Find the eigenvalues and corresponding eigenvectors of the matrix
Find All the characteristic values and characteristic vectors of the matrix
{ \color {Blue} A= \begin {bmatrix}
1& 1 &1 \\
1& 1 &1 \\
1& 1 & 1
\end {bmatrix}}Solution
the characteristic equation of the matrix is
\left |A - \lambda I \right |=0
{\color {Blue} or \; \; \begin {vmatrix}
1 - \lambda & 1 & 1 \\
1 & 1 - \lambda & 1\\
1 & 1 & 1 - \lambda
\end {vmatrix}=0}{\color{Blue} or \;\; \begin{vmatrix}
3 - \lambda & 1 &1 \\
3 - \lambda & 1 - \lambda & 1\\
3 - \lambda & 1 & 1 - \lambda
\end{vmatrix}=0 }\;\; {\color{Red} on \; C_1 \rightarrow C_ 1 + C_ 2 + C_ 3}{\color {Blue} or \; \; \; (3 - \lambda )\begin{vmatrix}
1 & 1 & 1 \\
1 & 1 - \lambda & 1\\
1 & 1 & 1 - \lambda
\end {vmatrix}=0 }{\color {Blue} or \; \; \; (3 - \lambda )\begin{vmatrix}
1 & 1 & 1 \\
0 & - \lambda & 0 \\
0 & 0 & - \lambda
\end {vmatrix}=0 } \; \; \; {\color{Red} on \; \; R_ 2 \rightarrow R_ 2 - R_1\; , \; R_ 3 \rightarrow R_ 3 - R_1}or \; \; (3- \lambda) \lambda ^ 2 = 0 \\ \\ \therefore \lambda=0,0, \lambda=3
The characteristic values are 0, 0, 3
{\color{DarkBlue} Let \; \; \;X=\begin{bmatrix}
x_1\\ x_2
\\x_3
\end{bmatrix} \; \;be \; \;the \; \;characteristic \;vector \;corresponding \\ \;to \; the \; characteristic \;value \;{\color{DarkOrange} \lambda=0 } \;then}(A-0I)X=0
{\color {Blue} or \; \; \; \begin{vmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end {vmatrix}\begin{bmatrix}
x_1\\ x_2
\\x_3
\end{bmatrix}= \begin{bmatrix}
0\\ 0
\\0
\end{bmatrix}} {\color {Blue} or \; \; \; \begin{vmatrix}
1 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end {vmatrix}\begin{bmatrix}
x_1\\ x_2
\\x_3
\end{bmatrix}= \begin{bmatrix}
0\\ 0
\\0
\end{bmatrix}} {\color{Red}\; \; on \; R_ 2 \rightarrow R_ 2 - R_1 \; ,\; R_ 3 \rightarrow R_ 3 - R_1}The coefficient matrix is of rank 1 and equation have 3-1=2 linearly independent solutions.these equation reduced to a single equation
x_1+x_2+x_3=0
The \; solutions \; are\; {\color{DarkBlue} x_1=1\; , \; x_2=0 \; , \; x_3=-1} \;and \; {\color{DarkBlue} x_1=0,x_2=1,x_3=-1
}Hence \; \; X_1= \begin {bmatrix}
1 \\ 0
\\ -1
\end {bmatrix} \; and \; \; X_2= \begin{bmatrix}
0 \\ 1
\\ -1
\end {bmatrix}Are characteristics vectors corresponding to the characteristic value 0
For \; \lambda=3 \;,we \,have
(A-3I)X=0
{\color {Blue} or \; \; \; \begin{vmatrix}
-2 & 1 & 1 \\
1 & -2 & 1 \\
1 & 1 & -2
\end {vmatrix}\begin{bmatrix}
x_1\\ x_2
\\x_3
\end{bmatrix}= \begin{bmatrix}
0\\ 0
\\0
\end{bmatrix}} {\color {Blue} or \; \; \; \begin{vmatrix}
1 & 1 & -2 \\
1 & -2 & 1 \\
-2 & 1 & 1
\end {vmatrix}\begin{bmatrix}
x_1\\ x_2
\\x_3
\end{bmatrix}= \begin{bmatrix}
0\\ 0
\\0
\end{bmatrix}} {\color{Red}\; \; on \; R_ 1 \leftrightarrow R_ 3}{\color {Blue} or \; \; \; \begin{vmatrix}
1 & 1 & -2 \\
0 & -3 & 3 \\
0 & 0 & 0
\end {vmatrix}\begin{bmatrix}
x_1\\ x_2
\\x_3
\end{bmatrix}= \begin{bmatrix}
0\\ 0
\\0
\end{bmatrix}} {\color{Red}\; \; on \; R_ 2 \rightarrow R_ 2 - R_1 \; ,\; R_ 3 \rightarrow R_ 3 + R_2 +R_1}Rank of the coefficient matrix is 2 and so these equation have 3-2=1 independent solution corresponding to the value λ=3
These \; Equations \;reduces \;to \\ \\ x_1+x_2+x_3=0 \\ -3x_2+3x_3=0
{\color{Teal} Solving \; these \; equations \; we \; get \ x_1=x_2=x_3=k
} \\ \\ \therefore X=\begin{bmatrix}
k\\ k
\\ k
\end{bmatrix}, where \;k\ne0 X=\begin{bmatrix}
1\\ 1
\\ 1
\end{bmatrix}, is \; the \; characteristic\; vector \; on \; taking \; k =1Hence characteristic vectors are
\begin {bmatrix}
1 \\ 0
\\ -1
\end {bmatrix}, \begin{bmatrix}
0 \\ 1
\\ -1
\end {bmatrix},\begin{bmatrix}
1\\ 1
\\ 1
\end{bmatrix}Keywords
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