Let
"A=\\begin{bmatrix}\n 1 & 1 & 2 \\\\\n 1 & 2 & 1 \\\\\n 2 & 1 & 1\n\\end{bmatrix}"
"A-\\lambda I=\\begin{bmatrix}\n 1-\\lambda & 1 & 2 \\\\\n 1 & 2-\\lambda & 1 \\\\\n 2 & 1 & 1-\\lambda\n\\end{bmatrix}"
"det(A-\\lambda I)=\\begin{vmatrix}\n 1-\\lambda & 1 & 2 \\\\\n 1 & 2-\\lambda & 1 \\\\\n 2 & 1 & 1-\\lambda\n\\end{vmatrix}="
"= (1-\\lambda)\\begin{vmatrix}\n 2-\\lambda & 1 \\\\\n 1 & 1-\\lambda\n\\end{vmatrix}-(1)\\begin{vmatrix}\n 1 & 1 \\\\\n 2 & 1-\\lambda\n\\end{vmatrix}+(2)\\begin{vmatrix}\n 1 & 2-\\lambda \\\\\n 2 & 1\n\\end{vmatrix}="
"=(1-\\lambda)(2-2\\lambda-\\lambda+\\lambda^2-1)-(1-\\lambda-2)+"
"+2(1-4+2\\lambda)=1-3\\lambda+\\lambda^2-\\lambda+3\\lambda^2-\\lambda^3+5\\lambda-5="
"=-\\lambda^3+4\\lambda^2+\\lambda-4"This is a characteristic polynomial.
Solve "det(A-\\lambda I)=0"
"-\\lambda^3+4\\lambda^2+\\lambda-4=0""-\\lambda^2(\\lambda-4)+(\\lambda-4)=0""(\\lambda-4)(1-\\lambda)(1+\\lambda)=0"The roots are: "\\lambda_1=4,\\lambda_2=1,\\lambda_3=-1."
These are the eigenvalues.
Find the eigenvectors.
"\\lambda=4"
"\\begin{bmatrix}\n 1-\\lambda & 1 & 2 \\\\\n 1 & 2-\\lambda & 1 \\\\\n 2 & 1 & 1-\\lambda\n\\end{bmatrix}=\\begin{bmatrix}\n -3 & 1 & 2 \\\\\n 1 & -2 & 1 \\\\\n 2 & 1 & -3\n\\end{bmatrix}" Perform row operations to obtain the rref of the matrix:
"R_2=R_2+(1\/3)R_1"
"\\begin{bmatrix}\n -3 & 1 & 2 \\\\\n 0 & -5\/3 & 5\/3 \\\\\n 2 & 1 & -3\n\\end{bmatrix}" "R_3=R_3+(2\/3)R_1"
"\\begin{bmatrix}\n -3 & 1 & 2 \\\\\n 0 & -5\/3 & 5\/3 \\\\\n 0 & 5\/3 & -5\/3\n\\end{bmatrix}" "R_3=R_3+R_2"
"\\begin{bmatrix}\n -3 & 1 & 2 \\\\\n 0 & -5\/3 & 5\/3 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_2=(-3\/5)R_2"
"\\begin{bmatrix}\n -3 & 1 & 2 \\\\\n 0 & 1 & -1 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_1=R_1-R_2"
"\\begin{bmatrix}\n -3 & 0 & 3 \\\\\n 0 & 1 & -1 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_1=(-1\/3)R_1"
"\\begin{bmatrix}\n 1 & 0 & -1 \\\\\n 0 & 1 & -1 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" Now, solve the matrix equation
"\\begin{bmatrix}\n 1 & 0 & -1 \\\\\n 0 & 1 & -1 \\\\\n 0 & 0 & 0\n\\end{bmatrix}\\begin{bmatrix}\n v_1 \\\\\n v_2 \\\\\n v_3\n\\end{bmatrix}=\\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0\n\\end{bmatrix}"If we take "v_3=t," then "v_1=t, v_2=t,v_3=t."
Therefore
"\\mathbf {v}=\\begin{bmatrix}\n t \\\\\n t \\\\\n t\n\\end{bmatrix}=\\begin{bmatrix}\n 1 \\\\\n 1 \\\\\n 1\n\\end{bmatrix}t" "\\lambda=1"
"\\begin{bmatrix}\n 1-\\lambda & 1 & 2 \\\\\n 1 & 2-\\lambda & 1 \\\\\n 2 & 1 & 1-\\lambda\n\\end{bmatrix}=\\begin{bmatrix}\n 0 & 1 & 2 \\\\\n 1 & 1 & 1 \\\\\n 2 & 1 & 0\n\\end{bmatrix}"Perform row operations to obtain the rref of the matrix:
Swap rows 1 and 2
"\\begin{bmatrix}\n 1 & 1 & 1 \\\\\n 0 & 1 & 2 \\\\\n 2 & 1 & 0\n\\end{bmatrix}" "R_3=R_3-(2)R_1"
"\\begin{bmatrix}\n 1 & 1 & 1 \\\\\n 0 & 1 & 2 \\\\\n 0 & -1 & -2\n\\end{bmatrix}" "R_3=R_3+R_2"
"\\begin{bmatrix}\n 1 & 1 & 1 \\\\\n 0 & 1 & 2 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_1=R_1-R_2"
"\\begin{bmatrix}\n 1 & 0 & -1 \\\\\n 0 & 1 & 2 \\\\\n 0 & 0 & 0\n\\end{bmatrix}"Now, solve the matrix equation
"\\begin{bmatrix}\n 1 & 0 & -1 \\\\\n 0 & 1 & 2 \\\\\n 0 & 0 & 0\n\\end{bmatrix}\\begin{bmatrix}\n u_1 \\\\\n u_2 \\\\\n u_3\n\\end{bmatrix}=\\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0\n\\end{bmatrix}" If we take "u_3=t," then "u_1=t, u_2=-2t,u_3=t."
Therefore
"\\mathbf {u}=\\begin{bmatrix}\n t \\\\\n -2t \\\\\n t\n\\end{bmatrix}=\\begin{bmatrix}\n 1 \\\\\n -2 \\\\\n 1\n\\end{bmatrix}t" "\\lambda=-1"
"\\begin{bmatrix}\n 1-\\lambda & 1 & 2 \\\\\n 1 & 2-\\lambda & 1 \\\\\n 2 & 1 & 1-\\lambda\n\\end{bmatrix}=\\begin{bmatrix}\n 2 & 1 & 2 \\\\\n 1 & 3 & 1 \\\\\n 2 & 1 & 2\n\\end{bmatrix}"Perform row operations to obtain the rref of the matrix:
"R_3=R_3-R_1"
"\\begin{bmatrix}\n 2 & 1 & 2 \\\\\n 1 & 3 & 1 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_2=R_2-(1\/2)R_1"
"\\begin{bmatrix}\n 2 & 1 & 2 \\\\\n 0 & 5\/2 & 0 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_2=(2\/5)R_2"
"\\begin{bmatrix}\n 2 & 1 & 2 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_1=R_1-R_2"
"\\begin{bmatrix}\n 2 & 0 & 2 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" "R_1=(1\/2)R_1"
"\\begin{bmatrix}\n 1 & 0 & 1 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & 0\n\\end{bmatrix}" Now, solve the matrix equation
"\\begin{bmatrix}\n 1 & 0 & 1 \\\\\n 0 & 1 & 0 \\\\\n 0 & 0 & 0\n\\end{bmatrix}\\begin{bmatrix}\n w_1 \\\\\n w_2 \\\\\n w_3\n\\end{bmatrix}=\\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0\n\\end{bmatrix}" If we take "w_3=t," then "w_1=-t, w_2=0, w_3=t."
Therefore
"\\mathbf {w}=\\begin{bmatrix}\n -t \\\\\n 0 \\\\\n t\n\\end{bmatrix}=\\begin{bmatrix}\n -1 \\\\\n 0 \\\\\n 1\n\\end{bmatrix}t"Eigen value: "4," eigenvector:"\\begin{bmatrix}\n 1\/\\sqrt{3} \\\\\n 1\/\\sqrt{3} \\\\\n 1\/\\sqrt{3}\n\\end{bmatrix}"
Eigen value: "1," eigenvector: "\\begin{bmatrix}\n 1\/\\sqrt{6} \\\\\n -2\/\\sqrt{6} \\\\\n 1\/\\sqrt{6}\n\\end{bmatrix}"
Figenvalue: "-1," eigenvector: "\\begin{bmatrix}\n -1\/\\sqrt{2} \\\\\n 0 \\\\\n 1\/\\sqrt{2}\n\\end{bmatrix}"
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