If $\{0_v\} \neq T \leq S_1$ and $S_1 \oplus S_2 = S_1 \oplus S_3$, then $T+S_2 = T+S_3$?

I tried this way:

Let $S_1=span\{(1,0,0),(0,1,0)\}$ and $T$, it's subspace, $=span\{(1,0,0)\}$.

Also lets have $S_2=span\{(0,0,1)\}$ and $S_3=span\{(1,1,1)\}$

Therefore, $S_1 \cap S_2 = \{0_v\}$ and $S_1 \cap S_3 = \{0_v\}$ so we have the direct sum.

Thus, $T+S_2 = span\{(1,0,0)\} +span\{(0,0,1)\} \neq T+S_3 = span\{(1,0,0)\} +span\{(1,1,1)\}$

Since for example, the vector $(1,1,1)$ is not contained in $T+S_2$

How does it look?

  • $\begingroup$ Looks good to me! $\endgroup$ – zipirovich Nov 4 '18 at 20:17
  • 1
    $\begingroup$ Your counterexample looks good! There are even counterexamples with the $S_i$ subspaces of $\Bbb{R}^2$. $\endgroup$ – Servaes Nov 4 '18 at 20:57
  • $\begingroup$ By $$S_1 \oplus S_2 = S_2 \oplus S_3$$ did you mean $$S_1 \oplus S_2 = S_1 \oplus S_3\ ?$$ $\endgroup$ – bof Nov 4 '18 at 23:42
  • $\begingroup$ @bof yes I meant that thanks for figurint it out..! It's still correct ? $\endgroup$ – iggykimi Nov 5 '18 at 19:00

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