- Characteristic equation of A is $\mid A-\lambda I \mid = 0$
$\lambda^3-(8+7+3) \lambda^2 + [(56+21+24)-(36+4+16)]\lambda - \mid A \mid =0$
$\lambda^3-18\lambda^2+45\lambda-0=0$
$\lambda=0,3,15$ are eigen values of A
As all eigen values of A are distinct. A is diagonalisable.
Now to find eigen vector $X\neq0$ such that $[A-\lambda I]=0$
$ \left[ {\begin{array}{cc}
8-\lambda & -6 & 2 \\
-6 & 7-\lambda & -4 \\
2 & -4 & 3-\lambda
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$ (1)
Put $\lambda=0$ in (1)
$ \left[ {\begin{array}{cc}
8 & -6 & 2 \\
-6 & 7 & -4 \\
2 & -4 & 3
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
$8x-6y+2z=0$
$-6x+7y-4z=0$
By Cramer's rule,
$\frac{x}{\mid {\begin{array}{cc}
-6 & 2 \\
7 & -4 \\
\end{array} } \mid}$= $\frac{-y}{\mid {\begin{array}{cc}
8 & 2 \\
-6 & -4 \\
\end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc}
8 & -6 \\
-6 & 7 \\
\end{array} } \mid}$
$\frac{x}{10}=\frac{-y}{-20}=\frac{z}{20}$
$ X_1=$
$ \left[ {\begin{array}{cc}
1 \\
2 \\
2
\end{array} } \right]$
Put $\lambda=3$ in (1) we get,
$\left[ {\begin{array}{cc}
5 & -6 & 2 \\
-6 & 4 & -4 \\
2 & -4 & 0
\end{array} } \right]$$ \left[ {\begin{array}{cc}
x \
y \
z
\end{array} } \right]$=$ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
From 1st and 3rd row as they are distinct
$5x-6y+2z=0$
$2x-4y+0=0$
By Cramer's rule,
$\frac{x}{\mid {\begin{array}{cc}
-6 & 2 \\
-4 & 0 \\
\end{array} } \mid}$=$ \frac{x}{\mid {\begin{array}{cc}
5 & 2 \\
2 & 0 \\
\end{array} } \mid}$= $\frac{x}{\mid {\begin{array}{cc}
5 & -6 \\
2 & -4 \\
\end{array} } \mid}$
$\frac{x}{8}=\frac{-y}{-4}=\frac{z}{-8}$
$ X_2=$
$ \left[ {\begin{array}{cc}
2 \\
1 \\
-2
\end{array} } \right]$
Put $\lambda=15$ in (1), we get
$ \left[ {\begin{array}{cc}
-7 & -6 & 2\\
-6 & 8 & -4 \\
2 & -4 & -12
\end{array} } \right] $ $ \left[ {\begin{array}{cc}
x\\
y \\
z
\end{array} } \right] $= $ \left[ {\begin{array}{cc}
0\\
0 \\
0
\end{array} } \right]$
$-7x-6y+2z=0$
$-6x-8y-4z=0$
$\frac{x}{\mid {\begin{array}{cc}
-6 & 2 \\
-8 & -4\\
\end{array} } \mid}$ = $\frac{-y}{\mid {\begin{array}{cc}
-7 & 2 \\
-6 & -4 \\
\end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc}
-7 & -6 \\
-6 & -8 \\
\end{array} } \mid}$
$\frac{x}{2}=\frac{-y}{2}=\frac{z}{1} $
$ X_3=$
$ \left[ {\begin{array}{cc}
2\\
-2 \\
1
\end{array} } \right]$
Transforming matrix $M=$
$ \left[ {\begin{array}{cc}
1 & 2 & 2\\
2 & 1 & -2 \\
2 & -2 & 1
\end{array} } \right] $
Diagonal matrix $D=$
$ \left[ {\begin{array}{cc}
0 & 0 & 0\\
0 & 3 & 0 \\
0 & 0 & 15
\end{array} } \right] $
- Characteristic equation of A is $\mid A-\lambda I \mid = 0$
$\lambda^3-(-9+17+3) \lambda^2 + [(-27+21-63)-(-32-64+32)]\lambda - \mid A \mid =0$
$\lambda^3-\lambda^2-5\lambda-3=0$
$\lambda=3,-1, -1$ are eigen values of A
Now to find eigen vector $X\neq0$ such that $[A-\lambda I]=0$
$ \left[ {\begin{array}{cc}
9-\lambda & 4 & 4 \\
-8 & 3-\lambda & 4 \\
-16 & 8 & 7-\lambda
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$ (1)
Put $\lambda=3$ in (1)
$ \left[ {\begin{array}{cc}
-12 & 4 & 4 \\
-8 & 0 & 4 \\
-16 & 8 & 4
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$= $ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
$-8x-0y+4z=0$
$-16x+8y-4z=0$
By Cramer's rule,
$\frac{x}{\mid {\begin{array}{cc}
0 & 4 \\
8 & 4 \\
\end{array} } \mid}$= $\frac{-y}{\mid {\begin{array}{cc}
-8 & 4 \\
-16 & 4 \\
\end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc}
-8 & 0 \\
-16 & 8 \\
\end{array} } \mid}$
$\frac{x}{-1}=\frac{-y}{1}=\frac{z}{-2}$
$ X_1=$
$ \left[ {\begin{array}{cc}
-1 \\
-1 \\
-2
\end{array} } \right]$
Put $\lambda=-1$ in (1) we get,
$\left[ {\begin{array}{cc}
-8 & 4 & 4 \\
-8 & 4 & 4 \\
16 & 8 & 8
\end{array} } \right]$ $ \left[ {\begin{array}{cc}
x \\
y \\
z
\end{array} } \right]$=$ \left[ {\begin{array}{cc}
0 \\
0 \\
0
\end{array} } \right]$
From 1st and 3rd row as they are distinct
$-8x+4y+4z=0$
$-2x+y+z=0$
$x=0$, $x=1$.
$y=1$, $y=0$
$z=-1$, $z=2$
$ X_2=$
$ \left[ {\begin{array}{cc}
0 \\
1 \\
-1
\end{array} } \right]$
$ X_3=$
$ \left[ {\begin{array}{cc}
1\\
0 \\
2
\end{array} } \right]$
For $\lambda=1$ algebraic multiplicity=geometric multiplicity=2. Therefore, algebraic multiplicity=geometric multiplicity. Hence, A is diagonalisable.
Transforming matrix $M=$
$ \left[ {\begin{array}{cc}
-1 & 0 & 1\\
-1 & 1 & 0 \\
-2 & -1 & 2
\end{array} } \right] $
Diagonal matrix $D=$
$ \left[ {\begin{array}{cc}
3 & 0 & 0\\
0 & -1 & 0 \\
0 & 0 & -1
\end{array} } \right] $