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I have a "wrong" counterexample to the following statement in linear algebra but I don't see why it's wrong: let $T:V\to W$ be a linear map between vector spaces. Then, $V$ is the direct sum of $\textrm{im}(T)$ and $\ker(T)$.

Let $V$ be the space of polynomials over a field with degree less than or equal to n, and let $T \colon V \to V$ be the differential operator. Then $\textrm{im}(T)$ is the space of polynomials with degree less than or equal to $n-1$, and $\ker(T)$ is the base field. Their direct sum is a proper subspace of $V$.

msteve
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Vladimir
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  • Their sum isn't direct. For example, $;1=Tx;,;;T(1)=0\implies 1\in\ker T\cap\text{Im},T;$ ...but this is a counterexample to the fact that that sum is direct! – Timbuc Mar 13 '15 at 20:28
  • There is a typo: $T$ should be a map $V \to V$ – Crostul Mar 13 '15 at 20:29
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    It is not true that $V$ is that direct sum; in fact $\mathrm{Im}(T)$ is not even a subspace of $V$. – Marc van Leeuwen Mar 13 '15 at 20:38
  • I feel even more confused after reading all the comments and answers... – Vladimir Mar 13 '15 at 20:42
  • @Vladimir That may mean you need to go over the basics again. – Timbuc Mar 13 '15 at 20:42
  • @Vladimir we are having trouble reading your question. Do you believe the statement, or the counterexample? Why? – hunter Mar 13 '15 at 20:49
  • @hunter I now understand I gave a wrong counterexample to a wrong statement... Now I just wonder if someone could give me a good reference on this (a nice treatment of this in a linear algebra book). – Vladimir Mar 13 '15 at 20:55

2 Answers2

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The statement of the theorem should be the following: let $T \colon V \to V$ be a linear transformation such that $\textrm{ker}(T) \cap \textrm{im}(T) = 0$, then $$ V = \textrm{ker}(T) \oplus \textrm{im}(T). $$ As mentioned in the comments above, this theorem does not directly apply to the given linear transformation $T \colon V \to V$, because the subspaces $\textrm{ker}(T)$ and $\textrm{im}(T)$ intersect nontrivially (their intersection is the constants).

However, it is true that $V$ is isomorphic to $U \oplus W$ as vector spaces, where $U \simeq \textrm{ker}(T)$ and $W = \textrm{im}(T)$. The isomorphism is not equality, in the sense that $V$ is not the direct sum of the subspaces $\textrm{ker}(T) \subset V$ and $\textrm{im}(T)$.

msteve
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  • In the example given by the OP the sum is not direct at all to begin with. Read the comments below the question. – Timbuc Mar 13 '15 at 20:33
  • This is true, just to add something to this. It seems like OP may be confusing inner direct products with external direct products. Which is understandable as the notation is the same. – Bill Trok Mar 13 '15 at 20:33
  • I think this answer is wrong as it appears now, but can easily be turned into correct saying that what is true is that $;V\cong U\oplus W;$ , with $;U\cong\ker T;,;;W\cong\text{Im},T;$ . This follows at once from the dimensions theorem, and in fact one can choose $;U=\ker T;$ . – Timbuc Mar 13 '15 at 20:35
  • Is this not the same as saying $V \simeq \textrm{ker}(T) \oplus \textrm{im}(T)$? You're just giving a (possibly) different isomorphism on each factor. – msteve Mar 13 '15 at 20:41
  • @msteve It certainly can depend on agreed meanings, yet in this case it may say that the sum of $;\ker T;,;;\text{Im};T;$ is always direct, which is far from being true. – Timbuc Mar 13 '15 at 20:44
  • Good point @Timbuc, I've edited the post to try to avoid any such ambiguity. – msteve Mar 13 '15 at 21:09
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I think everyone else is reading your question backwards. You believe that a statement about direct sums is true, but you've written down what you know to be a counterexample to it. You want to know what is wrong with your counterexample.

However, the counterexample is correct, because the statement is wrong. First, as was pointed out in the comments, the statement $V$ is the direct sum of $\text{im}(T)$ and $\ker(T)$ doesn't even make sense unless $V = W$.

But even for this, as your counterexample shows, the statement is not true.

What is true is the following: $V$ is the direct sum of $\ker(T)$ and a subspace $V'$ which is mapped isomorphically to $\text{Im}(T)$ via $T$, i.e. $T|_V'$ is an isomorphism to $\text{Im}(T)$.

hunter
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