Answer to Question #319286 in Linear Algebra for raymundo

Question #319286

        [ 1 0 -1

3. Consider the matrix A =  0 3 0

                      -1 0 1 ]


  1. Find the eigenvalues of A.
  2. Find the eigenspaces corresponding to each eigenvalue from A.
1
Expert's answer
2022-03-28T17:46:31-0400

"A=\\begin{pmatrix}\n 1& 0&-1 \\\\\n 0 & 3&0\\\\-1&0&1\n\\end{pmatrix}"


"|\u03bbI-A|=\\begin{vmatrix}\n \u03bb-1& 0&1 \\\\\n 0 & \u03bb-3&0\\\\1&0&\u03bb-1\n\\end{vmatrix} = 0"


"=>|\u03bbI-A|=(\u03bb-3)[(\u03bb-1)\u00b2-1)]=0"


"=>|\u03bbI-A|=(\u03bb-3)(\u03bb\u00b2-2\u03bb)=\u03bb(\u03bb-3)(\u03bb-2)=0"


"\u2234 \u03bb=0,3,2"



when "\u03bb=0"


"MX=\\begin{pmatrix}\n -1& 0&1 \\\\\n 0 & -3&0\\\\1&0&-1\n\\end{pmatrix} \\begin{pmatrix}\n x_{1} \\\\\n x_{2}\\\\x_{3}\n\\end{pmatrix} =\\begin{pmatrix}\n 0 \\\\\n 0\\\\0\n\\end{pmatrix}"


"=>\\begin{pmatrix}\n -1& 0&1 \\\\\n 0 & -3&0\\\\1&0&-1\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&-1 \\\\0&-3&0\n \\\\-1&0&1\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&-1 \\\\\n 0&-3&0\\\\0&0&0\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&-1 \\\\\n 0&1&0\\\\0&0&0\n\\end{pmatrix}"


"=>x_{1}-x_{3}=0" "x_{2}=0"

Let "x_{3}=t"


"x_{1}=x_{3}=t," "x_{2}=0"


The eigenspace corresponding to the eigenvalue λ=0 is

"\\{\\begin{pmatrix}\n x_{1} \\\\\n x_{2}\\\\x_{3}\n\\end{pmatrix}\u2208\u211d\u00b3 : x_{1}=x\u2083, x\u2082=0\\}"


when "\u03bb=2"


"MX=\\begin{pmatrix}\n 1& 0&1 \\\\\n 0 & -1&0\\\\1&0&1\n\\end{pmatrix} \\begin{pmatrix}\n x_{1} \\\\\n x_{2}\\\\x_{3}\n\\end{pmatrix} =\\begin{pmatrix}\n 0 \\\\\n 0\\\\0\n\\end{pmatrix}"


"=>\\begin{pmatrix}\n 1& 0&1 \\\\\n 0 & -1&0\\\\1&0&1\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&1 \\\\0&-1&0\n \\\\0&0&0\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&1 \\\\\n 0&1&0\\\\0&0&0\n\\end{pmatrix}"


"=>x_{1}+x_{3}=0" , "x_{2}=0"


The eigenspace corresponding to the eigenvalue λ=2 is


"\\{\\begin{pmatrix}\n x_{1} \\\\\n x_{2}\\\\x_{3}\n\\end{pmatrix}\u2208\u211d\u00b3 : x_{1}+x\u2083=x\u2082=0\\}"



when "\u03bb=3"

"MX=\\begin{pmatrix}\n 2& 0&1 \\\\\n 0 & 0&0\\\\1&0&2\n\\end{pmatrix} \\begin{pmatrix}\n x_{1} \\\\\n x_{2}\\\\x_{3}\n\\end{pmatrix} =\\begin{pmatrix}\n 0 \\\\\n 0\\\\0\n\\end{pmatrix}"


"=>\\begin{pmatrix}\n 2& 0&1 \\\\\n 0 & 0&0\\\\1&0&2\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&2 \\\\2&0&1\n \\\\0&0&0\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&1 \\\\\n 0&0&-3\\\\0&0&0\n\\end{pmatrix}\\backsim\\begin{pmatrix}\n 1&0&1 \\\\\n 0&0&1\\\\0&0&0\n\\end{pmatrix}"


"=>x_{1}+x_{3}=0" , "x_{3}=0"


"\u2234x_{1}=0"


The eigenspace corresponding to the eigenvalue λ=3 is


"\\{\\begin{pmatrix}\n x_{1} \\\\\n x_{2}\\\\x_{3}\n\\end{pmatrix}\u2208\u211d\u00b3 : x_{1}=0=x\u2083\\}"




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