Linear Algebra

Properties of Matrix Transformations

Imron Rosyadi

Properties of Matrix Transformations

Composition, Invertibility, and Equivalence

Compositions of Matrix Transformations

Suppose \(T_{A}\) maps from \(R^{n}\) to \(R^{k}\) and \(T_{B}\) maps from \(R^{k}\) to \(R^{m}\).
The composition \(T_{B}\circ T_{A}\) maps \(\mathbf{x}\) in \(R^{n}\) to \(T_{B}(T_{A}(\mathbf{x}))\) in \(R^{m}\).

\[ (T_{B}\circ T_{A})(\mathbf{x}) = T_{B}(T_{A}(\mathbf{x})) \tag{1} \]

This composition is itself a matrix transformation:

\[ (T_{B}\circ T_{A})(\mathbf{x}) = B(A\mathbf{x}) = (B A)\mathbf{x} \] Thus, the standard matrix for \(T_{B}\circ T_{A}\) is \(BA\).

\[ T_{B}\circ T_{A} = T_{B A} \tag{2} \]

Note

The transformation \(T_A\) is performed first, then \(T_B\).
This is analogous to function composition \((f \circ g)(x) = f(g(x))\).

Emphasize that the order of operations in composition is from right to left, matching matrix multiplication. Explain the domain and codomain relationships. Mention that this property is fundamental for understanding how complex linear transformations can be built from simpler ones.

Generalizing Compositions

Compositions can extend to finite successions of matrix transformations. For \(T_{A}\colon R^{n}\to R^{k}\), \(T_{B}\colon R^{k}\to R^{l}\), and \(T_{C}\colon R^{l}\to R^{m}\):

\[ (T_{C}\circ T_{B}\circ T_{A})(\mathbf{x}) = T_{C}(T_{B}(T_{A}(\mathbf{x}))) \] The standard matrix for this composition is \(CBA\):

\[ T_{C}\circ T_{B}\circ T_{A} = T_{C B A} \tag{3} \]

We denote the standard matrix for a transformation \(T\) as \([T]\).
So, Formula (2) can be restated as:

\[ [T_{2} \circ T_{1}] = [T_{2}][T_{1}] \tag{4} \] And Formula (3) as:

\[ [T_{3} \circ T_{2} \circ T_{1}] = [T_{3}][T_{2}][T_{1}] \tag{5} \]

Reinforce the idea that the order of matrix multiplication matches the reverse order of transformation application. This is a common point of confusion.
Explain why the notation \([T]\) is useful when we don’t want to explicitly name the matrix.

Composition Is Not Commutative (Example 1)

Warning

WARNING: In general, \(T_{B} \circ T_{A} \neq T_{A} \circ T_{B}\). Order matters!

Let \(T_{1}\colon R^{2} \to R^{2}\) be reflection about \(y=x\).
Let \(T_{2}\colon R^{2} \to R^{2}\) be orthogonal projection onto the \(y\)-axis.

Standard matrices:
\([T_{1}] = \left[ \begin{array}{l l}{0} & {1}\\ {1} & {0} \end{array} \right]\)
\([T_{2}] = \left[ \begin{array}{l l}{0} & {0}\\ {0} & {1} \end{array} \right]\)

\[ [T_{1}\circ T_{2}] = [T_{1}][T_{2}] = \left[ \begin{array}{l l}{0} & {1}\\ {1} & {0} \end{array} \right]\left[ \begin{array}{l l}{0} & {0}\\ {0} & {1} \end{array} \right] = \left[ \begin{array}{l l}{0} & {1}\\ {0} & {0} \end{array} \right] \]

\[ [T_{2}\circ T_{1}] = [T_{2}][T_{1}] = \left[ \begin{array}{l l}{0} & {0}\\ {0} & {1} \end{array} \right]\left[ \begin{array}{l l}{0} & {1}\\ {1} & {0} \end{array} \right] = \left[ \begin{array}{l l}{0} & {0}\\ {1} & {0} \end{array} \right] \] Clearly, \([T_{2} \circ T_{1}] \neq [T_{1} \circ T_{2}]\).

Explain why these matrices represent the transformations. Walk through the matrix multiplication step-by-step. Emphasize that the resulting matrices are different, which means the transformations themselves are different. This is a crucial concept, as matrix multiplication is not generally commutative.

Visualizing Non-Commutativity

Let’s see the effect of \(T_1 \circ T_2\) vs \(T_2 \circ T_1\) on a vector \(\mathbf{x} = (1, 1)\).

\(T_1 \circ T_2\) (Projection then Reflection)

1. Project onto \(y\)-axis: \(T_2(1,1) = (0,1)\)
2. Reflect about \(y=x\): \(T_1(0,1) = (1,0)\)

Final result: \((1,0)\)

\(T_2 \circ T_1\) (Reflection then Projection)

1. Reflect about \(y=x\): \(T_1(1,1) = (1,1)\)
2. Project onto \(y\)-axis: \(T_2(1,1) = (0,1)\)

Final result: \((0,1)\)

Visualizing Non-Commutativity

Figure 4.10.2: Visual demonstration of non-commutative composition.

This interactive plot clearly shows how applying the transformations in different orders leads to different final vectors. Click on the points and lines to see their values. This visual reinforces the algebraic conclusion that order matters.

Composition of Rotations Is Commutative (Example 2)

Let \(T_{1}, T_{2}\) be rotations about the origin by \(\theta_{1}\) and \(\theta_{2}\). Standard matrices:

\[ [T_{1}] = \left[ \begin{array}{cc}\cos \theta_{1} & -\sin \theta_{1} \\ \sin \theta_{1} & \cos \theta_{1} \end{array} \right], \quad [T_{2}] = \left[ \begin{array}{cc}\cos \theta_{2} & -\sin \theta_{2} \\ \sin \theta_{2} & \cos \theta_{2} \end{array} \right] \]

The composition \(T_{2} \circ T_{1}\) rotates by \(\theta_{1} + \theta_{2}\). Its standard matrix should be:

\[ [T_{2} \circ T_{1}] = \left[ \begin{array}{cc}\cos (\theta_{1} + \theta_{2}) & -\sin (\theta_{1} + \theta_{2}) \\ \sin (\theta_{1} + \theta_{2}) & \cos (\theta_{1} + \theta_{2}) \end{array} \right] \]

Verify this with matrix multiplication and trigonometric identities:

\[ \begin{array}{r l}{[T_{2}][T_{1}] = \left[ \begin{array}{c c}{\cos \theta_{2}} & {-\sin \theta_{2}}\\ {\sin \theta_{2}} & {\cos \theta_{2}} \end{array} \right]\left[ \begin{array}{c c}{\cos \theta_{1}} & {-\sin \theta_{1}}\\ {\sin \theta_{1}} & {\cos \theta_{1}} \end{array} \right]} & {}\\ {= \left[ \begin{array}{c c}{\cos \theta_{2}\cos \theta_{1} - \sin \theta_{2}\sin \theta_{1}} & {-(\cos \theta_{2}\sin \theta_{1} + \sin \theta_{2}\cos \theta_{1})}\\ {\sin \theta_{2}\cos \theta_{1} + \cos \theta_{2}\sin \theta_{1}} & {-\sin \theta_{2}\sin \theta_{1} + \cos \theta_{2}\cos \theta_{1}} \end{array} \right]} & {}\\ {= \left[ \begin{array}{c c}{\cos (\theta_{1} + \theta_{2})} & {-\sin (\theta_{1} + \theta_{2})}\\ {\sin (\theta_{1} + \theta_{2})} & {\cos (\theta_{1} + \theta_{2})} \end{array} \right]} & {}\\ {= [T_{2}\circ T_{1}]} & {} \end{array} \]

This implies \(R_{\theta_{1}}R_{\theta_{2}} = R_{\theta_{1} + \theta_{2}}\).

Go through the trigonometric identities used: \(\cos(A+B) = \cos A \cos B - \sin A \sin B\) and \(\sin(A+B) = \sin A \cos B + \cos A \sin B\). Explain why rotation is an example where transformations do commute intuitively.

Visualizing Rotational Commutativity

viewof theta1 = Inputs.range([0, 2 * Math.PI], {label: "Angle 1 (θ₁)", step: 0.1, value: Math.PI / 4})
viewof theta2 = Inputs.range([0, 2 * Math.PI], {label: "Angle 2 (θ₂)", step: 0.1, value: Math.PI / 2})

Figure 4.10.3: Interactive demonstration of commutative rotations.

Adjust the angles using the sliders. Observe how the blue, red, and green final points always coincide, demonstrating commutativity. The blue path shows applying R1 then R2. The red path shows R2 then R1. The green path shows a single rotation by the sum of angles. All end at the same point.

Composition of Two Reflections (Example 3)

Let \(T_{1}\colon R^{2}\to R^{2}\) be reflection about the \(y\)-axis.
Let \(T_{2}\colon R^{2}\to R^{2}\) be reflection about the \(x\)-axis.

Standard matrices:
\([T_{1}] = \left[ \begin{array}{c c}{-1} & 0\\ 0 & 1 \end{array} \right]\)
\([T_{2}] = \left[ \begin{array}{c c}{1} & 0\\ 0 & {-1} \end{array} \right]\)

\[ [T_{1}\circ T_{2}] = [T_{1}][T_{2}] = \left[ \begin{array}{c c}{-1} & 0\\ 0 & 1 \end{array} \right]\left[ \begin{array}{c c}{1} & 0\\ 0 & {-1} \end{array} \right] = \left[ \begin{array}{c c}{-1} & 0\\ 0 & {-1} \end{array} \right] \]

\[ [T_{2}\circ T_{1}] = [T_{2}][T_{1}] = \left[ \begin{array}{c c}{1} & 0\\ 0 & {-1} \end{array} \right]\left[ \begin{array}{c c}{-1} & 0\\ 0 & 1 \end{array} \right] = \left[ \begin{array}{c c}{-1} & 0\\ 0 & {-1} \end{array} \right] \] In this special case, \(T_{1}\circ T_{2} = T_{2}\circ T_{1}\). Both result in reflection about the origin.
The operator \(T(\mathbf{x}) = - \mathbf{x}\) is reflection about the origin.

Explain why reflection about an axis and reflection about the origin are important geometric transformations. Show that applying reflections in different orders still leads to the same outcome, which is reflection about the origin. This is another example of commutative transformations, but for a different reason than rotations.

Composition of Three Transformations (Example 4)

Find the standard matrix for \(T\colon R^{3}\to R^{3}\) that:

1. Rotates a vector counterclockwise about the \(z\)-axis through an angle \(\theta\) (\(T_1\)).
2. Reflects the resulting vector about the \(yz\)-plane (\(T_2\)).
3. Projects that vector orthogonally onto the \(xy\)-plane (\(T_3\)).

The composition is \(T = T_{3}\circ T_{2}\circ T_{1}\).

Standard matrices for individual transformations:

\[ [T_{1}] = \left[ \begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right] \quad \text{(Rotation about z-axis)} \]

\[ [T_{2}] = \left[ \begin{array}{ccc} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \quad \text{(Reflection about yz-plane)} \]

\[ [T_{3}] = \left[ \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \quad \text{(Projection onto xy-plane)} \]

Review where these standard matrices come from (previous sections or common knowledge).
Emphasize the order of multiplication: \([T] = [T_3][T_2][T_1]\).

Composition of Three Transformations (Example 4 Cont.)

Now, we multiply the matrices in the correct order:

\[ [T] = [T_{3}][T_{2}][T_{1}] \]

\[ [T] = \left[ \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right]\left[ \begin{array}{ccc} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\left[ \begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right] \]

First, calculate \([T_2][T_1]\): \[ [T_{2}][T_{1}] = \left[ \begin{array}{ccc} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\left[ \begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right] = \left[ \begin{array}{ccc} - \cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right] \]

Then, multiply by \([T_3]\): \[ [T] = \left[ \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right]\left[ \begin{array}{ccc} - \cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right] \]

\[ [T] = \left[ \begin{array}{ccc} - \cos \theta & \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 \end{array} \right] \]

Composition of Three Transformations (Example 4 Cont.)

Walk through the matrix multiplication carefully. The Python code provides a concrete numerical example to verify the symbolic result. Highlight the effect of each transformation on the final matrix (e.g., the last row becoming zeros due to projection).

One-to-One Matrix Transformations

Important

DEFINITION 1: A matrix transformation \(T_{A} \colon R^{n} \to R^{m}\) is said to be one-to-one if \(T_{A}\) maps distinct vectors (points) in \(R^{n}\) into distinct vectors (points) in \(R^{m}\).

Alternative expressions of Definition 1:

1. \(T_{A}\) is one-to-one if for each vector \(\mathbf{b}\) in the range of \(A\), there is exactly one vector \(\mathbf{x}\) in \(R^{n}\) such that \(T_{A} \mathbf{x} = \mathbf{b}\).
2. \(T_{A}\) is one-to-one if the equality \(T_{A}(\mathbf{u}) = T_{A}(\mathbf{v})\) implies that \(\mathbf{u} = \mathbf{v}\).

Explain intuitively what “one-to-one” means: no two distinct inputs map to the same output. Discuss why the alternative expressions are equivalent. For example, if \(T_A(\mathbf{u}) = T_A(\mathbf{v})\) implies \(\mathbf{u}=\mathbf{v}\), it means there’s only one input for each output.

Examples of One-to-One Transformations

Rotation Operators

Rotation operators on \(R^{2}\) are one-to-one because distinct vectors rotated through the same angle have distinct images.

Orthogonal Projection

In contrast, orthogonal projection of \(R^{2}\) onto the \(x\)-axis is not one-to-one. It maps distinct points on the same vertical line into the same point on the \(x\)-axis.

Use the visuals to reinforce the definitions. For rotation, imagine two different arrows from the origin. After rotation, they will still be different arrows. For projection, imagine two points stacked vertically. They both collapse to the same point on the x-axis.

Kernel and Range

Tip

DEFINITION 2: If \(T_{A}\colon R^{n}\to R^{m}\) is a matrix transformation:

• The kernel of \(T_{A}\), denoted \(\ker (T_{A})\), is the set of all vectors in \(R^{n}\) that \(T_{A}\) maps into \(\mathbf{0}\).
• The range of \(T_{A}\), denoted \(R(T_{A})\), is the set of all vectors in \(R^{m}\) that are images under this transformation of at least one vector in \(R^{n}\).

In brief:

\[ \ker (T_{A}) = \mathrm{null~space~of~}A \tag{6} \]

\[ R(T_{A}) = \mathrm{column~space~of~}A \tag{7} \]

Connect these definitions to concepts already learned: null space and column space. Emphasize that the kernel is the set of vectors that “disappear” or are mapped to the origin. The range is the set of all possible outputs of the transformation.

Different Views of Subspaces

The key to solving mathematical problems often lies in adopting the right point of view.

Subspace of \(R^{n}\) (Kernel/Null Space)

• Matrix view: The null space of \(A\).
• System view: The solution space of \(A\mathbf{x} = \mathbf{0}\).
• Transformation view: The kernel of \(T_{A}\).

Subspace of \(R^{m}\) (Range/Column Space)

• Matrix view: The column space of \(A\).
• System view: All \(\mathbf{b}\) in \(R^{m}\) for which \(A\mathbf{x} = \mathbf{b}\) is consistent.
• Transformation view: The range of \(T_{A}\).

Discuss how these different perspectives (matrix, system, transformation) are all describing the same underlying mathematical object or concept. This reinforces the interconnectedness of linear algebra topics and helps students build a more comprehensive understanding.

Theorem 4.10.1: Equivalence for Operators

If \(A\) is an \(n\times n\) matrix and \(T_{A}\colon R^{n}\to R^{n}\) is the corresponding matrix operator, then the following statements are equivalent:

1. \(A\) is invertible.
2. The kernel of \(T_{A}\) is \(\{\mathbf{0}\}\).
3. The range of \(T_{A}\) is \(R^{n}\).
4. \(T_{A}\) is one-to-one.

Important

This theorem establishes fundamental relationships between the invertibility of a matrix \(A\) and properties of its corresponding matrix transformation \(T_A\).

Highlight that this theorem is very powerful because it links algebraic properties (invertibility) to geometric properties (one-to-one, full range, trivial kernel). Walk through the proof steps, clearly stating the reason for each implication (e.g., “from previous theorems”).

Theorem 4.10.1: Equivalence for Operators (Cont.)

Proof of \((b)\Rightarrow (c)\):
Assume the kernel of \(T_{A}\) is \(\{\mathbf{0}\}\).
By Formula (6), the null space of \(A\) is \(\{\mathbf{0}\}\).
This implies that the nullity of \(A\) is 0.
By the Rank-Nullity Theorem (\(n = \mathrm{rank}(A) + \mathrm{nullity}(A)\)), if nullity is 0, then \(\mathrm{rank}(A) = n\).
Since \(A\) is an \(n \times n\) matrix and its rank is \(n\), its column space spans \(R^{n}\) (i.e., the column space is all of \(R^{n}\)).
By Formula (7), the range of \(T_{A}\) is \(R^{n}\).

Explain the Rank-Nullity Theorem if students are not fully familiar.
Connect rank \(n\) to the column space spanning \(R^n\).

Example 5: The Rotation Operator on \(R^{2}\) Is One-to-One

As illustrated in Figure 4.10.6, the operator \(T\colon R^{2}\to R^{2}\) that rotates vectors through an angle \(\theta\) is one-to-one.
In accordance with parts (a) and (d) of Theorem 4.10.1, its standard matrix should be invertible.

The standard matrix for \(T\) is:

\[ [T] = \left[ \begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right] \tag{10} \]

This matrix is invertible because its determinant is nonzero:

\[ \operatorname *{det}[T] = \left| \begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right| = \cos^{2}\theta +\sin^{2}\theta = 1\neq 0 \]

Example 5: The Rotation Operator on \(R^{2}\) Is One-to-One

Reiterate the connection between a non-zero determinant and invertibility. The Python code provides a numerical verification, showing that for any angle, the determinant is 1.

Example 6: Projection Operators Are Not One-to-One

As illustrated in Figure 4.10.7, the operator \(T\colon R^{2}\to R^{2}\) that projects onto the \(x\)-axis in the \(xy\)-plane is not one-to-one.
In accordance with parts (a) and (d) of Theorem 4.10.1, its standard matrix should not be invertible.

The standard matrix for \(T\) is:

\[ [T] = \left[ \begin{array}{cc}1 & 0 \\ 0 & 0 \end{array} \right] \]

Since \(\operatorname *{det}[T] = (1)(0) - (0)(0) = 0\), the operator \(T\) is not one-to-one.

Example 6: Projection Operators Are Not One-to-One

Explain why a zero determinant implies non-invertibility. Connect this back to the geometric intuition: projection “collapses” information, making it impossible to uniquely reverse the operation. The Python code confirms the zero determinant.

Inverse of a One-to-One Matrix Operator

If \(T_{A}\colon R^{n}\to R^{n}\) is a one-to-one matrix operator, then \(A\) is invertible (by Theorem 4.10.1).
The matrix operator \(T_{A^{-1}}\colon R^{n}\to R^{n}\) corresponding to \(A^{-1}\) is called the inverse operator of \(T_{A}\).

This terminology is appropriate because \(T_{A}\) and \(T_{A^{-1}}\) cancel each other’s effects:

\[ \begin{array}{r}T_{A}(T_{A^{-1}}(\mathbf{x})) = A A^{-1}\mathbf{x} = I\mathbf{x} = \mathbf{x} \\ T_{A^{-1}}(T_{A}(\mathbf{x})) = A^{-1}A\mathbf{x} = I\mathbf{x} = \mathbf{x} \end{array} \]

Equivalently:

\[ \begin{array}{c}T_{A}\circ T_{A^{-1}} = T_{A A^{-1}} = T_{I} \\ T_{A^{-1}}\circ T_{A} = T_{A^{-1}A} = T_{I} \end{array} \]

Notational matters:

The standard matrices for these operators are related by:

\[ [T_{A^{-1}}] = [T_{A}]^{-1} \tag{8} \] Or, more generally:

\[ [T^{-1}] = [T]^{-1} \tag{9} \]

Emphasize the “undoing” nature of inverse transformations. Explain that \(T_I\) is the identity transformation, which maps a vector to itself. The notation \([T]^{-1}\) means the inverse of the matrix \([T]\).

Example 7: Standard Matrix for \(T^{-1}\) (Rotation)

Let \(T\colon R^{2}\to R^{2}\) be the operator that rotates each vector in \(R^{2}\) through the angle \(\theta\).
Its standard matrix is \([T] = \left[ \begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right]\).

Geometrically, to undo the effect of \(T\), one must rotate each vector in \(R^{2}\) through the angle \(-\theta\).
The standard matrix for a rotation by \(-\theta\) is:

\[ \left[ \begin{array}{cc}\cos (-\theta) & -\sin (-\theta) \\ \sin (-\theta) & \cos (-\theta) \end{array} \right] = \left[ \begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array} \right] \]

Let’s compute \([T]^{-1}\): For a \(2 \times 2\) matrix \(\left[ \begin{array}{cc}a & b \\ c & d \end{array} \right]\), its inverse is \(\frac{1}{ad-bc}\left[ \begin{array}{cc}d & -b \\ -c & a \end{array} \right]\).

\[ [T]^{-1} = \frac{1}{\cos^{2}\theta - (-\sin^{2}\theta)}\left[ \begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array} \right] = \frac{1}{1}\left[ \begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array} \right] \]

\[ [T]^{-1} = \left[ \begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array} \right] \] This confirms \([T^{-1}] = \left[ \begin{array}{cc}\cos (-\theta) & -\sin (-\theta) \\ \sin (-\theta) & \cos (-\theta) \end{array} \right]\).

Review the formula for the inverse of a \(2 \times 2\) matrix. Show how the algebraic inverse matches the geometric intuition of rotating by the negative angle.

Interactive Rotation and Its Inverse

Let’s rotate a vector and then apply its inverse.

// Define an input for the rotation angle
viewof rotation_angle = Inputs.range([0, 2 * Math.PI], {label: "Rotation Angle (rad)", step: 0.1, value: Math.PI / 3})

Adjust the Rotation Angle slider. Observe how the blue vector changes, and the red dotted vector always returns to overlap the original white vector. This visually confirms that the inverse transformation successfully “undoes” the original transformation.

Example 8: Finding \(T^{-1}\)

Show that the operator \(T\colon R^{2}\to R^{2}\) defined by the equations:

\[ \begin{array}{r}w_{1} = 2x_{1} + x_{2} \\ w_{2} = 3x_{1} + 4x_{2} \end{array} \] is one-to-one, and find \(T^{- 1}(w_{1},w_{2})\).

Solution: The matrix form of these equations is:

\[ \left[ \begin{array}{l}w_{1} \\ w_{2} \end{array} \right] = \left[ \begin{array}{ll}2 & 1 \\ 3 & 4 \end{array} \right]\left[ \begin{array}{l}x_{1} \\ x_{2} \end{array} \right] \] So the standard matrix for \(T\) is:

\[ [T] = \left[ \begin{array}{ll}2 & 1 \\ 3 & 4 \end{array} \right] \] First, check if \(T\) is one-to-one by checking if \([T]\) is invertible (i.e., \(\operatorname *{det}[T] \neq 0\)). \(\operatorname *{det}[T] = (2)(4) - (1)(3) = 8 - 3 = 5 \neq 0\). Since the determinant is non-zero, \([T]\) is invertible, and thus \(T\) is one-to-one.

Explain the process: convert equations to matrix form, calculate determinant to check invertibility (and thus one-to-one property).

Example 8: Finding \(T^{-1}\) (Cont.)

Now, find the standard matrix for \(T^{-1}\):

\[ [T^{-1}] = [T]^{-1} = \frac{1}{5}\left[ \begin{array}{cc}4 & -1 \\ -3 & 2 \end{array} \right] = \left[ \begin{array}{cc}\frac{4}{5} & -\frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{array} \right] \]

Thus, to find \(T^{-1}(w_1, w_2)\), we multiply the inverse matrix by the vector \(\left[ \begin{array}{l}w_{1} \\ w_{2} \end{array} \right]\):

\[ [T^{-1}]\left[ \begin{array}{c}w_{1} \\ w_{2} \end{array} \right] = \left[ \begin{array}{cc}\frac{4}{5} & -\frac{1}{5} \\ -\frac{3}{5} & \frac{2}{5} \end{array} \right]\left[ \begin{array}{c}w_{1} \\ w_{2} \end{array} \right] = \left[ \begin{array}{c}\frac{4}{5} w_{1} - \frac{1}{5} w_{2} \\ -\frac{3}{5} w_{1} + \frac{2}{5} w_{2} \end{array} \right] \]

From this, we conclude that:

\[ T^{-1}(w_{1},w_{2}) = \left(\frac{4}{5} w_{1} - \frac{1}{5} w_{2}, - \frac{3}{5} w_{1} + \frac{2}{5} w_{2}\right) \]

Example 8: Finding \(T^{-1}\) (Cont.)

Walk through the calculation of the inverse matrix. Show how to express the inverse transformation in coordinate form. The Python code verifies the inverse calculation and demonstrates how to use it to find the original vector from an image vector.

Theorem 4.10.2: Equivalent Statements

This theorem compiles many equivalent statements regarding an \(n \times n\) matrix \(A\). If \(A\) is an \(n \times n\) matrix, then the following statements are equivalent:

1. \(A\) is invertible.
2. \(A\mathbf{x} = \mathbf{0}\) has only the trivial solution.
3. The reduced row echelon form of \(A\) is \(I_{n}\).
4. \(A\) is expressible as a product of elementary matrices.
5. \(A\mathbf{x} = \mathbf{b}\) is consistent for every \(n \times 1\) matrix \(\mathbf{b}\).
6. \(A\mathbf{x} = \mathbf{b}\) has exactly one solution for every \(n \times 1\) matrix \(\mathbf{b}\).
7. \(\operatorname *{det}(A) \neq 0\).
8. The column vectors of \(A\) are linearly independent.
9. The row vectors of \(A\) are linearly independent.
10. The column vectors of \(A\) span \(R^{n}\).

11. The row vectors of \(A\) span \(R^{n}\).
12. The column vectors of \(A\) form a basis for \(R^{n}\).
13. The row vectors of \(A\) form a basis for \(R^{n}\).
14. \(A\) has rank \(n\).
15. \(A\) has nullity 0.
16. The orthogonal complement of the null space of \(A\) is \(R^{n}\).
17. The orthogonal complement of the row space of \(A\) is \(\{\mathbf{0}\}\).
18. The kernel of \(T_{A}\) is \(\{\mathbf{0}\}\).
19. The range of \(T_{A}\) is \(R^{n}\).
20. \(T_{A}\) is one-to-one.

Tip

This comprehensive theorem is a cornerstone of linear algebra, connecting concepts from systems of equations, matrix properties, vector spaces, and transformations. Understanding these equivalences deepens your comprehension of matrix behavior.

This theorem summarizes many key concepts. You don’t need to memorize the letter labels, but understand that if one of these is true, all are true. Highlight the newly added parts (r), (s), (t) from Theorem 4.10.1, showing how transformations fit into this larger picture. Encourage students to review and connect these statements, as they are frequently used in problem-solving and proofs.