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Let $X$ be a vector space over $\mathbb{C}$ and $T:X\to X$ a linear transformation with $T=T^2$. $I:X\to X$ is the identity map. Prove that $\text{Ker}(T) = \text{Im}(I-T)$ and that $\text{Im}(T) = \text{Ker}(I-T)$.

I figured if $x \in \text{Ker}(T)$ then $T(x)=0$ so $(I-T)(x)=I(x)=x$, thus $x \in \text{Im}(I-T)$. But I couldn't figure it out the other way around. With the second equality I don't even know where to start. Could someone show me how to do this?

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For the first equality in the other direction to the one mentioned. Let $x \in Im(I - T)$ then $\exists y \in X$ s.t. $x = (I - T)(y) = y - T(y)$ but then $T(x) = T(y - T(y)) = T(y) - T^2(y) = 0$. And so $x \in Ker(T)$.

The second equality is very similar. Let $x \in Im(T)$ then $\exists y \in X$ s.t. $x = T(y)$. Then, $(I - T)(x) = x - T(x) = T(y) - T(x) = T(y) - T^2(y) = T(y) - T(y) = 0$ and so $x \in Ker(I - T)$. If $x \in Ker(I - T)$ then $0 = (I - T)(x) = x - T(x)$. So $x = T(x) \in Im(T)$.

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for the first one, $$x \in \Im (I-T) \implies \exists y \in X , (I-T)(y)=x$$ equivalently $$T((I-T)(y))=T(x)$$ and then $$T(y)-T^2(y)=T(x) \implies T(x)=0$$ so $x \in Ker T$

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You proved that $Ker(T)\subset Im(I-T)$.

Let $y\in Im(I-T)$ so there is a $x\in X$ such that $(I-T)(x)=y$.

We have $y=(I-T)(x)=I(x)-T(x)=x-T(x)$.

So $T(y)= T(x-T(x)) = T(x)-T^2(x) = T(x)-T(x)=0$.

Therefore $y\in Ker(T)$. So $Im(I-T)\subset Ker(T)$

Can you figure out the other equality?

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