Question

A, P and D are nn matrices.

Check the true statements below:

A. If A is diagonalizable, then A is invertible.
B. A is diagonalizable if A=PDP−1 for some diagonal matrix D and some invertible matrix P.
C. A is diagonalizable if and only if A has n eigenvalues, counting multiplicities.
D. None of the above

Answer 1

Any ideas on which may be true or false?

Answer 2

well i think B

Answer 3

B is true for sure, that is correct.

Answer 4

what about the others?

Answer 5

To disprove A, do you think you could come up with a diagonal matrix that is not invertible?

Answer 6

doesnt have to be huge, a 2x2 or a 3x3 would do.

Answer 7

sure we can have one

Answer 8

0,0 and 0,3

Answer 9

yep, that wouldnt be invertible. so A is out the window.

Answer 10

what about C

Answer 11

Have you heard the terms "geometric multiplicity" and "arithmetic multiplicity"?

Answer 12

Ive heard multiplicity ( some numbers with the same value)

Answer 13

hmm...let me see if i can find a counter example for C. I want to find a matrix where the characteristic polynomial has a repeated root, but there is only one eigenvector for that eigenvalue.

Answer 14

i.e where the arithmetic multiplicity of an eigenvalue doesnt match the geometric multiplicity.

Answer 15

that first line is not true. A diagonal matrix could have an eigenvalue of 0.

Answer 16

just think of the null space/kernel of a matrix. Its the set of all vectors such that:\[Ax=0\Longrightarrow Ax=0x\] So if a matrix has a non-trivial null space, then 0 will be an eigenvalue.

Answer 17

well thats true

Answer 18

oh ok....

Answer 19

about C i think A is diagonalizable if and only if A has an eigenvector corresponding to each eigenvalue

Answer 20

hmm...thats half right. Lets say I have a 3x3 matrix A, and I have:\[\det (A-\lambda I)=(\lambda-2)^2(\lambda-1)=0\]This tells me that my eigenvalues are 2 (arithmetic multiplicity 2) and 1 (arithmetic multiplicty 1). Now lets say I look for the eigenvectors associated with the eigenvalue 2, and i only come up with 1. In other words, when i looked for a basis for the null space of A-2I, it was one-dimensional. Then this matrix would not be diagonalizable, because i need 3 eigenvectors to form a basis, and I would only be able to find 2. One from eigenvalue 2, and one from eigenvalue 1.

Answer 21

So a matrix is diagonalizable if you can find enough eigenvectors to match the multiplicity of the eigenvalue.

Answer 22

so C is not true

Answer 23

its not, im having trouble coming up with an example though lol.

Answer 24

do not worry I have one in the book, but I came here just to make
thanks a lot :)

Answer 25

and thanks you @anonymoustwo44