Eigenvalues of a matrix $A$ such that $ A^2=0$.

Suppose the matrix $A$ is a $2 \times 2$ non-zero matrix with entries in $\Bbb C$. Which of the following statements must be true?

  1. $PAP^{-1}$ is a diagonal matrix for some invertible matrix $P$ with entries in $\Bbb R$.

  2. $A$ has only one distinct eigenvalue in $\Bbb C$ with multiplicity $2$.

  3. $A$ has two distinct eigenvalues in $\Bbb C$.

  4. $Av = v$ for a non-zero $v$.

Please suggest which of the possibilities hold. It seems to me that the characteristic poly is $ f(t) = t^2$, which means option (2) holds that is only one eigenvalue zero with multiplicity $2$.

2022-07-25 20:45:20
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Answers: 2

Assuming the condition $A^2=0$ of the title is indeed part of the question, none of the given statements must be true. Moreover statements 4., 3., and (given that $A\neq0$) 1. must in fact be false. For 4. this is clear, since it would imply $A^2v=Av=v\neq0$, for 3. it is similar because this implies there must be a non-zero eigenvalue $\lambda$, and for a corresponding eigenvector $v$ one gets $A^2v=A\lambda v=\lambda^2v\neq0$, and this shows for 1. that the diagonal matrix in questoin can only be the null matrix, but $PAP^{-1}=0$ would imply $A=0$ (even without assumping $P$ to have real entries), contrary to the hypothesis.

I need to explain why I consider it wrong to say that 2., which is clearly intended to be the correct answer, must be true. This is because 2. uses the notion "multiplicity of an eigenvalue" without saying what that means, and this is rather ambiguous. If you look at the definition of eigenvalue (a scalar $\lambda$ such that there exists a vector $v\neq0$ with $Av=\lambda v$) then it's a true-or-false matter, with nothing there to suggest a multiplicity. If one would count the number of different possible choices for $v$, then one would get infinite mutliplicity. Given that, it would be more natural to take instead the dimension of the set of possibilities for $v$ (after throwing in the non-eigenvector value $0$ to make it a subspace), which is called the geometric multiplicity of $\lambda$ as eigenvalue, and with this definition the multiplicity of the unique eigenvalue $0$ will be $1$ rather than $2$ (being $2$ would mean $Av=0$ for a $2$-dimensional space of $v$'s, which means all $v$'s, and this is false), which makes statement 2. false.

Now I can guess that you probably intend another definition of the multiplicity of an eigenvalue, namely its multiplicity as root of the characteristic polynomial, and with this definition the statement 2. indeed becomes true, as $A^2=0$ implies that the characteristic polynomial is $X^2$. But there is no reason that one must use the characteristic polynomial to find eigenvalues, so it is not reasonable to call this the (unqualified) multiplicity of an eigenvalue. Indeed one could equally well determine eigenvalues as the roots of the minimal polynomial, and this would lead to yet another notion of multiplicity (in the current case the minimal polynomial can also, and even more easily, be seen to be $X^2$, but in general it can be different; it has exactly the same roots as the characteristic polynomial, but their multiplicities may be smaller).

2022-07-25 21:50:33

Hint: $\lambda$ eigenvalue of $A\Rightarrow \lambda^2$ eigenvalue of $A^2$. What are the eigenvalues of $A^2$, and also what is therefore the only feasible eigenvalue of $A$? Just how can $A$ resemble (approximately basechange)?

2022-07-25 21:20:15