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Let $A$ be a $2\times2$ real square matrix of rank $1$. If $A$ is not diagonalizable, then which of the following is true.
(a) $A$ is nilpotent
(b) $A$ is not nilpotent
(c) the characteristic polynomial of $A$ is linear.
(d) $A$ has a non-zero eigenvalue.

I can say that d is false.

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Given the context that you have stated, you should not think of trying to brute force the coordinates. Instead, think about what facts you know, and how to apply them to solve this problem. –  Calvin Lin Dec 30 '12 at 6:08

4 Answers 4

Think about the Jordan normal form. Since it is not diagonalizable, it must be a 2-block. Since it has rank 1, the eigenvalues must be 0. Hence, $A^2=0$, so the matrix is nilpotent, it has minimal polynomial = characteristic polynomial which is $A^2=0$.

Hence only (a) is true.

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2  
A non-diagonalizable matrix need not always be in Jordan normal form. –  user17762 Dec 30 '12 at 6:10
    
Jordan blocks, with 1's off the diagonal. I stated that it must be a 2-block. –  Calvin Lin Dec 30 '12 at 6:12
    
@Marvis - Does it matter ? I believe every part in this answer is correct –  Belgi Dec 30 '12 at 6:13
    
@Belgi I can't speak for Marvis, but I think the point is, if we use Jordan forms, we are essentially extending the ground field to $\mathbb{C}$. And some people may think that if a result can be proved over the original field, we'd better do so, otherwise the true reason why the result holds may be concealed. Pambos' answer, for instance, is "purer" than Lin's. I am fine with Lin's answer, but different people have different views. –  user1551 Dec 30 '12 at 9:44
    
@user1551 I am using the Jordan form for real matrices. The 'complex' jordan blocks of the form $\begin{matrix} a b \\ -b a\\ \end{matrix}$ do not have rank 1. Hence, we must have a 2-block with real eigenvalues. –  Calvin Lin Dec 30 '12 at 16:08

Since $A$ is of rank $1$, we have $$A = \begin{bmatrix}1 \\ u_2 \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix} = \begin{bmatrix} v_1 & v_2\\ u_2 v_1 & u_2 v_2\end{bmatrix}$$ The eigen values are $v_1 + u_2v_2$ and $0$. Given that the matrix is non-diagonalizable, a necessary condition is that the two eigenvalues must be equal. Hence, we have the other eigenvalue also to be zero i.e. $v_1 + u_2v_2 = 0 \implies u_2 = -\dfrac{v_1}{v_2}$. Hence, $$A = \begin{bmatrix} v_1 & v_2\\ -\dfrac{v_1^2}{v_2} & -v_1\end{bmatrix}$$ Hence, $$A^2 = \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix} \times \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix}= \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}$$

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The characteristic polynomial of $A$ is $\chi_A(x)=x^2+bx+c$ for some $b,c$.
Since $A$ has rank $1\Rightarrow \det A=0 \Rightarrow c=0$. Thus $\chi_A(x)=x(x+b)$.
Since $A$ is non-diagonalizable it has only one eigenvalue $\Rightarrow b=0 \Rightarrow\chi_A(x)=x^2 \ldots$

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Hint: Consider $$\left(\begin{array}{cc}0&1\\ 0&0\end{array}\right).$$

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This is diagnoalizable. You want 1 in the upper right. –  Calvin Lin Dec 30 '12 at 6:02
    
Apologies, typo. –  Alexander Gruber Dec 30 '12 at 6:03
    
And of course, this doesn't answer the question, unless it is a Multiple choice with only 1 answer, then he has to pick (a) –  Calvin Lin Dec 30 '12 at 6:09
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I am simply pointing out that one can swiftly reduce the problem to considering (a) only. Sometimes it is helpful to back up a second and consider an obvious example. As I said this is a hint, not a full solution. –  Alexander Gruber Dec 30 '12 at 6:15

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