3.6 Diagonalization
1-CharacteristicValues-AnnihilatingPolynomials.pdf
Ex 1
Find all \(K\) in \(\mathbb{R}\) such that \(A\) is diagonalizable. \(A = \begin{bmatrix} K & 1 & -K \\ 0 & K & 1 \\ 0 & K+1 & 0 \end{bmatrix}\)
Then \(\chi_{A}(x) = \det (xI - A) = \begin{vmatrix} x - K & -1 & K \\ 0 & x - K & -1 \\ 0 & -k-1 & x \end{vmatrix}\)
\(= (x - K) \cdot \left[(x - K)x - (-1)(-k-1)\right]\)\(=(x-K)\cdot\left(x+1\right)\left(x-1-k\right)\)
The eigenvalues of \(A\) are: \(\lambda_1 = K,\ \lambda_2 = -1,\ \lambda_3 = 1 + K\)
\(A\) is diagonalizable if all eigenvalues are distinct.
Case 1: \(\lambda_1 = \lambda_2\)
\(K = -1 \implies \lambda_3 = 1 + (-1) = 0\), \(\chi_A(x) = (x+1)(x+1)x = (x+1)^2 x\)
Then \(A = \begin{bmatrix} -1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\) and \(E_{-1} = \ker (A + I_3) = \ker \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}\) and \(\mathrm{rank}(A + I_3) = 2\), then \(\dim E_{-1} = 1\)
\(\dim E_{-1} = 1 \neq \text{mult}(\chi_A, -1) = 2\)
Thus\(A\) is non-diagonalizable.
Case 2: \(\lambda_1 = \lambda_3\)
\(K = 1 + K \implies 0 = 1 \implies \text{impossible!}\)
Case 3: \(\lambda_2 = \lambda_3\)
\(-1 = 1 + K \implies K = -2\)
\(\chi_A(x) = (x+2)(x+1)(x+1) = (x+2)(x+1)^2\)
\(A = \begin{bmatrix} -2 & 1 & 2 \\ 0 & -2 & 1 \\ 0 & -1 & 1 \end{bmatrix},E_{-1} = \ker (A + I_3) = \ker \begin{bmatrix} -1 & 1 & 2 \\ 0 & -1 & 1 \\ 0 & -1 & 2 \end{bmatrix}\)
Then \(\mathrm{rank}(A + I_3) = 2\)\(\dim E_{-1} = 1 \neq 2 = \text{mult}(\chi_A, -1)\)
\(A\) is non-diagonalizable.
Final Conclusion
\(A\) is diagonalizable if and only if \(K \in \mathbb{R} \setminus \{-1, -2\}\)
Ex2 If \((a_n)_{n \in \mathbb{N}_0}\) is defined by:
\(a_0 = 0,\quad a_1 = 1,\quad a_2 = 1\)
\(a_{n+3} = 6a_{n} - 11a_{n+1} + 6a_{n+2} \ \forall n \in \mathbb{N}_0\)
Find the general formula for \(a_n\), \(\forall n \in \mathbb{N}_0\)
Main idea: Consider the matrix and construct a matrix equation with indices differing by one, then it can be inductively simplified to lower coefficients.
#Method tool: Use change of basis to diagonal matrix to calculate the higher-order matrix of A.#
\(\begin{cases} a_{n+1} = a_{n+1} \\ a_{n+2} = a_{n+2} \\ 6a_n - 11a_{n+1} + 6a_{n+2} = a_{n+3} \end{cases},\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{bmatrix} \begin{bmatrix} a_{n} \\ a_{n+1} \\ a_{n+2} \end{bmatrix}= \begin{bmatrix} a_{n+1} \\ a_{n+2} \\ a_{n+3} \end{bmatrix}\)
\(\equiv A \begin{bmatrix} a_{n} \\ a_{n+1} \\ a_{n+2} \end{bmatrix}= \begin{bmatrix} a_{n+1} \\ a_{n+2} \\ a_{n+3} \end{bmatrix}\) where \(A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{bmatrix}\)
Thus \(\begin{bmatrix} a_{n} \\ a_{n+1} \\ a_{n+2} \end{bmatrix}=A \begin{bmatrix} a_{n-1} \\ a_{n} \\ a_{n+1} \end{bmatrix}=A\left(A \begin{bmatrix} a_{n-2} \\ a_{n-1} \\ a_{n} \end{bmatrix}\right)=\ldots=A^{n} \begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \end{bmatrix}\)
Thus \(\begin{bmatrix} a_{n} \\ a_{n+1} \\ a_{n+2} \end{bmatrix}=A^{n} \begin{bmatrix} a_{0} \\ a_{1} \\ a_{2} \end{bmatrix}.\text{ That is }a_{n}=\left\lbrack r_{1}\right\rbrack \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\)
If \(A\) is diagonalizable, there exists an invertible matrix \(P\) such that \(A = P D P^{-1}\)Where \(D\) is a diagonal matrix.
\(A^n = (PDP^{-1})^n = P D^n P^{-1}\)
Is \(A\) diagonalizable?
\(\chi_A(x) = \begin{vmatrix} x & -1 & 0 \\ 0 & x & -1 \\ 6 & -11 & x - 6 \end{vmatrix} = x^3 - 6x^2 + 11x - 6=(x-1)(x-2)(x-3)\), \(A \text{ is diagonalizable}.\)
\(E_{1}=\text{Ker}(A-I)=\ker \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 6 & -11 & 5 \end{bmatrix}=\text{span}\{(1,1,1)\}\)
\(E_{2}=\text{Ker}(A-2I)=\ker \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 6 & -11 & 4 \end{bmatrix}=\text{span}\{(\frac{1}{4},\frac{1}{2},1)\}\)
\(E_3 = \text{Ker}(A - 3I) = \text{Ker} \begin{bmatrix} -3 & 1 & 0 \\ 0 & -3 & 1 \\ 6 & -11 & 3 \end{bmatrix} = \text{span} \left\{ ( 1 , \frac{1}{3} ,0) \right\}\)
\(B=\{(1,1,1),(\frac{1}{4},\frac{1}{2},1),(\frac{1}{9},\frac{1}{3},1)\}\) then \(P= \begin{bmatrix} 1 & \frac{1}{4}& \frac{1}{9}\\ 1 & \frac{1}{2}& \frac{1}{3}\\ 1 & 1 & 1 \end{bmatrix},D= \begin{bmatrix} 1 & & \\ & 2 & \\ & & 3 \end{bmatrix}\), \(P^{-1} = \begin{bmatrix} 3 & -\frac{5}{2} & \frac{1}{2} \\ -12 & 16 & -4 \\ 9 & -\frac{27}{2} & \frac{9}{2} \end{bmatrix}\)
geometric \(\leq\) algebraic multiplicity (the degree of \((x-c)\))
Method tool: Use minimal polynomial with order to simplify the higher-order matrix#
Main idea: Consider \(\chi_A(x)\) and relate to \(\det A\)
