Find the eigenvalues and corresponding vectors of the following matrix, and diagonalise if possible. Find the matrix which diagonolise it.  A= 1 -1 2 2 -2 4 3 -3 6 

We are given a matrix \( A \) and tasked with finding its eigenvalues, eigenvectors, and whether it can be diagonalized. If it can be diagonalized, we will also find the matrix that diagonalizes it.

The matrix \( A \) is:

\[
A = \begin{pmatrix}
1 & -1 & 2 \\
2 & -2 & 4 \\
3 & -3 & 6
\end{pmatrix}
\]

### Step 1: Find Eigenvalues
The eigenvalues \( \lambda \) are found by solving the characteristic equation:

\[
\det(A - \lambda I) = 0
\]

Where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue. Let’s compute the determinant of \( A - \lambda I \).

\[
A - \lambda I = \begin{pmatrix}
1 - \lambda & -1 & 2 \\
2 & -2 - \lambda & 4 \\
3 & -3 & 6 - \lambda
\end{pmatrix}
\]

We will now compute the determinant:

\[
\det(A - \lambda I) = \begin{vmatrix}
1 - \lambda & -1 & 2 \\
2 & -2 - \lambda & 4 \\
3 & -3 & 6 - \lambda
\end{vmatrix}
\]

Let's calculate this determinant.The characteristic polynomial is:

\[
\det(A - \lambda I) = -\lambda^3 + 5\lambda^2 = \lambda^2(\lambda - 5)
\]

### Step 2: Find the Eigenvalues
The characteristic equation is:

\[
\lambda^2(\lambda - 5) = 0
\]

Thus, the eigenvalues are:

\[
\lambda_1 = 0 \quad \text{(with multiplicity 2)}, \quad \lambda_2 = 5
\]

### Step 3: Find the Eigenvectors
To find the eigenvectors corresponding to each eigenvalue, we solve \( (A - \lambda I) \mathbf{v} = 0 \) for each \( \lambda \).

#### Eigenvalue \( \lambda_1 = 0 \):
We solve \( A \mathbf{v} = 0 \).

The matrix \( A - 0I = A \) is:

\[
A = \begin{pmatrix}
1 & -1 & 2 \\
2 & -2 & 4 \\
3 & -3 & 6
\end{pmatrix}
\]

Let’s solve this system to find the eigenvectors corresponding to \( \lambda_1 = 0 \).

#### Eigenvalue \( \lambda_2 = 5 \):
We solve \( (A - 5I) \mathbf{v} = 0 \).

Let’s now find the eigenvectors for both eigenvalues.### Step 4: Eigenvectors

#### Eigenvalue \( \lambda_1 = 0 \):
The eigenvectors corresponding to \( \lambda_1 = 0 \) are:

\[
\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}
\]

#### Eigenvalue \( \lambda_2 = 5 \):
The eigenvector corresponding to \( \lambda_2 = 5 \) is:

\[
\mathbf{v}_3 = \begin{pmatrix} \frac{1}{3} \\ \frac{2}{3} \\ 1 \end{pmatrix}
\]

### Step 5: Diagonalization
To diagonalize \( A \), we construct matrices \( P \) and \( D \), where:

- \( P \) is the matrix with the eigenvectors as columns.
- \( D \) is the diagonal matrix with the corresponding eigenvalues on the diagonal.

The matrix \( P \) is:

\[
P = \begin{pmatrix}
1 & -2 & \frac{1}{3} \\
1 & 0 & \frac{2}{3} \\
0 & 1 & 1
\end{pmatrix}
\]

The diagonal matrix \( D \) is:

\[
D = \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 5
\end{pmatrix}
\]

Thus, \( A \) can be diagonalized as:

\[
A = P D P^{-1}
\]

Let me know if you'd like the inverse of \( P \) or any other details!

### Follow-up Questions:
1. How do you find the eigenvalues from the characteristic equation?
2. Why is the eigenvalue \( 0 \) repeated twice?
3. What is the geometric interpretation of the eigenvectors?
4. Can a matrix with zero eigenvalues be invertible?
5. What is the significance of the matrix \( P \) in diagonalization?

**Tip:** Eigenvalues and eigenvectors are critical for understanding how linear transformations behave along certain directions.

Find the eigenvalues and corresponding vectors of the following matrix, and diagonalise if possible. Find the matrix which diagonalises it. A= 1 -1 2 2 -2 4 3 -3 6

Discover how to find the eigenvalues and eigenvectors of matrix A = 1 -1 2, 2 -2 4, 3 -3 6, and learn how to diagonalize the matrix. This solution covers linear algebra concepts, including the characteristic equation, eigenvalue theorem, and diagonalization. Suitable for college-level students.

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