Diagonalise $\begin{pmatrix}2 & 1\\0 & 2\end{pmatrix}$. The characteristic polynomial is $(x-2)^2$ hence there's only one eigenvalue that is $2$.
Asked
Active
Viewed 222 times
0
-
Are you sure it is diagonalizible? – user1337 Aug 04 '13 at 10:59
-
Your matrix's already in Jordan Canonical Form, so it can't be diagonalized. Another way to see this is to observe that its minimal polynmomial is not the product of different linear factors. – DonAntonio Aug 04 '13 at 11:04
-
If all eigenvalues are equal, then either the matrix is already diagonal, or it is not diagonalisable. Show this as a simple exercise :) – Mårten W Aug 04 '13 at 11:07
-
@MårtenW without using that a matrix is diagonalizable iff the minimal polynomial a product of distinct factors how would one prove it ? – Dominic Michaelis Aug 04 '13 at 11:10
-
2@Grobber Would you be so kind to post an answer containing the results, so that the question doesn't remain unanswered ? – Dominic Michaelis Aug 04 '13 at 11:11
-
@DominicMichaelis Compute the eigen space for the eigen value $2$. – knsam Aug 04 '13 at 11:11
-
@kan Here it is an easy computation, but in the general case I don't see it via the eigenspaces – Dominic Michaelis Aug 04 '13 at 11:12
-
In the general case too, the same argument works. – knsam Aug 04 '13 at 11:14
-
@DominicMichaelis: Assume $A=SDS^{-1}$. If all eigenvalues are equal (say, equal to $\lambda$), then $D = \lambda I$, and hence $A = \lambda I$. – Mårten W Aug 04 '13 at 11:21
-
@MårtenW oh right it commutes thanks – Dominic Michaelis Aug 04 '13 at 11:27
-
In case this is of interest here: I just added a question for the case of infinite size. see http://math.stackexchange.com/questions/459454/with-infinite-size-we-can-have-p-cdot-m-m-cdot-d-d-diagonal-where-m – Gottfried Helms Aug 04 '13 at 13:02
2 Answers
1
Yeah. My solution goes as follows. Thanks to the comments. Let $A$ be a matrix with characteristic polynomial $(x-\lambda)^n$. Then it is diagonalisable iff $A=\lambda I$.
Suppose on the contrary it is similar to a non-scalar diagonalisable matrix. Then clearly characteristic polynomial is different. Hence we are done.
In my case we see its not a scalar matrix. So, not diagonalisable.
Grobber
- 3,336
0
You should also note that your matrix is currently in its Jordan Canonical Form, so it wouldn't be possible to diagonalise this matrix.
