2

Let $A$ be a $3\times 3$ matrix with characteristic polynomial $p(x)=-x^3$ and suppose $w\in \Bbb R^3$ is a vector such that $A^2w\neq0$. I am supposed to show that $B=\{w,Aw,A^2w\}$ is a basis for $\Bbb R^3$.

To show that this set is linearly independent, $A^3w$ must be equal to $0$ but I cannot see why.

I am so sorry for this very easy question.

I appreciate any help

user519955
  • 1,335

2 Answers2

1

The Cayley-Hamilton Theorem says that any matrix satisfies its characteristic polynomial; this means that $-A^3 = 0$ which implies $A^3 = 0$. This, of course, means that $A^3w=0$.

If $aw + bAw + cA^2w =0$, then applying $A^2$ to both sides we obtain $aA^2w= 0$. Since $A^2w\neq0$, this means $a=0$. Now replacing $a$ by zero and applying $A$ to both sides of the original equation (which is now $bAw + cA^2w=0$) we get $bA^2w=0$, which again implies $b=0$. Then you're left with $cA^2w=0$ which implies $c=0$. Thus $a=b=c=0$ and the set is linearly independent.

If you're not allowed to use Cayley-Hamilton, by definition of $A$ you know that all eigenvalues of $A$ are zero. This post can show you how to prove that $A^3=0$ from this information, but it requires you to know about Jordan normal form. Otherwise, I know of no other ways to show that $A^3=0$ (besides potentially directly, using the fact that you know both the trace and determinant of $A$ are zero (the cofficient of $x^2$ and the constant term in the characteristic polynomial are both zero!), but this seems like an algebra nightmare), and perhaps someone else has a solution without using Cayley-Hamilton/canonical forms.

kamills
  • 2,252
  • 13
  • 16
  • This is a nice answer. Regarding how to show $A^3w=0$, what you wrote nearly gives you the proof. Consider the set ${w, Aw, A^2w, A^3w}$. Now recall that $A$ is $3\times 3$. Along the same lines you can show that for any $v$ we have $A^kv=0$ for $k$ no larger than $2$. – Adam Zalcman Jan 08 '21 at 00:11
  • @AdamZalcman I must be missing something, but I actually don't see how it follows that this implies $A^3 w = 0$. – kamills Jan 08 '21 at 04:03
  • I had an argument like this in mind. This avoids the use of Cayley-Hamilton, but requires one to know that if a matrix's sole eigenvalue is zero then it is nilpotent, which AFAIK requires the use of canonical forms (possibly under the guise of generalized eigenvectors), so I suppose this also isn't what OP had in mind. – Adam Zalcman Jan 08 '21 at 05:40
  • 1
    Right! I was trying to follow my line of proof to show that $A^3 w=0$ but it would at some point require knowing that some power $A^k w=0$. I just read your answer, though, which is really nice. – kamills Jan 08 '21 at 16:51
  • Thanks! I came to the same realization as you after I actually got down to doing it. – Adam Zalcman Jan 08 '21 at 17:35
1

I have found a proof that does not use Cayley-Hamilton theorem.

The roots of $p(x) = -x^3$ are precisely the eigenvalues of $A$, so $A$ has only one eigenvalue: zero. We will prove that $B=\{w,Aw,A^2w\}$ is a basis for $\mathbb{R}^3$ by first showing that $Aw$ is not a multiple of $w$ and then that $A^2w$ is not a linear combination of $w$ and $Aw$.

If $Aw$ was a multiple of $w$ then $w$ would be an eigenvector of $A$ and $Aw$ would be zero, contradicting assumption. Thus, $w$ and $Aw$ are linearly independent.

Let $V = \mathrm{span}(w, Aw)$. For a proof by contradiction, suppose that $A^2w$ is a linear combination of $w$ and $Aw$. In this case, $Av \in V$ for every $v \in V$. Thus, we can consider the restriction $B$ of $A$ to $V$. Every eigenvalue of $B$ is an eigenvalue of $A$. Therefore, the only eigenvalue of $B$ is zero. Let $aw + bAw$ be the eigenvector of $B$ associated with the eigenvalue. Then

$$ aAw + bA^2w = 0.\tag1 $$

Now, $Aw \ne 0$ and $A^2w \ne 0$ so $a \ne 0$ and $b \ne 0$. Therefore, we can write $(1)$ as

$$ A^2w = -\frac{a}{b}Aw, $$

but this means that $\frac{a}{b} \ne 0$ is an eigenvalue of $A$ contradicting the fact that its characteristic polynomial is $p(x)=-x^3$. Thus, $A^2w$ is not a linear combination of $w$ and $Aw$.

Adam Zalcman
  • 3,524