# Math Help - prove or disprove direct sum question

1. ## prove or disprove direct sum question

Prove or disprove

$W, U_1, U_2$ are subspaces of $V$.

If $V=U_1 \oplus U_2$

then $W = (U_1 \cap W) \oplus (U_2 \cap W)$

Attempt:

False

let's say $dimV = 8, dimU_1 = dimU_2 = 4 (dimV = dimU_1 + dimU_2)$
and $dimW = 5$

then $dim(U_1 \cap W)$ may equal 4
and $dim(U_2 \cap W)$ may equal 4

and then we get $5 \neq 4+4$

2. **Is the question understandable? Please let me know if there is something I need to explain better.

Thanks!

3. Originally Posted by jayshizwiz
Prove or disprove

$W, U_1, U_2$ are subspaces of $V$.

If $V=U_1 \oplus U_2$

then $W = (U_1 \cap W) \oplus (U_2 \cap W)$

Attempt:

False

let's say $dimV = 8, dimU_1 = dimU_2 = 4 (dimV = dimU_1 + dimU_2)$
and $dimW = 5$

then $dim(U_1 \cap W)$ may equal 4
and $dim(U_2 \cap W)$ may equal 4

and then we get $5 \neq 4+4$

You haven't disproved anything: "may" it's not existence. You have to come up with a particular example that shows

clearly that the claim is false.

Hint: Take a look at $\mathbb{R}^2\,,\,\,U_1=\left\{\,\binom{x}{x}\in\ma thbb{R}^2\right\}\,,\,U_2=\left\{\,\binom{x}{-x}\in\mathbb{R}^2\right\}\,,\,\,W=\left\{\binom{x} {0}\in\mathbb{R}^2\right\}$

Tonio

4. Originally Posted by tonio
You haven't disproved anything: "may" it's not existence. You have to come up with a particular example that shows

clearly that the claim is false.

Hint: Take a look at $\mathbb{R}^2\,,\,\,U_1=\left\{\,\binom{x}{x}\in\ma thbb{R}^2\right\}\,,\,U_2=\left\{\,\binom{x}{-x}\in\mathbb{R}^2\right\}\,,\,\,W=\left\{\binom{x} {0}\in\mathbb{R}^2\right\}$

Tonio
I truly hope you aren't grading my exam (;

i sohld get at least partial credit

5. The misconception that some students have, and that this problem is attempting to dispel, is that the direct sum is like a (set) union. But in fact, the direct sum of U1 and U2 contains a whole lot of elements that are not in U1 or U2. In case Tonio's excellent example wasn't clear, think about this and read it again.