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I diagonal matrix is obviously diagonalizable since I can conjugate it with the identity. ...(1)

Besides, a matrix 2x2 is diagonalizable iff it has two distinct eigenvalues....(2)

For example the matrix $\begin{bmatrix}4&0\\0&4\end{bmatrix}$ has only one eigenvalue :4 of algebraic multiplicity 2, then it shouldn't be diagonaliz zable, should it? but it obviously is diagonalizable (because of (1)) What am doing wrong?

I am not very sure of (2), but in an exercise we were interested in characterizing the 2x2 non- diagonalizable matrices, and the professor said that the characteristic polynomial should have a double root, so only one eigenvalue of algebraic multiplicity 2, that's why I believed that to have instead a diagonalizable matrix, the eigenvalues should be distinct.

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    $\begingroup$ If it has distinct eigenvalue, the matrix is diagonizable, but the reverse is not always true. $\endgroup$
    – Paul
    Jul 5, 2020 at 23:29
  • $\begingroup$ If it is not diagonalisable, it has one eigenvalue of multiplicity 2. The converse does not hold, since you’ve given an obvious counterexample. You have basically done a mistake in stating the contronominale of your professor‘s statement $\endgroup$
    – tommy1996q
    Jul 5, 2020 at 23:30
  • $\begingroup$ But it is true that a 2 x 2 non-diagonal matrix is diagonalizable IFF it has two distinct eigenvalues (which is what the OP was asking, I believe), since a scalar matrix is similar only to itself. $\endgroup$
    – Ned
    Jul 5, 2020 at 23:43
  • $\begingroup$ @tommy1996q what should the correct negation be? I understand my negation is not correct but can't tell why $\endgroup$ Jul 5, 2020 at 23:45
  • $\begingroup$ Irrespective of the equality of the eigenvalues we can always say that a 2*2 matrix is diagonalisable if only if it has two linearly independent eigenvectors. $\endgroup$ Jul 5, 2020 at 23:52

1 Answer 1

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A typical 2 x 2 non-diagonalizable matrix is $$\pmatrix{ 1 & 1 \\ 0 & 1} $$ Its characteristic polynomial has one double-root, but its minimal polynomial is also $(x-1)^2$, which makes it different from the identity, whose char. poly has a double root, but whose minimal polyonomial is $(x-1)$.

What your prof. said was correct, but you negated it incorrectly. :)

By the way, I applaud your questioning this. Asking questions like this, even ones that seem stupid, is part of how you learn to recognize certain classes of errors and learn not to make them again. Go, you!

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  • $\begingroup$ What should the correct negation of my prof.'s statement be? $\endgroup$ Jul 5, 2020 at 23:36
  • $\begingroup$ So if I am correct the positive statement should be: If A is diagonalisable, then it has one eigenvalue of multiplicity 2, that is it is just an if not an iff. $\endgroup$ Jul 5, 2020 at 23:41
  • $\begingroup$ The OP is correct in saying that a 2x2 NON-DIAGONAL matrix is diagonalizable IFF it has two distinct eigenvalues, because a 2x2 diagonal matrix with a repeated eigenvalue is a scalar matrix and is not similar to any non-diagonal matrix. $\endgroup$
    – Ned
    Jul 5, 2020 at 23:51
  • $\begingroup$ @ned I didn't said non-diagonal, but non-diagonalizable, that is the professor said that in order to be non-diagonalizable, it should have two disctinct eigenvalues, and I asked if a matrix in general was diagonalizable iff it had 2 distict eigenvalues. From your statement I deduce that what the professor said was not entirely correct because he didn't said non-diagonal matrices, and my counterexampe shows we can have a matrix with a double eigenvalue that is diagonalizzable, if the matrix is already diagonal $\endgroup$ Jul 6, 2020 at 0:04
  • $\begingroup$ @J.C.VegaO my guess was that the professor meant "if you are looking at a 2x2 matrix, either you can see that it's diagonal already, or if it's not, it can be diagonalized IFF it has two distinct eigenvalues." $\endgroup$
    – Ned
    Jul 6, 2020 at 0:26

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