Let $A\in \mathbb{R}^{2\times 2}$, $A^2=0$. Prove that $\det(tI-A)=t^2$ for any scalar $t$ (Without the use of Cayley Hamilton theorem).

Question

I tried alot of stuff and went far as taking a general $A$ matrix and trying to apply it and I feel that I'm missing something and I'm not in the right direction.
Note that I would be happy to get all types of answers but this question is intended to be solved using determinant properties.
Any help is appreciated, thanks in advance.

EDIT: I have not learnt Cayley-Hamliton theorem yet, and looking for a solution without it.

Do you know the Cayley-Hamilton theorem? Anyway, show that $tI-A$ is always invertible if $t \neq 0$. — Aphelli, Jan 12 '21 at 14:20
All you've told us is "solve using determinant properties". Everything else you say is not telling us anything in the way of context. Show us what you claim you tried (Everything? then you should have a lot of work to show us!) — amWhy, Jan 12 '21 at 14:23
Okay let me clear myself more, seems like this can be trivially solved using Cayley Hamilton theorem and I got it as a question before I even learnt it, so what ways would be to proof it without knowing the theorem? is it just by going far and taking a general matrix? — Pwaol, Jan 12 '21 at 14:25
@Pwaol It's a bit involved without Cayley Hamilton. I can't imagine this problem would be assigned to novice linear algebraists without Cayley Hamilton. — Rushabh Mehta, Jan 12 '21 at 14:26
@DonThousand I see, this might be the reason that it's only for $2\times 2$ matrices. — Pwaol, Jan 12 '21 at 14:28
I can push you in the right direction: name $A$ as $\begin{bmatrix}a & b\c & d\end{bmatrix}$. Then $A^2=0$ is 4 equations for $a,b,c,d$ and write $\det(tI-A)$ explicitly in $a,b,c,d$. Analyse. — A.Γ., Jan 12 '21 at 14:29
@Pwaol can you edit your question to say that you haven't learned C-H yet and want a solution without it? — Trevor Gunn, Jan 12 '21 at 14:45
@TrevorGunn sure I can, should I edit the title? or the body or it doesn't matter? — Pwaol, Jan 12 '21 at 14:46
The title might be a bit crowded but if it will fit then sure. Definitely in the body though. — Trevor Gunn, Jan 12 '21 at 14:48

score 2 · Answer 1 · 2021-01-12T15:05:31.143

You can take advantage of the small size of the matrix $A$.

All you need is to work out the following cases by hand: $$ \begin{pmatrix} 0&0\\0&0 \end{pmatrix},\quad \begin{pmatrix} 0&1\\0&0 \end{pmatrix}. $$

Once you work out the simple cases, find a relation between $\det(tI-J)$ and $\det(tI-BJB^{-1})$ and show that you can write $A$ as $A=BJB^{-1}$ for some matrix $B$.

score 2 · Accepted Answer · answered Jan 12 '21 at 14:35

$$\det \begin{pmatrix} t - a & -b \\ -c & t - d \end{pmatrix} = t^2 - (a + d)t + (ad - bc)$$

Since $A^2 = 0$, we have $ad - bc = 0$.

If $a + d \neq 0$ then setting $t = a + d$ we have that $\det(tI - A) = 0$. On the other hand,

$$ (tI - A)(\alpha I + \beta A) = I$$

has a solution (try to find it). So $tI - A$ is invertible if $t \neq 0$. But then $a + d$ cannot be non-zero.

score 2 · Answer 3 · answered Jan 12 '21 at 14:51

Here is yet another way to show it:
In some extension field (in this case we can just take $\Bbb C$ of course) the characteristic polynomial $\det(xI-A)$ will split into linear factors. Let $\lambda\in\Bbb C$ be any zero of it, i.e. $\det(\lambda I-A)=0$. Then $\lambda I-A$ is not invertible, hence there is a $v\in\Bbb C^2$ (an 'eigenvector') with $v\ne0$ and $(\lambda I-A)v=0$. So we get $Av=\lambda v$. Applying $A$ again gives us $$0=A^2v=A\lambda v=\lambda^2 v$$ Therefore $\lambda^2=0\Rightarrow \lambda=0$. Hence $\det(xI-A)$ has only one zero in $\Bbb C$, namely $0$ (not counting multiplicity). Therefore $\det(xI-A)=x^2$.
Of course this proof generalizes immediately to higher dimensions.

score 1 · Answer 4 · answered Jan 12 '21 at 15:18

If $A^2=\pmatrix{a&b\\c&d}\pmatrix{a&b\\c&d}=\pmatrix{a^2+bc&b(a+d)\\c(a+d)&bc+d^2}=\pmatrix{0&0\\0&0},$

then $b=0$, in which case $a=d=0$, or $b\ne0$, in which case $a=-d$ and $c=\dfrac{-a^2}b$,

so $A=\pmatrix{0&0\\c&0}$ or $A=\pmatrix{a&b\\\dfrac{-a^2}b&-a}$.

Can you take it from here and compute $\det(tI-A)$ now?

Let $A\in \mathbb{R}^{2\times 2}$, $A^2=0$. Prove that $\det(tI-A)=t^2$ for any scalar $t$ (Without the use of Cayley Hamilton theorem).

4 Answers4