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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 $2$.

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Your question title says that $A^2=0$, does the definition of $A$ follow in the question also ?? – Theorem May 13 '12 at 18:17
preeti: You are right that (2) holds, but you should also show that (1), (3) and (4) do not. – Did May 13 '12 at 18:18
@NicoBellic: I don't think thats true. – froggie May 13 '12 at 18:18
@NicoBellic: I am confused as in this case A seems to be similar to a zero matrix. That is PAP^(-1) is a zero matrix. Does this mean that if entries belong to R then it should be diagonalizable ? – preeti May 13 '12 at 18:23
If a $nxn$ matrix has n distinct eigen values then its always diagonalizable. You may not need any extra conditions. – Theorem May 13 '12 at 18:40
up vote 2 down vote accepted

Hint: $\lambda$ eigenvalue of $A\Rightarrow \lambda^2$ eigenvalue of $A^2$. What are the eigenvalues of $A^2$, and what is hence the only possible eigenvalue of $A$? How can $A$ look like (up to basechange)?

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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).

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The characteristic polynomial is not quite as necessary as you make it, to introduce the algebraic multiplicity of an eigenvalue. To wit, for every given scalar $x$, consider the sequence of vector subspaces $E_k(x)=\text{Ker}((x I-A)^k)$. This sequence is nondecreasing and ultimately constant. Either $E_1(x)=\{0\}$, then $E_k(x)=\{0\}$ for every $k$ and $x$ is not an eigenvalue. Or $E_1(x)\ne\{0\}$, then $x$ is an eigenvalue and the dimension of $E_k(x)$ for every $k$ large enough is what one calls the algebraic multiplicity of $x$. .../... – Did May 15 '12 at 9:25
.../... In the end, the algebraic multiplicity may be defined through the characteristic polynomial, and in this sense it is indeed very much algebraic, but one can also define it through some more geometrical considerations such as linear subspaces and their dimensions. – Did May 15 '12 at 9:26
@Didier: I quite agree. But this also illustrates that one has many ways to define multiplicities: apart from the dimension of the eigenspace $E_1(x)$ and the dimension of the generalized eigenspace $E_k(x)$ for $k$ sufficiently large, another parameter of interest is the number of distinct nonzero spaces $E_i(x)$, i.e., the minimal $k$ for which $E_k(x)$ attains its ultimate value. This gives the multiplicity of $x$ as root of the minimal polynomial. For $x$ one has for $x$ a unique Young diagram whose first $i$ rows total $\dim E_i(x)$ boxes, and many "multiplicities" can be derived from it. – Marc van Leeuwen May 15 '12 at 17:17

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