Let $V$ be a finite linear subspace and $T$ be a linear transformation defined like this: $T:V \to V$ such that $\ker T^2 \subseteq \ker T$

Prove that: $V = \ker T \oplus \operatorname{im}T$

What I did is:

It's known that: $V = \operatorname{im}(T) + \ker(T)$ so all I need to prove is that:

$$\operatorname{im}(T) \cap \ker(T) = \{0\}$$

So I said that because $V$ is finite:

$$\dim(V) =N$$ $$\dim(\operatorname{im}(T)) + \dim(\ker(T)) = N$$

According to the dimensions theorem:

$$\begin{align*} \dim(\operatorname{im}(T) + \ker(T)) &= \dim(\operatorname{im}(T)) + \dim(\ker(T)) - \dim(\operatorname{im}(T) \cap \ker(T))\\\\ \dim(\operatorname{im}(T) \cap \ker(T)) &= \dim(\operatorname{im}(T)) + \dim(\ker(T)) - \dim(\operatorname{im}(T) + \ker(T)) \\\\ \dim(\operatorname{im}(T) \cap \ker(T)) &= 0 \end{align*}$$

But for some reason I didn't use the fact that $\ker T^2 \subseteq \ker T$, so I must have been wrong here.

Can someone please help me understand how to prove it?


  • 1
    $\begingroup$ Why do you think you know that $V=\ker(T)+\mathrm{im}(T)$? $\endgroup$ – Zev Chonoles Jul 17 '15 at 16:25
  • 2
    $\begingroup$ @Omnomnomnom: I'm pretty sure it's your inclusion that always holds: $T(x) = 0 \implies T(T(x)) = 0$ $\endgroup$ – john Jul 17 '15 at 16:30
  • $\begingroup$ @ZevChonoles, can't I say that a subspace is consisted from it's image and kernel? I thought that's trivial, if not, how else would I prove it? $\endgroup$ – FigureItOut Jul 17 '15 at 16:47
  • 2
    $\begingroup$ @FigureItOut: Well, you accepted the answer posted below, so I assume you are now familiar with the example of the map $\newcommand{\im}{\text{im}}$$T:\mathbb{R}^2\to\mathbb{R}^2$ defined by $$T\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} y\\ 0 \end{bmatrix}$$ which is represented by the matrix $$\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$$ The kernel and image of this $T$ are both $$\ker(T)=\im(T)=\left\{\begin{bmatrix} t\\0\end{bmatrix}:t\in\mathbb{R}\right\}\subset\mathbb{R}^2$$ and therefore $$\ker(T)+\im(T)=\ker(T)=\im(T)$$ is strictly smaller than the entire space $\mathbb{R}^2$. $\endgroup$ – Zev Chonoles Jul 17 '15 at 16:51

From your response you seem to think that $$\dim(\operatorname{im}(T)+\ker(T))=\dim(\operatorname{im}(T))+\dim(\ker(T));$$ but as pointed out in the comments this is not necessarily the case (consider for example the linear map $T:\mathbb{R}^2\to \mathbb{R}^2$ defined by the matrix $\begin{pmatrix} 0 & 1\\0 &0 \end{pmatrix}$, where $$\operatorname{im}(T)=\ker(T)=\mathbb{R}\cdot\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

To prove that $\operatorname{im}(T)\cap \ker(T)=\{0\}$, consider an element $x\in \operatorname{im}(T)\cap \ker(T)$; then $T(x)=0$ and $x=T(y)$ for some $y\in V$. Thus $T^2(y)=T(x)=0$, which means $y\in \ker(T^2)$; since $\ker(T^2)\subset \ker(T)$, $y\in \ker(T)$, and thus $x=T(y)=0$.

  • $\begingroup$ Yes thank you. It's fixed now. $\endgroup$ – Arnaud D. Jul 17 '15 at 16:41

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