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What is the relationship between $S \oplus T$ and $T \oplus S$? Is the direct sum operation commutative? Formulate and prove a similar statement concerning associativity. Is there an "identitiy" for direct sum? What about "negatives"?

$S \oplus T = T \oplus S$ because elements of $S \oplus T$ are of the form $s+t$ where $s \in S$ and $t \in T$, and we know that $s + t = t + s$. Associativity is valid with the same argument. The identity is $\{0\}$. However, we cannot have negatives, because a negative subspace doesn't make sense, since if $s \in S$ then $-s \in S$.

Let V be a finite-dimensional vector space over an infinite field F. Prove that if $S_1, ... ,S_k$ are subspaces of $V$ of equal dimension, then there is a subspace $T$ of $V$ for which $V = S_i \oplus T$ for all $I = 1, ... ,k$. In other words, $T$ is a common complement of the subspaces $S_i$.

We know that every subspace has a complement. So let's say that the complement of $S_m$ where $1 \leq m \leq k$ is $T$. Then $$\dim(V) = \dim(S_m) + \dim(T).$$

For any $S_n$ where $1 \leq n \leq k$, we know that $$\dim(S_n) + \dim(T) = \dim(S_n + T) + \dim(S \cap T).$$

Since $\dim(S_n) = \dim(S_m)$,

$$(\dim(V) - \dim(T)) + \dim(T) = \dim(S_n + T) + \dim(S \cap T)$$

$$\implies \dim(V) = \dim(S_n + T) + \dim(S \cap T) = \dim(S_n) + \dim(T)$$

So $T$ must also be a complement of $S_n$ for all $1 \leq n \leq k$.

Do you think my answer is correct?

Thank you in advance

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    $\begingroup$ In the general algebraic sense of direct sum, the two are (nicely) isomorphic, but not same. $\endgroup$ – André Nicolas Jun 28 '13 at 2:25
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    $\begingroup$ To prove that negatives with respect to $\oplus$ do not exist, suppose that $S\oplus T = \{0\}$ for some nontrivial $S$, and derive a contradiction; since $0\in T$, $S\subseteq \{0\}$. $\endgroup$ – vadim123 Jun 28 '13 at 2:25
  • $\begingroup$ In general, you should ask separate questions in separate threads. $\endgroup$ – Qiaochu Yuan Jun 28 '13 at 2:39
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    $\begingroup$ @Artus: in the first question, does $\oplus$ denote the internal or the external direct sum? $\endgroup$ – Qiaochu Yuan Jun 28 '13 at 2:39
  • $\begingroup$ @QiaochuYuan Well, it doesn't really specify whether it's external or internal, but I'm guessing that it's internal... $\endgroup$ – user58289 Jun 28 '13 at 2:47
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$$\implies \dim(V) = \dim(S_n + T) + \dim(S \cap T) = \dim(S_n) + \dim(T)$$

So $T$ must also be a complement of $S_n$ for all $1 \leq n \leq k$.

This conclusion is incorrect. For example, consider $V=M_2(\mathbb{R})$, two-by-two matrices with real coefficients. Set $$S_1=\left\{\left(\begin{smallmatrix}a&b\\c&0\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$

$$S_2=\left\{\left(\begin{smallmatrix}a&b\\0&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$

$$S_3=\left\{\left(\begin{smallmatrix}a&0\\b&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$

$$S_4=\left\{\left(\begin{smallmatrix}0&a\\b&c\end{smallmatrix}\right):a,b,c\in\mathbb{R}\right\}$$

Then $T=\left\{\left(\begin{smallmatrix}0&0\\0&d\end{smallmatrix}\right):d\in\mathbb{R}\right\}$ has $S_1\oplus T=V$, but $S_2\oplus T\neq V$.

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  • $\begingroup$ Thanks a lot, but I don't really understand why it is incorrect...because I just used the formula in the textbook... $\endgroup$ – user58289 Jun 28 '13 at 7:39
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    $\begingroup$ The dimension formula is correct, but the conclusion "So $T$ must be a complement of $S_n$" is incorrect. The reason is that $S_n+T$ might not equal $V$, as my example above shows. $\endgroup$ – vadim123 Jun 28 '13 at 13:06

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