L ̊at A =

1 1 1
1 1 1
1 1 1

. Best ̈am alla egenv ̈arden och motsvarande egenrum
till A. Finns det en bas f ̈or R3 som best ̊ar av egenvektorer till A? Best ̈am, om m ̈ojligt
en matris P och en diagonal matris D s ̊a att D = P −1AP 

We are given the matrix \( A \):

\[
A = \begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix}
\]

We need to:
1. Determine the eigenvalues and corresponding eigenspaces of \( A \).
2. Check if there is a basis for \( \mathbb{R}^3 \) that consists of eigenvectors of \( A \).
3. If possible, find a matrix \( P \) and a diagonal matrix \( D \) such that \( D = P^{-1} A P \).

### Step 1: Finding the Eigenvalues
To find the eigenvalues, we solve the characteristic equation:

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

Where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue. The matrix \( A - \lambda I \) is:

\[
A - \lambda I = \begin{bmatrix}
1 - \lambda & 1 & 1 \\
1 & 1 - \lambda & 1 \\
1 & 1 & 1 - \lambda
\end{bmatrix}
\]

The determinant of this matrix is:

\[
\det(A - \lambda I) = \det\begin{bmatrix}
1 - \lambda & 1 & 1 \\
1 & 1 - \lambda & 1 \\
1 & 1 & 1 - \lambda
\end{bmatrix}
\]

Expanding this determinant, we get:

\[
(1 - \lambda) \left( (1 - \lambda)^2 - 1 \right) - 1 \left( (1 - \lambda) - 1 \right) + 1 \left( (1 - \lambda) - 1 \right)
\]

Simplifying this expression gives:

\[
(1 - \lambda)\left((1 - \lambda)^2 - 1 \right) = (1 - \lambda) \left( \lambda^2 - 2 \lambda \right) = \lambda (\lambda^2 - 2 \lambda)
\]

This simplifies to:

\[
\lambda(\lambda - 3) = 0
\]

Thus, the eigenvalues are \( \lambda = 0 \) and \( \lambda = 3 \).

### Step 2: Finding the Eigenspaces

#### Eigenvalue \( \lambda = 3 \):

We solve \( (A - 3I) \mathbf{v} = 0 \) for the eigenvectors:

\[
A - 3I = \begin{bmatrix}
-2 & 1 & 1 \\
1 & -2 & 1 \\
1 & 1 & -2
\end{bmatrix}
\]

Solving this system, we find that the eigenspace corresponding to \( \lambda = 3 \) is spanned by the vector:

\[
\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}
\]

#### Eigenvalue \( \lambda = 0 \):

We solve \( A \mathbf{v} = 0 \) for the eigenvectors. The matrix \( A \) becomes:

\[
A - 0I = A = \begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix}
\]

The null space of this matrix corresponds to the eigenspace for \( \lambda = 0 \). Solving this system, we find that the eigenspace corresponding to \( \lambda = 0 \) is spanned by two vectors:

\[
\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}
\]

### Step 3: Basis of Eigenvectors for \( \mathbb{R}^3 \)

Since we have three linearly independent eigenvectors \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \), these form a basis for \( \mathbb{R}^3 \).

### Step 4: Diagonalization

We can form the matrix \( P \) from the eigenvectors:

\[
P = \begin{bmatrix} 
1 & 1 & 0 \\
1 & -1 & 1 \\
1 & 0 & -1 
\end{bmatrix}
\]

The diagonal matrix \( D \) will have the eigenvalues on its diagonal:

\[
D = \begin{bmatrix}
3 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
\]

Finally, we can verify that:

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

This completes the diagonalization of the matrix \( A \).

Would you like further details or have any questions?

### 5 Related Questions:
1. How do you calculate the inverse of a matrix?
2. Can all matrices be diagonalized?
3. What does it mean for a matrix to be defective?
4. How does the Jordan canonical form differ from diagonalization?
5. What are some real-world applications of eigenvalue decomposition?

### Tip:
Diagonalization is useful in simplifying matrix functions, like computing powers or exponentials of matrices.

Låt A = [1 1 1; 1 1 1; 1 1 1]. Bestäm alla egenvärden och motsvarande egenrum till A. Finns det en bas för R3 som består av egenvektorer till A? Bestäm, om möjligt, en matris P och en diagonal matris D så att D = P −1AP.

Explore the eigenvalues and eigenvectors of a 3x3 matrix A, including the process of diagonalization. Learn how to find the matrix P and diagonal matrix D such that D = P^{-1}AP. Suitable for university students studying Linear Algebra.

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