Which of the following sets of vectors \(\alpha = (a_1, \ldots, a_n)\) in \(\RR^n\) (\(n \geq 3\)) are subspaces of \(\RR^n\)?
all \(\alpha\) such that \(a_1 \geq 0\);
all \(\alpha\) such that \(a_1 + 3a_2 = a_3\);
all \(\alpha\) such that \(a_2 = a_1^2\);
all \(\alpha\) such that \(a_1 a_2 = 0\);
all \(\alpha\) such that \(a_2\) is rational.
Solution
Only (b) is a subspace.
1. The set in (a) is not a subspace.
It contains \((1,0,\ldots,0)\) but not its negative \((-1,0,\ldots,0)\).
2. The set in (b) is a subspace.
It is the kernel of the linear functional \[ (a_1,\ldots,a_n)\longmapsto a_1+3a_2-a_3. \] Equivalently, the defining homogeneous linear equation is preserved under vector addition and scalar multiplication.
3. The set in (c) is not a subspace.
The vectors \((1,1,0,\ldots,0)\) and \((-1,1,0,\ldots,0)\) satisfy \(a_2=a_1^2\), but their sum \((0,2,0,\ldots,0)\) does not.
4. The set in (d) is not a subspace.
Both \((1,0,0,\ldots,0)\) and \((0,1,0,\ldots,0)\) satisfy \(a_1a_2=0\), but their sum \((1,1,0,\ldots,0)\) does not.
5. The set in (e) is not a real vector subspace.
The vector \((0,1,0,\ldots,0)\) belongs to it, but multiplying by the irrational scalar \(\sqrt2\) produces second coordinate \(\sqrt2\), which is not rational.