Answer to Question #201572 in Linear Algebra for Darshan

Question #201572

Show that the matrix [ 4 6 6 1 −1 3 2 −5 −2 ] is not diagonalizable


1
Expert's answer
2021-06-02T11:27:50-0400
"\\begin{pmatrix}\n 4& 6 & 6 \\\\\n 1 & -1 & 3 \\\\\n2 & -5 & -2 \\\\\n\\end{pmatrix}"

Start from forming a new matrix by subtracting "\\lambda"  from the diagonal entries of the given matrix:


"\\begin{pmatrix}\n 4-\\lambda & 6 & 6 \\\\\n 1 & -1-\\lambda & 3 \\\\\n2 & -5 & -2-\\lambda \\\\\n\\end{pmatrix}"

"\\begin{vmatrix}\n 4-\\lambda & 6 & 6 \\\\\n 1 & -1-\\lambda & 3 \\\\\n2 & -5 & -2-\\lambda \\\\\n\\end{vmatrix}"

"=(4-\\lambda)\\begin{vmatrix}\n -1-\\lambda & 3 \\\\\n -5 & -2-\\lambda\n\\end{vmatrix}-6\\begin{vmatrix}\n 1 & 3 \\\\\n 2 & -2-\\lambda\n\\end{vmatrix}"

"+6\\begin{vmatrix}\n 1 & -1-\\lambda \\\\\n 2 & -5\n\\end{vmatrix}"


"=(4-\\lambda)(1+\\lambda)(2+\\lambda()+15(4-\\lambda)"

"+6(2+\\lambda)+36-30+12(1+\\lambda)"

"=8+12\\lambda+4\\lambda^2-2\\lambda-3\\lambda^2-\\lambda^3+60-15\\lambda"

"+12+6\\lambda+6+12+12\\lambda"

"=-\\lambda^3+\\lambda^2+13\\lambda+98"

Solve the equation 


"-\\lambda^3+\\lambda^2+13\\lambda+98=0"

Only approximate roots can be found.

The roots are 

"\\lambda_1\\approx5.9513233156912"


"\\lambda_2\\approx\u22122.4756616578456+3.21527996418223i"


"\\lambda_3\\approx\u22122.4756616578456-3.21527996418223i"


Since there are complex roots, the matrix is not diagonalizable.




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