First of all, I know that $\dim(\text{ker}(T)) =\dim(\text{Im}(T))= 1$ and that $T( \text{ker}(T))= T( \text{Im}(T))= 0$. But now I don't know how to proceed.
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1For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – Wolgwang Dec 18 '21 at 13:43
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@Wolgwang Ok but I am pretty sure that you got me – FunnyCallsw Dec 18 '21 at 13:44
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2This time I edited the question for you. Next time, be more careful ;) – Kandinskij Dec 18 '21 at 13:49
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@FrancescoScavella appreciate that – FunnyCallsw Dec 18 '21 at 13:49
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Does this answer your question https://math.stackexchange.com/questions/3529005/give-an-example-of-a-linear-transformation-tr2-rightarrow-r2-such-that-n – Kishalay Sarkar Dec 18 '21 at 13:51
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@KishalaySarkar no. this is the opposite way – FunnyCallsw Dec 18 '21 at 13:56
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See the title of your question – Kishalay Sarkar Dec 18 '21 at 13:58
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Another useful thread might be https://math.stackexchange.com/questions/393880/is-it-possible-that-kert-operatornameimt-for-some-linear-transformat @FunnyCallsw – Irving Rabin Dec 18 '21 at 17:44
1 Answers
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You are essentially looking for an endomorphism $T$, such that:
$T \circ T = T^2 = 0$
If we consider the matrix representation of $T$, we can see that what we need is a 2 by 2 matrix that is the zero matrix when raised to the power of $2$. Thus we arrive at the formula:
$\begin{bmatrix}a & b\\c & d\end{bmatrix}^2 = \begin{bmatrix}a^2 + bc & b(a+d)\\c(a+d) & d^2+bc\end{bmatrix} = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$
From this you can extract the following relations:
$a^2 = -bc = d^2$
$(b = 0 \land c = 0) \lor a = -d $
Assuming $T \neq 0$, we have for $a, c \in \mathbb{R}\setminus\{0\}$:
$d = -a$
$b = \frac{-a^2}{c}$
That means for any two non-zero $a, c$, you can construct a $T$ with the property you are looking for. That is:
$T = \begin{bmatrix}a & \frac{-a^2}{c}\\c & -a\end{bmatrix}$
Ata Keskin
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1$b = 0 \wedge c = 0 ;\vee; a = -d$ is ambiguous. I also think you are using $\land$ where you should use $\lor$ and vice versa. (This is easier to get right if you actually use the commands
\lorand\land.) – Arthur Dec 18 '21 at 18:21 -
Well afaik, $\land$ always binds stronger than $\lor$. Apart from that, the condition for $c(a+d)$ and $b(a+d)$ being both zero is the same as $c = 0 = b$ or $(a+d)=0$. We have: $c(a+d) = 0 \implies c = 0 \lor (a+d) = 0$. Assuming $(a+d) \neq 0$, we can use the equality $b(a+d) = 0$, and arrive at $b = 0$. I'm extremely certain that what I wrote is true. Can you provide a counter-example perhaps? – Ata Keskin Dec 18 '21 at 21:10
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You're right that you didn't mix them up. I was wrong there. But I am pretty certain that the vast majority of times I've seen $\land$ and $\lor$ in the same statement, they have been disambiguated with parentheses. People wouldn't do that if they felt there was universal consensus about the order of the two operations. – Arthur Dec 18 '21 at 23:15
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Alright then, I am updating my answer accordingly. Thanks for the advice! – Ata Keskin Dec 19 '21 at 14:03