The matrix $A=\left[\begin{array}{lll}2 & 2 & -6 \\ 2 & 2 & -6 \\ 2 & 2 & -6\end{array}\right]$ has two real eigenvalues, one of multiplicity 1 and one of multiplicity 2 . Find the eigenvalues and a basis of each eigenspace. \[ \begin{array}{l} \lambda_{1}=\text { has multiplicity } 1, \text { with a basis of } \\ \lambda_{2}=\quad \text { has multiplicity } 2, \text { with a basis of } \end{array} \]

Step 1 ：First, we need to find the eigenvalues by solving the characteristic equation, which is given by \(\text{det}(A - \lambda I) = 0\), where I is the identity matrix and \(\lambda\) are the eigenvalues.

Step 2 ：The matrix \(A - \lambda I\) is: \[\begin{array}{ccc} 2 - \lambda & 2 & -6 \\ 2 & 2 - \lambda & -6 \\ 2 & 2 & -6 - \lambda \end{array}\]

Step 3 ：Setting the determinant equal to zero gives us the characteristic equation: \((2 - \lambda)[(2 - \lambda)(-6 - \lambda) - (-12)] - 2[2(-6 - \lambda) - (-12)] + 6[4 - (2 - \lambda)] = 0\)

Step 4 ：Solving this equation gives us the eigenvalues. Solving the characteristic equation, we get: \(\lambda^3 - 8\lambda^2 + 16\lambda = 0\)

Step 5 ：Factoring out \(\lambda\), we get: \(\lambda(\lambda - 4)^2 = 0\)

Step 6 ：\(\boxed{\text{So, the eigenvalues are } \lambda_1 = 0 \text{ (with multiplicity 1) and } \lambda_2 = 4 \text{ (with multiplicity 2)}}\)

Step 7 ：Now, to find the eigenvectors, we substitute each eigenvalue back into the equation \((A - \lambda I)v = 0\) and solve for v.

Step 8 ：For \(\lambda_1 = 0\), we have: \[\begin{array}{ccc} 2 & 2 & -6 \\ 2 & 2 & -6 \\ 2 & 2 & -6 \end{array}\]v = 0

Step 9 ：This gives us the system of equations: \[\begin{array}{l} 2x + 2y - 6z = 0 \\ 2x + 2y - 6z = 0 \\ 2x + 2y - 6z = 0 \end{array}\]

Step 10 ：From this, we can see that any vector of the form \((x, y, z)\) where \(x + y = 3z\) is an eigenvector. A basis for this eigenspace is \{(1, 2, 1)\}.

Step 11 ：For \(\lambda_2 = 4\), we have: \[\begin{array}{ccc} -2 & 2 & -6 \\ 2 & -2 & -6 \\ 2 & 2 & -2 \end{array}\]v = 0

Step 12 ：This gives us the system of equations: \[\begin{array}{l} -2x + 2y - 6z = 0 \\ 2x - 2y - 6z = 0 \\ 2x + 2y - 2z = 0 \end{array}\]

Step 13 ：From this, we can see that any vector of the form \((x, y, z)\) where \(x = y\) and \(z = 0\) is an eigenvector. A basis for this eigenspace is \{(1, 1, 0)\}.

Step 14 ：\(\boxed{\text{So, the eigenvalues are } \lambda_1 = 0 \text{ with a basis of } \{(1, 2, 1)\} \text{ and } \lambda_2 = 4 \text{ with a basis of } \{(1, 1, 0)\}}\)