3
$\begingroup$

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.

  • 0
    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.2012-12-30

4 Answers 4

3

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.

  • 0
    @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.2012-12-30
2

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, $A^2 = \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix} \times \begin{bmatrix}1 \\ -\dfrac{v_1}{v_2} \end{bmatrix}\begin{bmatrix}v_1 & v_2 \end{bmatrix}= \begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}$

2

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 $\Rightarrow b=0 \Rightarrow\chi_A(x)=x^2 \ldots$

1

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

  • 1
    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.2012-12-30