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I am currently reading Linear Algebra Done Right by Sheldon Axler, and I have stumbled upon some proposition that I have trouble verifying.

Excerpt from the book:

Proposition. Suppose $\boldsymbol V$ is finite dimensional and $U_1,...,U_m$ are subspaces of $\boldsymbol V$ such that: $\boldsymbol V = U_1 + ··· + U_m$

and

$\dim \boldsymbol V = \dim U_1 +\cdots+ \dim U_m$.

Then $\boldsymbol V = U_1 \oplus \cdots \oplus U_m$.


So I tried to verify this with an actual example from the book itself that was given in a chapter before.

Say we use these three subspaces:

\begin{align} U_1 =& \big\{(x,y,0) ∈ F^3 : x,y ∈ F\big\},\\ U_2 =& \big\{(0,0,z) ∈ F^3 : z ∈ F\big\},\\ U_3 =& \big\{(0,y,y) ∈ F^3 : y ∈ F\big\}. \end{align}

Then:

$\boldsymbol V = U_1 + U_2 + U_3$ (ok),

$\dim \boldsymbol V = \dim U_1 + \dim U_2 + \dim U_3 = 1 + 1 + 1 = 3$ (ok)

But:

$\boldsymbol V ≠ U_1 ⊕ U_2 ⊕ U_3$ (not ok)


Could somebody explain me what I am confusing here?

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    $\begingroup$ check the dimension of U1 $\endgroup$
    – DGRasines
    Mar 8 '14 at 23:54
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The mistake is that $\dim U_1\ne 1$.

In particular, $$ U_1=\{(x,y,0):x,y\in F\}=\mathrm{span}\,\{(1,0,0),(0,1,0)\}, $$ and hence $$ \dim U_1=2. $$

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  • $\begingroup$ Thanks! Makes sense now :) $\endgroup$ Mar 9 '14 at 14:21

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