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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.

Rodrigo de Azevedo
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some_math_guy
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    If it has distinct eigenvalue, the matrix is diagonizable, but the reverse is not always true. – Paul Jul 05 '20 at 23:29
  • 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 – tommy1996q Jul 05 '20 at 23:30
  • 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. – Ned Jul 05 '20 at 23:43
  • @tommy1996q what should the correct negation be? I understand my negation is not correct but can't tell why – some_math_guy Jul 05 '20 at 23:45
  • 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. – Lawrence Mano Jul 05 '20 at 23:52
  • @J.C.VegaO If you have a statement like “hypothesis implies thesis”, the correct way to state the contronominale is “negation of thesis implies negation of hypothesis”. – tommy1996q Jul 06 '20 at 08:37

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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!

John Hughes
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  • What should the correct negation of my prof.'s statement be? – some_math_guy Jul 05 '20 at 23:36
  • 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. – some_math_guy Jul 05 '20 at 23:41
  • 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. – Ned Jul 05 '20 at 23:51
  • @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 – some_math_guy Jul 06 '20 at 00:04
  • @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." – Ned Jul 06 '20 at 00:26
  • @ned Actualy we had a matrix with unknown entries a11,a12,a21,a22 an we had to find the conditions for it not to be diagonalizable, he said the characteristic polynomial should have a double root, that is the matrix should have only a double eigenvalue, and wrote the equation ,but he said nothing about the matrix not being diagonal nor write it down. This adds an extra condition: a12 and a21 should not be both zero, don't you agree then that the statement was incomplete? – some_math_guy Jul 06 '20 at 00:39
  • @J.C.VegaO yes you are right -- that IFF statement is only correct if you eliminate the scalar matrices $cI_2$ from consideration. – Ned Jul 06 '20 at 01:49