Prove: Let $ A\in\mathsf{Mat}(2\times 2, \mathbb{R})$. If $ \det(A)$ is zero, then there exist a $ \mu\in\mathbb{R}$, so that $ \mu A = A^2$.

Question

Could you help me with this task? I should decide which of the two following statements in (i) and (ii) are true. I would appreciate it, if you would explain to me the solution in detail, because I want to try to understand the task.

(i) There is $ A\in\textsf{Mat}(6\times 6, \mathbb{C})$, so $ \dim_{\mathbb{C}}([\{A^i : i\geq0\}]) = 15$, where $ [\{A^i: i\geq0\}]$ is the linear span of $ \{A^i:i\geq0\}$.

(ii) Let $ A\in\mathsf{Mat}(2\times 2, \mathbb{R})$. If $ \det(A)$ is zero, then there exist a $ \mu\in\mathbb{R}$, so that $ \mu A = A^2$.

Thanks in advance.

Answer 1

Think about the characteristic polynomial $P(X)$ of $A$. There are three things that are needed here:

• Its degree: if $A$ is an $n\times n$ matrix, $P(X)$ has degree $n$.
• Its factorization: if $\lambda$ is an eigenvalue of $A$, then $X-\lambda$ divides $P(X)$.
• The Cayley-Hamilton theorem: $P(A)=0$.

Try to think yourself how to use these facts to answer the two questions (and make sure that you knew about these facts!). I give you a detailed solution below that you can check afterwards.

For question (i):

We have that $P(X)$ is a polynomial of degree $6$ such that $P(A)=0$, i.e., $a_6A^6 + a_5A^5 + \dots + a_1A + a_0A^0 = 0$ for some $a_0,\dots,a_6\in\Bbb C$. Thus, $A^6\in [\{A^0,A^1,\dots,A^5\}]$ and you can prove that also $A^n\in[\{A^0,A^1,\dots,A^5\}]$ for all $n>6$. What do you conclude about the dimension of $[\{A^i : i\geq 0\}]$?

For question (ii):

If $\det(A)=0$ then there exists a nonzero vector $v$ such that $Av=0$, so that $0$ is an eigenvalue of $A$. Thus, $P(A)$ is divisible by $X$. Since $P(A)$ is a polynomial with real coefficients and with degree $2$, it can be written in the form $X(X-\mu)$ for some $\mu\in\mathbb R$. Since $P(A)=0$, we obtain $A^2 - \mu A=0$, i.e., $A^2 = \mu A$.