Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : MAY 2014

Let $λ$ be Eigen value and x be corresponding Eigen vector of matrix A. $$\text {Characteristic Equation is } |A- λI|=0$$

$$\therefore \begin{vmatrix} -9-\lambda & 4 & 4 \\ -8 & 3-\lambda & 4 \\ -16 & 8 & 7-\lambda \\ \end{vmatrix}=0$$

On solving we get

$$\lambda^3-(-9+3+7)\lambda^2+(-11-1+5)\lambda-3=0$$

$\therefore \lambda^3-\lambda^2-5\lambda -3=0\\ \therefore \text { Eigen value $(\lambda)$ are } 3,-1,1$

Case 1: $\lambda=-1\\ \therefore [A-\lambda I]x=0$

$$\therefore \begin{bmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}$$

$$R_2-R_1 ;R_3-2R_1;R_1/4;$$

$$\therefore \begin{bmatrix} -2 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}.... (1)$$

Number of unknown $(n)=3$

Rank (r) = Number of non-zero rows=1

Here, Algebraic multiplicity (A.M.)= No. of times $(x=-1)$ is repeated=2

Geometric multiplicity $(G.M.)=n-r=3-1=2$

$\therefore$ AM=GM for $\lambda=-1$

Expanding (1)

$$-2X_1+x_2+x_3=0..... (2)$$

Put $x_1=0;x_2=1$

$$\therefore 0+1+x_3=0$$

$$\therefore x_3=-1$$

$\therefore $ Eigen Vector $x_1=[0\space \space 1-1]$

If K is non- zero scalar then $KX_1$ is also an Eigen vector

Put $x_1=1;x_2=0$ in $(2)$

$$\therefore -2+0+x_3=0$$

$$\therefore x_3=2$$

$\therefore $ Eigen over $x_2=[1\space \space 0\space \space 2]$

If K is non-zero scalar then $KX_2$ is also an Eigen Vector.

Case 2: $\lambda=3\\ \therefore [A-\lambda I]X=0$

$$\therefore \begin{bmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 4 & 4 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}$$

$$R_1-R_2 ;R_3-R_2;R_{2/4};$$

$$\therefore \begin{bmatrix} -4 & 4 & 0 \\ -2 & 0 & 1 \\ -8 & 8 & 0 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}$$

$$R_3-2R_1; R_{1/4}$$

$$\therefore \begin{bmatrix} -1 & 1 & 0 \\ -2 & 0 & 1 \\ 2 & 0 & 0 \\ \end{bmatrix}.\begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=\begin{bmatrix}0\\ 0\\0\end{bmatrix}.... (3)$$

Here , $n=3$ and $r=1$

A.M.= No. of times $λ =3$ is repeated =1

G.M.= $n-r=3-2=1$

$\therefore$ A.M=G.M for $\lambda =3$

Expanding (3),

$$-2x_1+x_3=0.... (4)$$

$$-x_1+x_2=0.....(5)$$

Put $x_1=1$

From (4) $-2(1)+x_3=0\\ \therefore x_3=2$

From (5) $-1+x_2=0\\ \therefore x_2=1$

$\therefore $ Eigen vector $x_3=[1\space \space 1\space \space 2]$

If K non-zero scalar then $KX_3$ is also an Eigen vector.

Since, A.M. = G.M. for all Eigen values, matrix A is diagonalizable.

$$\therefore M^{-1}AM=D$$

So, the given matrix a is diagonalizable to diagonal matrix D by the transforming matrix M, where

$$D= \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix}$$ and $$M= \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ -1 & 2 & 2 \\ \end{bmatrix}$$