A is a 7 x 7 matrix with three eigenvalues. One eigenspace is two-dimensional, and one of the other eigenspaces is three dimensional. Is it possible that A is not diagonalizable? Justify your answer

it is possible

Step-by-step explanation:

There are 3 eigenspaces, since A has 3 eigenvalue the dimension of each of each eigenspace is at least 1 (since there is at least 1 non zero eigenvector). It is given that one of the eigenspace has dimension 2 and other one has dimension 3. Since a 7x7 matrix his diagonalizable if and only if the sum of the eigenspaces' dimension is 7, the given matrix is diagonalizable if and only if the third eigenspace has dimension 2. We cannot be sure that this is the case, so it is possible that A is no diagonalizable. For example , consider the matrix A (attachment)

has three eigenvalue λ = 1,2,3, the eigenspace corresponding to λ = 1 has dimension 2. the eigenspace corresponding to λ = 2 has dimension 3, and the eigenspace corresponding to λ = 3 has dimension 1. Sum of the dimension is 6 =, so the matrix is not diagonalizable.

step-by-step explanation:

descending order means to order the polynomials from the term with the largest exponent to the term with the smallest exponent.  

therefore, the correct answer is d.  

y = 3x+4

step-by-step explanation:

to find the slope

slope = (y2-y1)/ (x2-x1)

    = (-8-4)/ (-4-0)

    = -12/-4

  = 3

using the slope intercept form of a line

y= mx+b

we know the slope is 3

we know the y intercept is 4 ( the point where it crosses the y axis or x=0)

y = 3x+4