# Harrison Chapman

## Linear Algebra I: Homework 9

Due: Friday, October 27, 2017
1. Find the eigenvalues and their corresponding eigenspaces for the matrix,

$\DeclareMathOperator{\span}{span}$ The characteristic polynomial of $C$ is $(\lambda - 2)^2(\lambda - 3)$, so the eigenvalues are $2, 2, 3$.

The eigenspace for eigenvalue 2 is two-dimensional and is spanned by the vectors found by solving the equation $(C - 2I) \vec v = \vec 0$; it is,

The eigenspace for eigenvalue 3 is one-dimensional and is spanned by the solution to the equation $(C - 3I) \vec v = \vec 0$; it is,

2. Remember that the eigenvalues of the matrix,

are $\lambda = 4, -2, -2$, with eigenvector for $\lambda = 4$,

and eigenvectors for $\lambda = -2$,

Find an invertible matrix $P$ and a diagonal matrix $D$ so that $A = PDP^{-1}$.

Based on the known eigenvalues, one choice for $D$ is,

Based on this choice of $D$, one valid choice for $P$ is,

3. Can the matrix

be diagonalized? Explain why or why not.

The characteristic polynomial of $M$ is, $(\lambda - 9)(\lambda - 11) + 1 = \lambda^2 - 20\lambda + 100 = (\lambda - 10)^2$, so the eigenvalues are $10, 10$. However, the only eigenvector of $M$ is

As there is only one eigenvector for the $2\times 2$ matrix $M$, it is not diagonalizable.

4. Can the matrix

be diagonalized? Explain why or why not.

The characteristic polynomial of $N$ is, $\lambda^2 - 4\lambda + 1$. This polynomial has discriminant “$b^2 - 4ac$” of $12$. As the discriminant is greater than zero there are two unique real roots to the polynomial, and hence two different real eigenvalues for $N$. As each eigenvalue has at least one eigenvector, $N$ has at least two eigenvectors. So, as $N$ is a $2\times 2$ matrix, this means that yes, $N$ is diagonalizable.

5. For $n$ a positive integer and $B$ the matrix,

find a formula for $B^n$.

$B$ diagonalizes as the matrix product,

Since $B^n = (PDP^{-1})^n = PD^n P^{-1}$, a formula for $B^n$ is,