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Recitation 11B

Problem: Suppose MM is the matrix [3−24−262423]\begin{bmatrix}3&-2&4\\-2&6&2\\4&2&3\end{bmatrix}. Follow the steps to find the orthogonal diagonalization of MM
(a) Find the eigenvalues of MM and their correspondent eigenspaces.
(b) Find the orthonormal basis of each eigenspace.
(c) Find the orthogonal diagonalization of MM.

Solution:

(a) The characteristic equation is given by ∣3−λ−24−26−λ2423−λ∣=−(λ+2)(λ−7)2\begin{vmatrix}3-\lambda & -2 & 4 \\ -2 & 6-\lambda & 2 \\ 4 & 2 & 3-\lambda\end{vmatrix}=-(\lambda+2)(\lambda-7)^2 Therefore the eigenvalues are λ1=−2,λ2=7\lambda_1=-2, \lambda_2=7. The eigenspace associated to λ=−2\lambda=-2 is the solution set of [5−24−282425][x1x2x3]=[000]  ⟹  [x1x2x3]=x2[21−2],\begin{bmatrix}5 & -2 & 4 \\ -2 & 8 & 2 \\ 4 & 2 & 5\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}\implies \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}=x_2\begin{bmatrix}2 \\ 1 \\ -2\end{bmatrix}, that is, the subspace V1V_1 spanned by [21−2]\begin{bmatrix}2 \\ 1 \\ -2\end{bmatrix}. The eigenspace associated to λ=7\lambda=7 is the solution set of [−4−24−2−1242−4]  ⟹  [x1x2x3]=x1[1−20]+x3[021],\begin{bmatrix}-4 & -2 & 4 \\ -2 & -1 & 2 \\ 4 & 2 & -4\end{bmatrix}\implies \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}=x_1\begin{bmatrix}1 \\ -2 \\ 0\end{bmatrix}+x_3\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}, that is the subspace V2V_2 spanned by [1−20]\begin{bmatrix}1 \\ -2 \\ 0\end{bmatrix} and [021]\begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix}. Remark: The dimension of each eigenspace is equal to the algebraic multiplicity of each eigenvalue in the characteristic equation. The eigenspaces are mutually orthogonal.

(b) The orthonormal basis for V1V_1 is formed by [21−2]∼[2/31/3−2/3].\begin{bmatrix}2 \\ 1 \\ -2\end{bmatrix}\sim\begin{bmatrix}2/3 \\ 1/3 \\ -2/3\end{bmatrix}. The orthonormal basis for V2V_2 is formed by [1−20]∼[1/5−2/50],[021]−(−4/5)[1/5−2/50]=[4/5−2/51]∼[4/452/455/45].\begin{bmatrix}1 \\ -2 \\ 0\end{bmatrix}\sim \begin{bmatrix}1/\sqrt{5} \\ -2/\sqrt{5} \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 2 \\ 1\end{bmatrix} - (-4/\sqrt{5})\begin{bmatrix}1/\sqrt{5} \\ -2/\sqrt{5} \\ 0\end{bmatrix} = \begin{bmatrix}4/5 \\ -2/5 \\ 1\end{bmatrix}\sim \begin{bmatrix}4/\sqrt{45} \\ 2/\sqrt{45} \\ 5/\sqrt{45}\end{bmatrix}.

(c) Put P=[2/31/54/451/3−2/52/45−2/305/45]P=\begin{bmatrix}2/3 & 1/\sqrt{5} & 4/\sqrt{45}\\1/3 & -2/\sqrt{5} & 2/\sqrt{45}\\-2/3 & 0 & 5/\sqrt{45}\end{bmatrix} and D=[−200070007]D=\begin{bmatrix}-2 & 0 & 0\\ 0 & 7 & 0\\ 0 & 0 & 7\end{bmatrix}. Then M=PDPTM=PDP^T.