Directions: Write <> or = between each pair of rational numbers

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0

3 years ago

2 + yx; use x = 6, and y = 5

Brrunno [24]

Answer:

answer: 32

Step-by-step explanation:

2+5*6

=32

4 0

3 years ago

jane hilman went to her bank. she has a balance of $1009.88 in her savings account. she withdrew $130.00 and the teller credited

blagie [28]

First substract $130.00 from $1009.88 like this:
$\$1009.88-\$130.00=\$879,88$
Then substract $6.19 we get:
$\$879,88-\$6.19=\$873,69$
$873.69 is the remaining amount in <span>Jane hitman's account. </span>

3 0

2 years ago

Problem 4. (3 points) Given the following matrix A = 1 1 0 1 . Which of the following statements must be TRUE? (I) The rank of t

shepuryov [24]

Answer:

C) II only

Step-by-step explanation:

Given the 2×2 matrix A = [1 1, 0 1]

The only statement true there is (II) only i.e the matrix is not diagonalizable but invertible. A matrix is invertible if the product of the matrix and its inverse is equal to an identity matrix

The rank of the matrix is not 1 but 2 because rank of a matrix is the number of non zero rows of a matrix and the number of non zero rows of this matrix is 2 thereby making I incorrect.

Also note that the sum of nullity and rank is equal to the number of columns of the given matrix

Nullity + rank = number of columns

Nullity = number of columns - rank

Nullity = 2-2 = 0

The nullity is therefore 0 not 1 making option III also false

8 0

3 years ago

Determine whether the series is convergent or divergent. 1 4 + 3 16 + 1 64 + 3 256 + 1 1024 + 3 4096 +

Alla [95]

This series converges.

$S=\dfrac14+\dfrac3{16}+\dfrac1{64}+\dfrac3{256}+\dfrac1{1024}+\cdots$