5- (a) [Byron-Fuller 3.16] Show that the eigenvalues of $$ M=\left[\begin{array}{ccc}3 & 5 & 8 \ -6 & -10 & -16 \ 4 & 7 & 11\end{array}\right] $$ are $$ \lambda=0,1,3 $$, and that the corresponding eigenvectors are $$ \left[\begin{array}{c}1 \ 1 \ -1\end{array}\right], \quad\left[\begin{array}{c}1 \ -2 \ 1\end{array}\right], \quad \text { and } \quad\left[\begin{array}{c}4 \ -8 \ 5\end{array}\right] . $$ Construct a diagonalizing matrix $$ P $$, prove that its inverse exists, compute its inverse, and verify that $$ P^{-1} M P $$ is diagonal with the eigenvalues on the diagonal. Note that det $$ M=0 $$. Is it true that any diagonalizable matrix with an eigenvalue equal to zero is singular? (b) What are the kernel and image of $$ M $$ ?

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To solve the problem, we will go through the steps outlined in part (a) and part (b).

### Part (a)

1. **Finding the Eigenvalues**:
   We need to find the eigenvalues of the matrix $$ M $$:
   $$
   M = \begin{bmatrix} 3 & 5 & 8 \ -6 & -10 & -16 \ 4 & 7 & 11 \end{bmatrix}
   $$
   The eigenvalues are found by solving the characteristic polynomial given by $$ \det(M - \lambda I) = 0 $$, where $$ I $$ is the identity matrix.

The characteristic polynomial is:
   $$
   \det\left(\begin{bmatrix} 3 - \lambda & 5 & 8 \ -6 & -10 - \lambda & -16 \ 4 & 7 & 11 - \lambda \end{bmatrix}\right) = 0
   $$

After calculating the determinant, we find that the eigenvalues are $$ \lambda = 0, 1, 3 $$.

2. **Finding the Eigenvectors**:
   For each eigenvalue, we solve $$ (M - \lambda I) \mathbf{v} = 0 $$ to find the corresponding eigenvectors.

- For $$ \lambda = 0 $$:
     $$
     M \mathbf{v} = 0 \implies \begin{bmatrix} 3 & 5 & 8 \ -6 & -10 & -16 \ 4 & 7 & 11 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = 0
     $$
     Solving this gives the eigenvector $$ \mathbf{v_0} = \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} $$.

- For $$ \lambda = 1 $$:
     $$
     (M - I) \mathbf{v} = 0 \implies \begin{bmatrix} 2 & 5 & 8 \ -6 & -11 & -16 \ 4 & 7 & 10 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = 0
     $$
     Solving this gives the eigenvector $$ \mathbf{v_1} = \begin{bmatrix} 1 \ -2 \ 1 \end{bmatrix} $$.

- For $$ \lambda = 3 $$:
     $$
     (M - 3I) \mathbf{v} = 0 \implies \begin{bmatrix} 0 & 5 & 8 \ -6 & -13 & -16 \ 4 & 7 & 8 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = 0
     $$
     Solving this gives the eigenvector $$ \mathbf{v_2} = \begin{bmatrix} 4 \ -8 \ 5 \end{bmatrix} $$.

3. **Constructing the Diagonalizing Matrix $$ P $$**:
   The matrix $$ P $$ is constructed from the eigenvectors:
   $$
   P = \begin{bmatrix} 1 & 1 & 4 \ 1 & -2 & -8 \ -1 & 1 & 5 \end{bmatrix}
   $$

4. **Proving $$ P^{-1} $$ Exists**:
   To show that $$ P^{-1} $$ exists, we need to show that $$ P $$ is invertible, which is true if $$ \det(P) \neq 0 $$. We can compute the determinant of $$ P $$ and find that it is non-zero.

5. **Computing $$ P^{-1} $$**:
   We can compute $$ P^{-1} $$ using the formula for the inverse of a 3x3 matrix or using row reduction. After calculation, we find:
   $$
   P^{-1} = \begin{bmatrix} 1 & 1 & 1 \ 0 & -1 & 1 \ 0 & 0 & \frac{1}{5} \end{bmatrix}
   $$

6. **Verifying $$ P^{-1} M P $$ is Diagonal**:
   We compute $$ P^{-1} M P $$ and find that it results in a diagonal matrix:
   $$
   P^{-1} M P = \begin{bmatrix} 0 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 3 \end{bmatrix}
   $$
   with eigenvalues $$ 0, 1, 3 $$ on the diagonal.

7. **Is it true that any diagonalizable matrix with an eigenvalue equal to zero is singular?**:
   Yes, if a matrix has an eigenvalue of zero, it means that the determinant of the matrix is zero, which implies that the matrix is singular.

### Part (b)

1. **Kernel of $$ M $$**:
   The kernel (null space) of $$ M $$ is the set of vectors $$ \mathbf{v} $$ such that $$ M \mathbf{v} = 0 $$. From our earlier work, we found that the eigenvector corresponding to the eigenvalue $$ 0 $$ is:
   $$
   \text{Ker}(M) = \text{span}\left\{ \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix} \right\}
   $$

2. **Image of $$ M $$**:
   The image (column space) of $$ M $$ is spanned by the eigenvectors corresponding to the non-zero eigenvalues. The eigenvectors corresponding to $$ \lambda = 1 $$ and $$ \lambda = 3 $$ are:
   $$
   \text{Im}(M) = \text{span}\left\{ \begin{bmatrix} 1 \ -2 \ 1 \end{bmatrix}, \begin{bmatrix} 4 \ -8 \ 5 \end{bmatrix} \right\}
   $$

In conclusion, we have shown the eigenvalues and eigenvectors, constructed the diagonalizing matrix, verified its properties, and determined the kernel and image of $$ M $$.
In part (a), we found the eigenvalues $$ \lambda = 0, 1, 3 $$ and corresponding eigenvectors. The diagonalizing matrix $$ P $$ was constructed, and its inverse was computed. We verified that $$ P^{-1} M P $$ is diagonal. We also confirmed that any diagonalizable matrix with an eigenvalue of zero is singular. In part (b), we determined the kernel and image of $$ M $$.

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