This solution involves finding matrices $D$ and $P$ for the matrix $A$, such that $A = P D P^{-1}$, where $A$ is given and has eigenvalues $\lambda_1 = 2$, $\lambda_2 = 4$, and $\lambda_3 = 6$.

The steps in the solution involve:

1. Finding Eigenvectors for Each Eigenvalue:

   • For $\lambda_1 = 2$, the matrix $(A - 2I)$ is created and row-reduced to find the eigenvector associated with $\lambda_1$.
   • For $\lambda_2 = 4$, the matrix $(A - 4I)$ is created, row-reduced, and its eigenvector is found.
   • For $\lambda_3 = 6$, the matrix $(A - 6I)$ is similarly handled to determine its eigenvector.

2. Constructing $P$ and $D$:

   • $P$ is formed by placing each eigenvector as columns, corresponding to $\lambda_1$, $\lambda_2$, and $\lambda_3$.
   • $D$ is constructed as a diagonal matrix with eigenvalues $\lambda_1$, $\lambda_2$, and $\lambda_3$ on the diagonal.

The final matrices obtained are:

• $P = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{bmatrix}$
• $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Would you like a detailed breakdown of each step or have specific questions?

Related Questions

1. How can I verify that $A = P D P^{-1}$ holds for these matrices?
2. What are some properties of diagonalizable matrices?
3. How are eigenvalues and eigenvectors calculated for larger matrices?
4. How does the choice of basis affect the diagonalization process?
5. Can all matrices be diagonalized, and why or why not?

Tip: When constructing $P$ from eigenvectors, ensure they are linearly independent for $P$ to be invertible.