Which is a SUBSPACE of $\mathbb{R}^3$?
A$\{(x, y, z) : x + y + z = 1\}$
B$\{(x, y, z) : x = y = 1\}$
C$\{(x, y, z) : x \geq 0\}$
D$\{(x, y, z) : x + y + z = 0\}$
Answer & Solution
Correct answer: D. $\{(x, y, z) : x + y + z = 0\}$
1. A SUBSPACE must (a) contain $0$, (b) be closed under addition, (c) be closed under scalar mult.
2. Option B: $\{x + y + z = 0\}$. Contains $0 = (0,0,0)$ since $0 + 0 + 0 = 0$. Closed: if $x+y+z=0$ and $x'+y'+z'=0$, then $(x+x')+(y+y')+(z+z')=0$. Scalar: $\alpha(x+y+z) = 0$. ✓ Subspace.
3. Option A: $x+y+z=1$. Does NOT contain $(0,0,0)$. Not a subspace.
4. Option C: $x \geq 0$. Not closed under scalar mult (multiplying by $-1$ violates the constraint).
5. Option D: $x=y=1$. Doesn't contain origin.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §1.2 (Subspaces)._
Related questions
Let $A$ be a $3 \times 3$ matrix with eigenvalues 1, 2, 3. Then $\det(A)$ equalsIf $A$ is invertible, the eigenvalues of $A^{-1}$ areThe TRACE of $A = \begin{pmatrix} 2 & 5 \\ 7 & 3 \end{pmatrix}$ isGRAM-SCHMIDT process converts a basis intoTwo vectors $u, v$ are ORTHOGONAL iffAn ORTHOGONAL matrix $Q$ satisfiesFor a SYMMETRIC real matrix $A$, all eigenvalues areCAYLEY-HAMILTON THEOREM states that for any $n \times n$ matrix $A$ with characteristic po