Subject: Applied Mathematics 2

Topic: Matrices

Difficulty: High

Matrix A will be diagonalizable or it is similar to diagonal matrix if corresponding to each eigen value there are distinct eigen vectors i.e. geometric multiplicity of each eigen value must be equal to its algebraic multiplicity diagonalizable If matrix A is diagonalizable then D = M-1 A M ,D is the diagonal matrix whose diagonal elements are the eigen values of A & M is called transforming matrix whose columns are eigen vectors of A

Sol. To find eigen values of A ,characteristic eqation is $ |A-\lambda I| =0$

i.e. $ \left| {\begin{array}{cc} 2-\lambda & 3 & 4 \\ 0 & 2-\lambda & -1 \\ 0 & 0 & 1-\lambda \end{array} } \right|=0$ expanding we get a cubic equation in $ \lambda $

as $ \lambda^3 - 5\lambda^2 - 8\lambda - 4 = 0 i.e. \lambda = 1, 2, 2$

To find eigen vectors corresponding to each eigen value we write matrix equation

$[A-\lambda I]\times0$, where X is eigen vector (which is a column matrix)

Let X=$ \left[ {\begin{array}{cc} x_1\\ x_2\\ x_3 \end{array} } \right]$,

Therefore matrix equation can be written as

$ \left| {\begin{array}{cc} 2-\lambda & 3 & 4 \\ 0 & 2-\lambda & -1 \\ 0 & 0 & 1-\lambda \end{array} } \right|$ $ \left[ {\begin{array}{cc} x_1\\ x_2\\ x_3 \end{array} } \right]=0$..........(1)

{i}For$ \lambda= 1$ ,equation (1) becomes

$ \left[ {\begin{array}{cc} 1 & 3 & 4 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array} } \right]$ $ \left[ {\begin{array}{cc} x_1\\ x_2\\ x_3 \end{array} } \right]=0$

To find $x_1 ,x_2 , x_3$ expanding to different rows R1 & R2 from above $eq^n$ & applying cramer’s rule

,we get $ x_1 +3x_2 +4x_3 =0 \\ 0x_1 +x_2 - 1x_3 =0 $

By cramer’s rule $\frac {x_1}{ \left[ {\begin{array}{cc} 3 & 4 \\ 1 & -1 \end{array} } \right] } = \frac {-x_2}{ \left[ {\begin{array}{cc} 1 & 4 \\ 0 & -1 \end{array} } \right] } = \frac {x_3}{ \left[ {\begin{array}{cc} 1 & 3 \\ 0 & 1 \end{array} } \right] } $

or

$\frac {x_1}{-7}=\frac{-x_2}{-1}=\frac{x_3}{1}$

Therefore for $\lambda =1, X= \left[ {\begin{array}{cc} -7\\ 1\\ 1 \end{array} } \right]$

For $\lambda= 2$ ,equation (1) becomes

$ \left[ {\begin{array}{cc} 0 & 3 & 4 \\ 0 & 0 & -1 \\ 0 & 0 & -1 \end{array} } \right]$ $ \left[ {\begin{array}{cc} x_1\\ x_2\\ x_3 \end{array} } \right]=0$

To find $x_1 ,x_2 , x_3$ expanding to different rows R1 &R2 from above $eq^n $& applying cramer’s rule

,we get $0x_1 +3x_2 + 4x_3 =0 \\ 0x_1 +0x_2 - x_3 =0 $

By cramer’s rule $\frac {x_1}{ \left[ {\begin{array}{cc} 3 & 4 \\ 0 & -1 \end{array} } \right] } = \frac {-x_2}{ \left[ {\begin{array}{cc} 0 & 4 \\ 0 & -1 \end{array} } \right] } = \frac {x_3}{ \left[ {\begin{array}{cc} 0 & 3 \\ 0 & 0 \end{array} } \right] } $

or $\frac {x_1}{-7}=\frac{-x_2}{-1}=\frac{x_3}{1}$

Therefore for $ \lambda = 2 , X= \left[ {\begin{array}{cc} 3\\ 0\\ 0 \end{array} } \right]$

For repeated $\lambda = 2$ ,again we get the same matrix equation ,

therefore it has same eigen vector i.e. $X= \left[ {\begin{array}{cc} 3\\ 0\\ 0 \end{array} } \right]$

Since corresponding to each eigen value ,eigen vectors are not distinct or corresponding to each eigen value algebraic multiplicity is not equal to geometric multiplicity ,therefore above matrix is not diagonalizable