T = x ( − 1 1 3 ) + y ( 0 − 1 2 ) + z ( 0 0 1 ) T=x\begin{pmatrix}
-1 \\
1\\
3
\end{pmatrix}+y\begin{pmatrix}
0 \\
-1\\
2
\end{pmatrix}+z\begin{pmatrix}
0 \\
0\\
1
\end{pmatrix} T = x ⎝ ⎛ − 1 1 3 ⎠ ⎞ + y ⎝ ⎛ 0 − 1 2 ⎠ ⎞ + z ⎝ ⎛ 0 0 1 ⎠ ⎞
T = ( − 1 0 0 1 − 1 0 3 2 1 ) T=\begin{pmatrix}
-1&0 & 0 \\
1&-1 & 0\\
3&2&1
\end{pmatrix} T = ⎝ ⎛ − 1 1 3 0 − 1 2 0 0 1 ⎠ ⎞
T − 1 = ( − 2 0 0 1 − 2 0 3 2 0 ) T-1=\begin{pmatrix}
-2&0& 0 \\
1&-2 & 0\\
3&2&0
\end{pmatrix} T − 1 = ⎝ ⎛ − 2 1 3 0 − 2 2 0 0 0 ⎠ ⎞
( T + 1 ) 2 = ( 0 0 0 1 0 0 3 2 2 ) 2 = ( 0 0 0 0 0 0 8 4 4 ) (T+1)^2=\begin{pmatrix}
0&0&0 \\
1&0&0\\
3&2&2
\end{pmatrix}^2=\begin{pmatrix}
0&0&0\\
0&0& 0\\
8&4&4
\end{pmatrix} ( T + 1 ) 2 = ⎝ ⎛ 0 1 3 0 0 2 0 0 2 ⎠ ⎞ 2 = ⎝ ⎛ 0 0 8 0 0 4 0 0 4 ⎠ ⎞
( T − 1 ) ( T + 1 ) 2 = ( − 2 0 0 1 − 2 0 3 2 0 ) ( 0 0 0 0 0 0 8 4 4 ) = ( 0 0 0 0 0 0 0 0 0 ) (T-1)(T+1)^2=\begin{pmatrix}
-2&0& 0 \\
1&-2 & 0\\
3&2&0
\end{pmatrix}\begin{pmatrix}
0&0&0 \\
0&0 & 0\\
8&4&4
\end{pmatrix}=\begin{pmatrix}
0&0& 0 \\
0&0& 0\\
0&0&0
\end{pmatrix} ( T − 1 ) ( T + 1 ) 2 = ⎝ ⎛ − 2 1 3 0 − 2 2 0 0 0 ⎠ ⎞ ⎝ ⎛ 0 0 8 0 0 4 0 0 4 ⎠ ⎞ = ⎝ ⎛ 0 0 0 0 0 0 0 0 0 ⎠ ⎞
T satisfies the polynomial ( x − 1 ) ( x + 1 ) 2 (x-1)(x+1)^2 ( x − 1 ) ( x + 1 ) 2
( T − 1 ) ( T + 1 ) = ( − 2 0 0 1 − 2 0 3 2 0 ) ( 0 0 0 1 0 0 3 2 2 ) = ( 0 0 0 − 2 0 0 2 0 0 ) ≠ ( 0 0 0 0 0 0 0 0 0 ) (T-1)(T+1)=\begin{pmatrix}
-2&0&0 \\
1&-2 & 0\\
3&2&0
\end{pmatrix}\begin{pmatrix}
0&0& 0\\
1&0&0 \\
3&2&2
\end{pmatrix}=\begin{pmatrix}
0&0 & 0\\
-2&0&0\\
2&0&0
\end{pmatrix}\ne\begin{pmatrix}
0&0 & 0 \\
0&0& 0\\
0&0&0
\end{pmatrix} ( T − 1 ) ( T + 1 ) = ⎝ ⎛ − 2 1 3 0 − 2 2 0 0 0 ⎠ ⎞ ⎝ ⎛ 0 1 3 0 0 2 0 0 2 ⎠ ⎞ = ⎝ ⎛ 0 − 2 2 0 0 0 0 0 0 ⎠ ⎞ = ⎝ ⎛ 0 0 0 0 0 0 0 0 0 ⎠ ⎞
∴ \therefore ∴ Minimal polynomial is ( x − 1 ) ( x + 1 ) 2 (x-1)(x+1)^2 ( x − 1 ) ( x + 1 ) 2
Comments