Linear Algebra In C
True or False?
For all n , if A ∈ S n ( C ) , then A is diagonalizable .
Notation : S n ( C ) denotes the space of complex symmetric matrices
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1 solution
Moderator note:
A better way to understand this concept (depending on how much you know), is that a matrix is diagonalizable if and only if all of its Jordan blocks have dimension 1. So, for n = 2 , we're looking for a matrix that looks like ( 0 0 1 0 ) . In your language, the minimal polynomial is X 2 = 0 . Thus, the matrix that we want has trace 0 and determinant 0. Thus, we can make the diagonal entries 1 , − 1 , which gives the off-diagonal entry of i . IE X = ( 1 i i − 1 ) .
Now, it is clear that this matrix satisfies the conditions that we're looking for, hence we're done. (In your language, it is clear satisfies X 2 = 0 and it doesn't satisfy X = 0 , hence the minimal polynomial is indeed X = 0 . )
Can you explain further?
@Calvin Lin Yes. I will give some hints for those who don't know how to start, instead of writing the full proof. Let's fix A = ( a b b c ) ∈ M ( 2 , C ) . The criteria for checking iff a matrix is diagonalizable over C is
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The eigenvalues must lie in C
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The algebraic multiplicity of any eigenvalue must to be equal to the geometric multiplicity.
In order to find a counterexample, observe that the first condition is always true, hence try to break the second one.
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A better way to understand this concept (depending on how much you know), is that a matrix is diagonalizable if and only if all of its Jordan blocks have dimension 1. So, for n = 2 , we're looking for a matrix that looks like ( 0 0 1 0 ) . In your language, the minimal polynomial is X 2 = 0 . Thus, the matrix that we want has trace 0 and determinant 0. Thus, we can make the diagonal entries 1 , − 1 , which gives the off-diagonal entry of i . IE X = ( 1 i i − 1 ) .
Now, it is clear that this matrix satisfies the conditions that we're looking for, hence we're done. (In your language, it is clear satisfies X 2 = 0 and it doesn't satisfy X = 0 , hence the minimal polynomial is indeed X = 0 . )
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I don't know anything about the Jordan blocks and the minimal polynomial. So, I decided to play only with multiplicities. Thanks for showing another method.
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@Victor Moreno Diaz – There's a lot of theory to linear algebra, and having the "right"/good perspective will help to frame the question in a clean-cut manner.
This is a great question to ponder about. A better way to motivate it is to state that "real symmetric matrices are diagonalizable, so what about complex ones?" That, in turn, offers you a chance to deepen your understanding about the differences between real and complex matrices.
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@Calvin Lin – And also what about the rational ones? ;)
If n = 2 , just take A = ( − 1 2 3 i 2 3 i 2 ) .