Definition. An \( n\times n\) matrix \( D \), with entries \( d_{ij} \), is diagonal if \( d_{ij} = 0\) whenever \( i\ne j \).
For example, the matrix \( \begin{bmatrix} \color{red} -4 & 0 & 0 & 0 \\ 0 & \color{red} 0 & 0 & 0 \\ 0 & 0 & \color{red} 1 & 0 \\ 0 & 0 & 0 & \color{red} 6 \end{bmatrix} \) is diagonal. Note that the diagonal entires themselves may also be zero.
The following theorems are easily verifiable:
Theorem (Product of Diagonal Matrices). If \( S \) and \( T \) are \( n\times n\) diagonal matrices, with entries \( s_{ij} \) and \( t_{ij} \), respectively, then the entries of \( ST \) are \( s_{ij} t_{ij} \).
Theorem (Powers of Diagonal Matrices). If \( D \) is a diagonal matrix with entries \( d_{ij} \) and \( k \) is a positive integer, then the entries of \( D^k \) are \( d_{ij}^k \).
Definition. An \(n\times n\) matrix \( A \) is diagonalizable if it is similar to a diagonal matrix. That is, \( A \) is diagonalizable if and only if there exist an invertible matrix \( P \) and a diagonal matrix \( D \) for which \( A = PDP^{-1} \).
Using the Powers of Diagonal Matrices Theorem, it is relatively easy to find powers of diagonalizable matrices: \[ A^k = (PDP^{-1})^k = (PDP^{-1})(PDP^{-1})\cdots (PDP^{-1}) = PD^k P^{-1}. \]
Example 1. Let \( A = \begin{bmatrix} -7 & 5 \\ -10 & 8 \end{bmatrix} \) and note that \( A = PDP^{-1} \), where \( P = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \) and \( D = \begin{bmatrix} -2 & 0 \\ 0 & 3 \end{bmatrix} \). Find a formula for \( A^k \), where \( k \) is a positive integer.
We compute \( P^{-1} = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} \) and \( D^k = \begin{bmatrix} (-2)^k & 0 \\ 0 & 3^k \end{bmatrix} \). Now, \[ A^k = PD^k P^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} (-2)^k & 0 \\ 0 & 3^k \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 2(-2)^k - 3^k & 3^k - (-2)^k \\ 2(-2)^k - 2\cdot 3^k & 2\cdot 3^k - (-2)^k \end{bmatrix}. \Box \]
The Diagonalization Theorem gives us a way to determine which matrices are diagonalizable, and helps us understand what the matrices \( P \) and \( D \) represent.
The Diagonalization Theorem. An \( n\times n\) matrix \( A \) is diagonalizable if and only if \( A \) has \( n \) linearly independent eigenvectors. In addition, whenever \( A = PDP^{-1} \) for a diagonal matrix \( D \), the columns of \( P \) are linearly independent eigenvalues of \( A \) and the diagonal entries of \( D \) are eigenvalues of \( A \) that correspond to the columns of \( P \).
The Diagonalization Theorem says that \( A \) is diagonalizable if and only if there are enough eigenvectors of \( A \) to form a basis for \( \mathbb R^n \). Such a basis is called an eigenvector basis for \( \mathbb R^n \). It is important to note that not every matrix is diagonalizable, and not every matrix has an eigenvector basis!
Proof of the Diagonalization Theorem. First, note that if \( P \) is any \( n\times n\) matrix with columns \( \bbm v_1, \ldots, \bbm v_n \) and \( D \) is any diagonal matrix with diagonal entries \( \lambda_1, \ldots, \lambda_n \), then \[ AP = A \begin{bmatrix} \bbm v_1 & \bbm v_2 & \cdots & \bbm v_n \end{bmatrix} = \begin{bmatrix} A\bbm v_1 & A\bbm v_2 & \cdots & A\bbm v_n \end{bmatrix} \quad \mbox{and} \quad PD = \begin{bmatrix} \bbm v_1 & \bbm v_2 & \cdots & \bbm v_n \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} = \begin{bmatrix} \lambda_1 \bbm v_1 & \lambda_2 \bbm v_2 & \cdots & \lambda_n \bbm v_n \end{bmatrix}. \]
Now, suppose that \( A \) is diagonalizable with \( A = PDP^{-1} \) for some invertible matrix \( P \) and some diagonal matrix \( D \). Now, \( AP = PD \), and thus \( A\bbm v_i = \lambda_i \bbm v_i \) for all \( i \). These equations show that the \( n\) columns of \( P \) are eigenvectors of \( A \). Since \( P \) is invertible, these vectors are linearly independent by the Invertible Matrix Theorem.
Conversely, suppose that \( A \) has \( n \) linearly independent eigenvectors \( \bbm v_1, \bbm v_2, \ldots, \bbm v_n \) corresponding to eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_n \), respectively. Let \( P \) be the matrix whose columns are the \( \bbm v_i \) and let \( D \) be the diagonal matrix whose diagonal entries are the \( \lambda_i \). Since \( A \bbm v_i = \lambda_i \bbm v_i \) for all \( i\), we have \( AP = PD \). Since the \( \bbm v_i \) are linearly independent, \( P \) is invertible by the Invertible Matrix Theorem. Multiplying both sides by \(P^{-1} \) on the right gives \( A = PDP^{-1} \), and so \( A \) is diagonalizable. \( \Box \)
To illustrate the usefulness of diagonalization, we will spend some time investigating the Fibonacci numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. These numbers are defined by a recursive formula \( F_0 = 1 \), \( F_1 = 1 \), and \( F_{n+2} = F_{n+1} + F_n \) for \( n \ge 0 \).
If we write \( \bbm f_n = \vectwo {F_{n+1}} {F_n} \), then we can obtain elements of the Fibonacci sequence by repeatly multiplying by the matrix \( M = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \), starting with the vector \( \bbm f_0 = \vectwo 11 \):
This gives a formula for the \( n^{\rm th} \) Fibonacci number: it is the second entry of \( \bbm f_n = M^n \bbm f_0 \). Our goal is now to find a formula for \( M^n \) like the one we found in Example 1. The good news is that \(M\) is diagonalizable.
If we write \( \phi = \frac{1+\sqrt 5}{2} \) and \( \hat{\phi} = \frac{1-\sqrt 5} 2 \), then \( M \) can be diagonalized as \( M = \begin{bmatrix} \phi & \hat{\phi} \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \phi & 0 \\ 0 & \hat{\phi} \end{bmatrix} \begin{bmatrix} \phi & \hat{\phi} \\ 1 & 1 \end{bmatrix}^{-1} \). Now, \[ M^n = \begin{bmatrix} \phi & \hat{\phi} \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \phi & 0 \\ 0 & \hat{\phi} \end{bmatrix}^n \begin{bmatrix} \phi & \hat{\phi} \\ 1 & 1 \end{bmatrix}^{-1} = \frac 1 {\sqrt 5} \begin{bmatrix} \phi^{n+1} - \hat{\phi}^{n+1} & \phi^n - \hat{\phi}^n \\ \phi^n - \hat{\phi}^n & \phi^{n-1} - \hat{\phi}^{n-1} \end{bmatrix} . \]
Using this formula and \( \bbm f_n = \vectwo {F_{n+1}} {F_n} = M^n \bbm f_0 \) gives the closed formula for the \( n^{\rm th} \) Fibonacci number: \( F_n = \frac 1 {\sqrt 5} ( \phi^{n+1} - \hat{\phi}^{n+1} ) \).
This technique can give us closed (i.e., non-recursive) formulas for many different recursively-defined sequences.
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