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.

answer: 230 miles ( vote ! )

step-by-step explanation:

180.39 minus 17.99 equals 162.4

then divide 162.4 by .80 to get 230

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet