15 & -5 & 0 & 1
\end{array}
\right) \rightarrow
\left(
\begin{array}{ccc|ccc}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & 4 & 0 & 1 & 0\\
0 & 0 & 1 & -5 & 4 & 1
\end{array}
\right) \\
\left(
\begin{array}{ccc|ccc}
1 & 2 & 0 & 16 & -12 & -3 \\
0 & 1 & 0 & 20 & -15 & -4\\
0 & 0 & 1 & -5 & 4 & 1
\end{array}
\right) \rightarrow
\left(
\begin{array}{ccc|ccc}
1 & 0 & 0 & -24 & 18 & 5 \\
0 & 1 & 0 & 20 & -15 & -4 \\
0 & 0 & 1 & -5 & 4 & 1
\end{array}
\right) \\
\text{Therefore, the inverse of }
\begin{pmatrix}
1 & 2 & 3 \\
0 & 1 & 4 \\
5 & 6 & 0
\end{pmatrix}
\text{ is }
\begin{pmatrix}
-24 & 18 & 5 \\
20 & -15 & -4 \\
-5 & 4 & 1
\end{pmatrix} Invert ⎝ ⎛ 1 0 5 2 1 6 3 4 0 ⎠ ⎞ ⎝ ⎛ 1 0 5 2 1 6 3 4 0 1 0 0 0 1 0 0 0 1 ⎠ ⎞ → ⎝ ⎛ 1 0 0 2 1 − 4 3 4 − 15 1 0 − 5 0 1 0 0 0 1 ⎠ ⎞ → ⎝ ⎛ 1 0 0 2 1 0 3 4 1 1 0 − 5 0 1 4 0 0 1 ⎠ ⎞ ⎝ ⎛ 1 0 0 2 1 0 0 0 1 16 20 − 5 − 12 − 15 4 − 3 − 4 1 ⎠ ⎞ → ⎝ ⎛ 1 0 0 0 1 0 0 0 1 − 24 20 − 5 18 − 15 4 5 − 4 1 ⎠ ⎞ Therefore, the inverse of ⎝ ⎛ 1 0 5 2 1 6 3 4 0 ⎠ ⎞ is ⎝ ⎛ − 24 20 − 5 18 − 15 4 5 − 4 1 ⎠ ⎞ Invert ( 1 2 0 3 − 1 2 − 2 3 − 2 ) ( 1 2 0 1 0 0 3 − 1 2 0 1 0 − 2 3 − 2 0 0 1 ) → ( 1 2 0 1 0 0 0 − 7 2 − 3 1 0 0 7 − 2 2 0 1 ) → ( 1 2 0 1 0 0 0 − 7 2 − 3 1 0 0 0 0 − 1 1 1 ) Since the RREF of the matrix does not have a pivot in every row and column, it is not invertible. \text{Invert }
\begin{pmatrix}
1 & 2 & 0 \\
3 & -1 & 2 \\
-2 & 3 & -2
\end{pmatrix} \\
\left(
\begin{array}{ccc|ccc}
1 & 2 & 0 & 1 & 0 & 0 \\
3 & -1 & 2 & 0 & 1 & 0\\
-2 & 3 & -2 & 0 & 0 & 1
\end{array}
\right) \rightarrow
\left(
\begin{array}{ccc|ccc}
1 & 2 & 0 & 1 & 0 & 0 \\
0 & -7 & 2 & -3 & 1 & 0\\
0 & 7 & -2 & 2 & 0 & 1
\end{array}
\right) \rightarrow
\left(
\begin{array}{ccc|ccc}
1 & 2 & 0 & 1 & 0 & 0 \\
0 & -7 & 2 & -3 & 1 & 0\\
0 & 0 & 0 & -1 & 1 & 1
\end{array}
\right) \\
\text{Since the RREF of the matrix does not have a pivot in every row and column, it is not invertible.} Invert ⎝ ⎛ 1 3 − 2 2 − 1 3 0 2 − 2 ⎠ ⎞ ⎝ ⎛ 1 3 − 2 2 − 1 3 0 2 − 2 1 0 0 0 1 0 0 0 1 ⎠ ⎞ → ⎝ ⎛ 1 0 0 2 − 7 7 0 2 − 2 1 − 3 2 0 1 0 0 0 1 ⎠ ⎞ → ⎝ ⎛ 1 0 0 2 − 7 0 0 2 0 1 − 3 − 1 0 1 1 0 0 1 ⎠ ⎞ Since the RREF of the matrix does not have a pivot in every row and column, it is not invertible.