Answer to Question #101766 in Linear Algebra for Pindad

Question #101766
1. Find the Eigen values of

0 1 0
A= 0 0 1
216k^3 -108k^2 18k

lambda 1 = ?
lambda 2= ?
lambda 3= ?

2. Using the fact that the matrix

0 0 0 ... 0 -c subscript 0
1 0 0 ... 0 -c subscript 1
0 1 0 ... 0 -c subscript 2
... ... ... ... ... ...
0 0 0 ... 1 -c subscript n-1

has the characteristic polynomial

p(λ) = c0 + c1λ + ⋯ + cn − 1λn − 1 + λn

find a matrix with the characteristic polynomial

p(λ) = 1 − 5λ + λ2 + 6λ3 + λ4
1
Expert's answer
2020-01-29T14:26:37-0500

1.

"\u2223 \u0410-\u03bb E \u2223=\n\\begin{vmatrix}\n -\u03bb & 1&0 \\\\\n 0&-\u03bb& 1\\\\\n216k^3&-108k^2&18k-\u03bb\n\\end{vmatrix}=0\\\\"

"\u03bb^2 (18\u043a-\u03bb )+2 1 6 \u043a^ 3 +0-0-0-1 0 8 \u043a^2 \u03bb=0"

"-\u03bb ^3 +1 8 k \u03bb^2-1 0 8 \u043a^2 \u03bb+2 1 6 \u043a^3 =0"

"\u03bb ^3 -1 8 k \u03bb^2 +1 0 8 \u043a ^2 \u03bb-2 1 6 \u043a ^3 =0"

"( \u03bb ^3 -2 1 6 \u043a ^3 )-( 1 8 k \u03bb ^2 -1 0 8 \u043a ^2 \u03bb)=0"

"( \u03bb-6 \u043a ) ( \u03bb ^2 +6 k \u03bb+3 6 \u043a ^2 )-1 8 k \u03bb ( \u03bb-6 \u043a )=0"

"( \u03bb-6 \u043a ) ( \u03bb ^2 -1 2 k \u03bb+3 6 \u043a ^2 )=0"

"\u03bb-6 \u043a=0 ,( \u03bb-6 \u043a )^2 =0"

"\u03bb _1 =6 \u043a ,\\\\\n\u03bb _2 =6 \u043a ,\\\\\n\u03bb _3 =6 \u043a"


2.

the required matrix with the characteristic polynomial

"p ( \u03bb )=1-5 \u03bb+\u03bb ^2 +6 \u03bb ^3 +\u03bb ^4"

in our case

"\\begin{pmatrix}\n 0 &0&0&-c_0 \\\\\n 1&0&0&-c_1\\\\\n0&1&0&-c_2\\\\\n0&0&1&-c_3\n\\end{pmatrix}"

in our case

"c _0 \u200b=1,c _1 =\u22125,c _2 =1,c _3 =6"

there

"\\begin{pmatrix}\n 0 &0&0&-1 \\\\\n 1&0&0&5\\\\\n0&1&0&-1\\\\\n0&0&1&-6\n\\end{pmatrix}"


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