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Definition: $Z$ is called an algebraic complement of $Y$ in $X$ if $X=Y\oplus Z$.

Example: $Y=\Bbb R$ is a subspace of the Euclidean plane $\Bbb R^2$. "Clearly, $Y$ has infinitely many algebraic complements in $\Bbb R^2$."

I don't find this entirely clear, unless this is it:

$Y = \Bbb R = \{(y,0): y\in \Bbb R\}$, $Z_a=\{(a,z):z\in \Bbb R\}$, then we obtain $X=Y\oplus Z_a, \forall a\in \Bbb R$, is that what they are getting at?

Functional Analysis
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    Is $Z_a$ a subspace of $\mathbb{R}^2$? Try to think about lines of different slopes instead. – Najib Idrissi Sep 09 '15 at 11:39
  • @NajibIdrissi Oh doh, that was stupid of me, of course we need $(0,0)$ to occur, which pins $a=0$ in that configuration, or alternatively we can set any $(az,bz)$ instead, this gives us an actual linear subspace. – Functional Analysis Sep 09 '15 at 11:42
  • @NajibIdrissi I would be thankful if you were still here to verify the above is correct – Functional Analysis Sep 09 '15 at 11:44
  • You say $\mathbb{R}$ is a subspace of $\mathbb{R}^2$... Maybe you should read this http://math.stackexchange.com/questions/897775/is-mathbbr-a-subspace-of-mathbbr2/897788#897788. – Amitai Yuval Sep 09 '15 at 11:45

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As Najib says, the set $Z_a$ is not a linear subspace whenever $a\neq0$.

However, yes, the space $Y$ in your question has infinitely many algebraic complements in $\mathbb{R}^2$. In fact, for any $v\in\mathbb{R}^2\setminus Y$, we have $\mathbb{R}^2=Y\oplus\mathrm{span}(v)$.

Amitai Yuval
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