DATA605 Linear Algebra # Alain Kuiete
For the matrix A = \(\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}\)
The eigenvalues are 4 and 1
The matrix I - xA = \(\begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{bmatrix}\)
Now we put into row echelon form. By applying 2row2 + row1 and 2row3 + row1,
we have: \(\begin{bmatrix} 2 & -1 & -1 \\ 0 & 3 & -3 \\ 0 & -3 & 3 \end{bmatrix}\)
Next by applying row3 + row2 we have: \(\begin{bmatrix} 2 & -1 & -1 \\ 0 & 3 & -3 \\ 0 & 0 & 0 \end{bmatrix}\)
To obtain the eigenvectors V(v1, v2, v3)
\(\begin{bmatrix} 2 & -1 & -1 \\ 0 & 3 & -3 \\ 0 & 0 & 0 \end{bmatrix}\quad \begin{bmatrix} v1 \\ v2 \\ v3 \end{bmatrix}\quad =\quad \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
It gives the system of equation \(\begin{cases} 2v1\quad -\quad v2\quad -\quad v3\quad =\quad 0 \\ 3v2\quad -\quad v3\quad =\quad 0 \end{cases}\Longrightarrow \quad v1\quad =\quad v2\quad =\quad v3\quad =\quad t\)
The eigenspace for eigenvector 4 is \(\begin{Bmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} & t,\quad t\quad \epsilon \quad R \end{Bmatrix}\)
For the eigenvalue 1 , xI - A matrix is: I - xA = \(\begin{bmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{bmatrix}\)
We can multiply each row by -1 and obtain I - xA = \(\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}\)
We apply row2 - row1 and row3 -row1 have the row echelon form
\(\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
With the vector V(v1, v2, v3), we have the only eqution v1 + V2 + v3 = 0 or V1 = -v2 -v3
Thus, the eigenspace of the eigenvalue 1 is \(\begin{Bmatrix} \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} & s\quad +\quad \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}t,\quad s,t\quad \epsilon \quad R \end{Bmatrix}\)