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In the space of $2\times 2$ matrices, find explicitly the sets of matrices with 1)a single zero eigenvalue, 2) a pair of pure imaginary eigenvalues. Show that each set is a submanifold of $\mathbb{R}^4$ and find its codimension. (Hint: Use the Implicit Function Theorem). I thought the first set is the following \begin{equation} \left\{\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)|a,b,c,d\in \mathbb{R}, \ \ ad-bc=0, \ \ a+d\neq0 \right\} \end{equation} and the second is:

\begin{equation} \left\{\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)|a,b,c,d\in \mathbb{R},\ \ (a+d)^2-4(ad-bc)<0\right\} \end{equation}

I do not know how to continue. Any suggestions please?

thank you very much

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    I would say that the sets you described are just subsets of the sets asked. For example suppose you want a matrix with a single $0$ eigenvalue. Then Its determinant must be $0$, so $ad=bc$. Then for this $0$ to be unique you should also require $a+d\neq 0$. Therefore there are lots of matrices missing in your description.2012-09-25
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    The second set of matrices should satisfy, looking at the characteristic polynomial $(a+d)^2-4(ad-bc)<0$.2012-09-25
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    What about $\begin{pmatrix} 0&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}\frac 1 2 & \frac {\sqrt 3} 2 \\ -\frac {\sqrt 3} 2 & \frac 1 2 \end{pmatrix}$?2012-09-25
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    @uforoboa, Yes it is true, I have been too superficial to conclude.2012-09-25
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    @Mark.. Check again your conditions... For the second set it is unnecessary that $a+d=0$, while it is ESSENTIAL that $a+d\neq 0$ for the first. Then you should start to guess what your maps should be.2012-09-25
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    Please, correct the inaccurate statements, instead of leaving them there followed by correct versions.2012-09-25
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    I have to correct any inaccuracies. Thank you all.2012-09-25
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    @uforoboa I do not understand the condition $a+d\neq0$. For a $2\times2$ matrix, to have precisely one zero eigenvalue is to have determinant $0$ without being the zero matrix. For example $\begin{bmatrix} 1 & 1\\-1&-1\end{bmatrix}$ should count, and yet $a+d=0$.2012-09-25
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    @alex.jordan If I am not completely blinfolded the characteristic polynomial of your matrix is $x^2$, therefore it has $0$ as an eigenvalue, but with multiplicity $2$. Therefore I wouldn't say that your matrix has a single $0$ eigenvalue. Am I misinterpretating?2012-09-25
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    @uforoboa. Yes. The _algebraic_ multiplicity of the eigenvalue is 2, but its _geometric_ multiplicity (the dimension of its eigenspace) is only 1. I'm fairly sure that is what is meant by "a single zero eigenvalue".2012-09-25
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    @alex.jordan. To me it is fine what you're saying, but I would ask the response of a third party to confirm this. Really... This is just a matter of interpretation and since I am Italian it is really likely that I am wrong. Thank you for the explanation.2012-09-25

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