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5.4 Eigenvectors and Linear Transformations

Let \( T: V \to W \) be a linear transformation. This is the same as \( T(\vec{x}) = A\vec{x} \) for an \( m \times n \) matrix \( A \), where \( V \) corresponds to \( \mathbb{R}^n \) and \( W \) corresponds to \( \mathbb{R}^m \).

If a basis for \( V \) is \( \{ \vec{b_1}, \vec{b_2}, \dots, \vec{b_n} \} \) and a basis for \( W \) is \( \{ \vec{c_1}, \vec{c_2}, \dots, \vec{c_m} \} \), then a vector \( \vec{x} \) in \( V \) is:

\[ \vec{x} = r_1 \vec{b_1} + r_2 \vec{b_2} + \dots + r_n \vec{b_n} \]

and the transformation of \( \vec{x} \) is:

\[ \begin{aligned} T(\vec{x}) &= T(r_1 \vec{b_1} + \dots + r_n \vec{b_n}) \\ &= r_1 T(\vec{b_1}) + r_2 T(\vec{b_2}) + \dots + r_n T(\vec{b_n}) \\ &= \left[ [T(\vec{b_1})]_C [T(\vec{b_2})]_C \dots [T(\vec{b_n})]_C \right] \begin{bmatrix} r_1 \\ \vdots \\ r_n \end{bmatrix} = M [\vec{x}]_B \end{aligned} \]

Note on Notation:

\( [T(\vec{b_1})]_C \): transformation of \( \vec{b_1} \) written using \( C \)-coordinates.
\( [\vec{x}]_B \): \( B \)-coordinate of \( \vec{x} \).
\( M \): matrix for \( T \) relative to \( B \) and \( C \).

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Example

\[ A = \begin{bmatrix} 1 & -4 & 8 & 1 \\ 0 & 2 & -1 & 3 \\ 0 & 0 & 0 & 5 \end{bmatrix} \quad T: V \to W, \quad V = \mathbb{R}^4, \quad W = \mathbb{R}^3 \]

A basis for \( V \) is \( \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} = \{ \vec{b_1}, \dots, \vec{b_4} \} \)

A basis for \( W \) is \( \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} = \{ \vec{c_1}, \vec{c_2}, \vec{c_3} \} \)

Column 1: \( T\left(\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\right) = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = 1 \cdot \vec{c_1} + 0 \cdot \vec{c_2} + 0 \cdot \vec{c_3} = [T(\vec{b_1})]_C \)

Column 2: \( T\left(\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\right) = \begin{bmatrix} -4 \\ 2 \\ 0 \end{bmatrix} = -4 \cdot \vec{c_1} + 2 \cdot \vec{c_2} + 0 \cdot \vec{c_3} = [T(\vec{b_2})]_C \)

and so on.

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Linear Transformation Matrix Example

Example: Find the matrix for \( T \) relative to \( B \) and \( D \) if \( T: \mathbb{R}^3 \to \mathbb{R}^2 \) and:

\[ \begin{aligned} T(\vec{b}_1) &= 7\vec{d}_1 - 8\vec{d}_2 \\ T(\vec{b}_2) &= -7\vec{d}_1 - 4\vec{d}_2 \\ T(\vec{b}_3) &= -\vec{d}_1 \end{aligned} \]

The resulting transformation matrix \( A \) is constructed by placing the coordinate vectors of the images of the basis vectors as columns:

\[ A = \begin{bmatrix} 7 & -7 & -1 \\ -8 & -4 & 0 \end{bmatrix} \]

Column 1 Analysis

Column 1 represents \( T(\vec{b}_1) \) in new coordinates (\( \vec{d} \)'s):

\[ T(\vec{b}_1) = 7\vec{d}_1 - 8\vec{d}_2 \to [T(\vec{b}_1)]_C = \begin{bmatrix} 7 \\ -8 \end{bmatrix} \]

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Finding the Transformation Matrix

Example: If \( T: V \to \mathbb{R}^2 \) and

\[ T(x_1\vec{b}_1 + x_2\vec{b}_2 + x_3\vec{b}_3) = \begin{bmatrix} -2x_1 + 8x_2 + 2x_3 \\ -4x_1 - 5x_2 \end{bmatrix} \]

Find the transformation matrix.

→ We need \( T(\vec{b}_1), T(\vec{b}_2), T(\vec{b}_3) \) as columns of the matrix.

Because \( T \) is linear:

\[ x_1 T(\vec{b}_1) + x_2 T(\vec{b}_2) + x_3 T(\vec{b}_3) = x_1 \begin{bmatrix} -2 \\ -4 \end{bmatrix} + x_2 \begin{bmatrix} 8 \\ -5 \end{bmatrix} + x_3 \begin{bmatrix} 2 \\ 0 \end{bmatrix} \]

By identifying the coefficients of \( x_1, x_2, \) and \( x_3 \), we obtain the columns of the matrix:

\[ \text{So, } A = \begin{bmatrix} -2 & 8 & 2 \\ -4 & -5 & 0 \end{bmatrix} \]

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Example: Transformation Matrix for Polynomials

Let \( T: \mathbb{P}_3 \to \mathbb{P}_3 \) be defined by:

\[ T(a_0 + a_1 t + a_2 t^2 + a_3 t^3) = a_1 + 2a_2 t + 3a_3 t^2 \]

Find the transformation matrix.

Step 1: Identify the Standard Basis

Basis: \( \{1, t, t^2, t^3\} \)

Step 2: Apply the Transformation to Basis Elements

\[ \begin{aligned} T(a_0) &= a_0 T(1) = a_0 \cdot 0 &\to T(1) &= 0 \\ T(a_1 t) &= a_1 T(t) = a_1 \cdot 1 &\to T(t) &= 1 \\ T(a_2 t^2) &= a_2 T(t^2) = a_2 \cdot 2t &\to T(t^2) &= 2t \\ T(a_3 t^3) &= a_3 T(t^3) = a_3 \cdot 3t^2 &\to T(t^3) &= 3t^2 \end{aligned} \]

Step 3: Represent Basis Elements as Vectors

Basis in vector form:

\[ \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} \]

Step 4: Transform the Basis Vectors

\[ \begin{aligned} T\left(\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\right) &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} & T\left(\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\right) &= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \\ T\left(\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}\right) &= \begin{bmatrix} 0 \\ 2 \\ 0 \\ 0 \end{bmatrix} & T\left(\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\right) &= \begin{bmatrix} 0 \\ 0 \\ 3 \\ 0 \end{bmatrix} \end{aligned} \]

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So, the transformation matrix \( A \) is:

\[ A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \quad \to \text{differentiation of } \mathbb{P}_3 \]

Check:

\( T(1 + 2t + 3t^2 + 4t^3) = 2 + 6t + 12t^2 \)

\[ \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \\ 12 \\ 0 \end{bmatrix} \]

Diagonalization and Transformations

If \( A \) is diagonalizable, then \( A = PDP^{-1} \)

\( P \) is the matrix with eigenvectors as columns.
\( D \) is the diagonal matrix with corresponding eigenvalues.

So, \( A\vec{x} = PDP^{-1}\vec{x} \)

How to interpret this as transformations?

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Understanding the Transformation \( P^{-1} \vec{x} \)

What does \( P^{-1} \vec{x} \) mean?

Given matrix \( A \):

\[ A = \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \]

Eigenvalues: \( \lambda = 1, 2 \)

Eigenvectors and Matrix \( P \):

\( \lambda = 1, \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \)
\( \lambda = 2, \vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)

\[ P = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \]

Change of Basis

\[ P [\vec{x}]_B = \vec{x} \]

Where \( [\vec{x}]_B \) is using eigenvectors as a basis.

Example:

\[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \implies (1) \begin{bmatrix} 1 \\ 0 \end{bmatrix} + 2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \]

Note: The vector \( \begin{bmatrix} 1 \\ 2 \end{bmatrix} \) represents coordinates using eigenvectors as basis.

\[ [\vec{x}]_B = P^{-1} \vec{x} \longrightarrow \text{coordinates of } \vec{x} \text{ in the eigenvector space} \]

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Diagonalization Process: \( P D P^{-1} \vec{x} \)

\[ P D P^{-1} \vec{x} \]

Step 1: \( P^{-1} \vec{x} \) — transform to the eigenvector coordinate from original basis.
Step 2: \( D (P^{-1} \vec{x}) \) — lengthens/shortens whatever follows it (where \( D \) is the diagonal matrix).
Step 3: Finally, \( P (D P^{-1} \vec{x}) \) — goes back to original basis (away from eigenvector basis).

So, if \( A \) is diagonalizable, \( A \vec{x} \) first goes to eigenvector frame, then lengthens/shortens according to eigenvalues, then leaves the eigenvector frame.

Matrix Trace Properties

\( \text{trace}(A) \longrightarrow \) if \( A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \) then \( \text{trace}(A) = 1 + 5 + 9 = 15 \)

\[ \text{trace}(ABC) = \text{trace}(BCA) = \text{trace}(CAB) \]