Suppose that a2 = 0 for some matrix a. prove that the only possible eigenvalues of a are then 0.

The eigenvector of A is not equal to zero, then we can say λ or λ = 0. Therefore,  the only possible eigenvalues of A are 0.

Step-by-step explanation:

If we assume that λ is the eigenvalue of the matrix A and the eigenvector of the matrix A is ⁻ˣ. Therefore:

For

we have:

⁻0 = [⁰₀⁰₀][⁻ˣ] = *[⁻ˣ] = Aλ[⁻ˣ] = λ[⁻ˣ]

In the expression above, ⁻ˣ is not equal to zero, then λ = 0 or  λ is = 0. This shows that the only possible eigenvalues of A are zero '0'

Given:

Consider the below figure attached with this question.

The given data set is:

66, 65, 66, 70, 66, 68, 63, 60, 66, 68, 63, 65

To find:

The correct box plot for the given data set.

Solution:

We have,

66, 65, 66, 70, 66, 68, 63, 60, 66, 68, 63, 65

Arrange the data set in ascending order.

60, 63, 63, 65, 65, 66, 66, 66, 66, 68, 68, 70

Divide the data set in 4 equal parts by using the parenthesis.

(60, 63, 63), (65, 65, 66), (66, 66, 66), (68, 68, 70)

Minimum value = 60

First quartile:

Median:

Third quartile:

Maximum value = 70

It means the end points of the box plot are 60 and 70. The box lies between 64 and 67. Line inside the box at 66.

The box plot in option A is the only box plot that satisfy the above conditions.

Therefore, the correct option is A.