Answer to Question #160246 in Linear Algebra for Nikhil

Question

Check whether or not the matrix

A=[1 2 3]

[0 2 3]

[0 0 3]

Is diagonalisable. If it is find a matrix P so that P^-1AP is a diagonal matrix. If A is not diagonalisable, obtain its minimal polynomial.

Answer 1

Check whether or not the matrix

$A=\begin{pmatrix} 1 & 2 &3\\ 0 &2&3\\0 &0&3 \end{pmatrix}$

Is diagonalisable. If it is find a matrix P so that P^-1AP is a diagonal matrix. If A is not diagonalisable, obtain its minimal polynomial.

find the charateristics polynomial

$f(\lambda) =det(A-\lambda\Iota)=0$

$\begin{vmatrix} 1-\lambda & 2 &3\\ 0 &2-\lambda&3\\0 &0&3-\lambda \end{vmatrix} = (1-\lambda)[(2-\lambda)(3-\lambda)]=0$

$\lambda=1 , \lambda-2, \lambda=3$

since we have three eigen values, then the matrix is diagonalisable

when $\lambda=1$

$\begin{pmatrix} 0 & 2 &3\\ 0 &1&3\\0 &0&2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\\x_3 \end{pmatrix}=0$

$x_2 +3x_3=0......(i)$

$x_2+3x_3=0.......(ii)$

$x_3=0$

$\implies x_1=x_2=x_3=0$

when $\lambda=2$

$\begin{pmatrix} -1 & 2 &3\\ 0 &0&3\\0 &0&1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\\x_3 \end{pmatrix}=0$

$-x_1+2x_2+3x_3=0$

$x_3 =0$

since $x_3=0\implies 2x_2=x_1$

since there is no restriction placed on x₁

x₁=r(arbitrary) where can be any number

let r=2

then x₁ = 2

x₂ = 1

when $\lambda=3$

$\begin{pmatrix} -2 & 2 &3\\ 0 &-1&3\\0 &0&0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\\x_3 \end{pmatrix}=0$

bringing out the augmented matrix

$\begin{pmatrix} -2 & 2 &3&0\\ 0 &-1&3&0\\0 &0&0&0 \end{pmatrix}R_1\mapsto R_1-R_2$ $\begin{pmatrix} -2 &3&0&0\\ 0 &-1&3&0\\0 &0&0&0 \end{pmatrix}$

$\therefore -2x_1+3x_2=0\implies x_1={3 \over 2}x_2$

$-1x_2 +4x_3=0\implies x_2 =4x_3$

since there is no restriction placed on x₁

x₂=s (arbitrary) where can be any number

let s=2

then x₁ = 3

x₃ = ${1 \over 2}$

Matrix P =$\begin{pmatrix} 0 & 2 &3\\ 0 & 1&2 \\0&0&{1 \over 2} \end{pmatrix}$