Corrected exercises: subspaces, span and linear independence (MATH 133)

a) Zero: $0 - 2(0) + 3(0) = 0$, so $(0, 0, 0) \in U_1$. Addition: take ANY $u = (x_1, y_1, z_1)$ and $v = (x_2, y_2, z_2)$ in $U_1$, so $x_1 - 2y_1 + 3z_1 = 0$ and $x_2 - 2y_2 + 3z_2 = 0$. Then $u + v = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$ and $(x_1 + x_2) - 2(y_1 + y_2) + 3(z_1 + z_2) = (x_1 - 2y_1 + 3z_1) + (x_2 - 2y_2 + 3z_2) = 0 + 0 = 0$, so $u + v \in U_1$. Scalars: for any real $c$, $cx_1 - 2cy_1 + 3cz_1 = c(x_1 - 2y_1 + 3z_1) = c \cdot 0 = 0$, so $cu \in U_1$. Hence $U_1$ is a subspace: a plane through the origin. Checking that $(2, 1, 0)$ and $(1, 2, 1)$ add up correctly proves nothing; the marks are for the letters.

b) Zero first, because it is the cheapest test: $(0, 0)$ would need $0 = 2(0) + 1 = 1$, which is false. So $(0, 0) \notin U_2$ and $U_2$ is not a subspace. That single line is a complete answer. If you want a second witness, $(0, 1)$ and $(1, 3)$ are in $U_2$ but their sum $(1, 4)$ is not, since $2(1) + 1 = 3 \neq 4$. Geometrically $U_2$ is a line that misses the origin: a line is a subspace of $\mathbb{R}^2$ only when it passes through $0$.

c) Zero: $(0, 0)$ satisfies both inequalities, so $0 \in U_3$. Addition: if $x_1, y_1, x_2, y_2 \ge 0$ then $x_1 + x_2 \ge 0$ and $y_1 + y_2 \ge 0$, so $U_3$ IS closed under addition, proved with letters. Scalars fail, and the failure needs numbers: $u = (1, 2) \in U_3$, but $(-1)u = (-1, -2)$ has negative coordinates, so it is not in $U_3$. The trap is to write that $U_3$ is not closed because negative numbers exist: name the vector and the scalar, or the mark is not given.

d) Zero: $0 \cdot 0 = 0$, so $(0, 0) \in U_4$. Scalars: if $xy = 0$ then $(cx)(cy) = c^2 xy = 0$, so $U_4$ is closed under scalar multiplication. Addition fails: $(1, 0)$ and $(0, 1)$ are in $U_4$ (each has a zero coordinate), but $(1, 0) + (0, 1) = (1, 1)$ and $1 \cdot 1 = 1 \neq 0$. On the figure, $U_4$ is the union of the two axes: each axis is a subspace, and the sum of a vector from one with a vector from the other lands off both.

e) $U_3$ passes zero and addition and fails scalars; $U_4$ passes zero and scalars and fails addition. So neither closure condition is a consequence of the other, and a proof that checks only one of them is incomplete. The zero test is a FILTER: failing it ends the question, as in b), but passing it decides nothing, as both $U_3$ and $U_4$ show. In fact, for a non-empty set, closure under scalars already forces $0 = 0 \cdot u \in U$; that is why the zero test is a quick first check rather than a separate hurdle.

a) Zero: $0 + 0 + 0 = 0$ and $2(0) - 0 = 0$. Addition: if $u = (x_1, y_1, z_1)$ and $v = (x_2, y_2, z_2)$ satisfy both equations, then $(x_1 + x_2) + (y_1 + y_2) + (z_1 + z_2) = 0 + 0 = 0$ and $2(x_1 + x_2) - (z_1 + z_2) = (2x_1 - z_1) + (2x_2 - z_2) = 0$. Scalars: $cx_1 + cy_1 + cz_1 = c \cdot 0 = 0$ and $2cx_1 - cz_1 = c \cdot 0 = 0$. So $W$ is a subspace; it is the intersection of two planes through the origin. To describe it, solve: the second equation gives $z = 2x$, then $y = -x - z = -3x$. With $x = t$: $W = \{t(1, -3, 2)\}$, a line through the origin. Check: $1 - 3 + 2 = 0$ and $2(1) - 2 = 0$.

b) Zero passes: $0 = 0^2$. That is exactly why this set is dangerous: it looks fine at the origin. Scalars fail: $(1, 0, 1) \in P$ since $1 = 1^2$, but $2(1, 0, 1) = (2, 0, 2)$ and $2^2 = 4 \neq 2$. The square is the culprit: multiplying $x$ by $c$ multiplies $x^2$ by $c^2$, not by $c$. $P$ is a parabolic surface, curved, and no curved set is a subspace.

c) Zero passes and scalars pass: $|cx| = |c||x| = |c||y| = |cy|$. Addition fails, and the witness must be chosen so that the signs disagree: $(1, 1, 0)$ and $(1, -1, 0)$ are both in $A$, but their sum is $(2, 0, 0)$ and $|2| \neq |0|$. Geometrically $A$ is the union of the two planes $y = x$ and $y = -x$, the same phenomenon as the two axes in exercise 1, one dimension higher.

d) The set is a plane with normal $(1, 2, -1)$. It contains $0$ exactly when $0 + 0 - 0 = k^2 - 4$, that is $k^2 = 4$. If $k^2 \neq 4$, the origin is missing and the answer is no. If $k^2 = 4$, the equation is $x + 2y - z = 0$, homogeneous, and the proof of exercise 1 a) applies word for word, so the set is a subspace. Answer: $k = 2$ or $k = -2$. Writing only $k = 2$ loses a mark on almost every paper: $k^2 = 4$ has two roots.

e) The trap is to see $k$ on the right-hand side and conclude that the right side is a non-zero constant. It is not a constant, it multiplies an unknown. Move it across: $x + y + z - kz = 0$, that is $x + y + (1 - k)z = 0$, a homogeneous linear equation for every $k$. Zero, addition and scalars then follow exactly as in exercise 1 a), with $3$ replaced by $1 - k$. For $k = 1$ the equation becomes $x + y = 0$, still a plane through the origin (it contains the whole $z$-axis). So the set is a subspace for every real $k$.

a) Zero: $A0 = 0$, so $0 \in \text{null}(A)$. Addition: if $Ax = 0$ and $Ay = 0$, then $A(x + y) = Ax + Ay = 0 + 0 = 0$, by distributivity of the matrix product. Scalars: $A(cx) = c(Ax) = c0 = 0$. Each line cites the property of matrix multiplication it uses, and that citation is what the marker looks for. Note that the proof never looks at an entry of $A$: it works for every matrix at once, which is what a proof with letters buys.

b) Here the elements are OUTPUTS, so each must be written as $A$ times something. Zero: $0 = A0 \in \text{im}(A)$. Addition: two elements of the image are $Ax$ and $Ay$ for some $x, y \in \mathbb{R}^n$, and $Ax + Ay = A(x + y)$ is again $A$ times a vector, so it is in the image. Scalars: $c(Ax) = A(cx) \in \text{im}(A)$. The common slip is to write 'let $b \in \text{im}(A)$, then $Ab \in \dots$': $b$ lives in $\mathbb{R}^m$ and $Ab$ usually does not even exist.

c) For $x = (x_1, \dots, x_n)$, the product $Ax$ is the linear combination $x_1c_1 + x_2c_2 + \cdots + x_nc_n$ of the columns. So the set of all $Ax$ is the set of all combinations of the columns: $\text{im}(A) = \text{span}\{c_1, \dots, c_n\}$. Translation: $b \in \text{im}(A)$ exactly when the system $Ax = b$ is consistent. Every span question in this chapter is therefore a system, and exercise 4 uses it.

d) Row reduce the coefficient matrix: $R_2 \to R_2 - R_1$ gives $\begin{pmatrix} 1 & 1 & -1 \\ 0 & 1 & 2 \end{pmatrix}$, then $R_1 \to R_1 - R_2$ gives $\begin{pmatrix} 1 & 0 & -3 \\ 0 & 1 & 2 \end{pmatrix}$. The variable $z = t$ is free, $y = -2t$, $x = 3t$. So $\text{null}(A) = \{t(3, -2, 1) : t \in \mathbb{R}\}$. Check: $3 - 2 - 1 = 0$ and $3 - 4 + 1 = 0$. Geometrically, it is the line through the origin where the two planes $x + y - z = 0$ and $x + 2y + z = 0$ meet.

e) Same row operations on $(A \mid b)$: $R_2 \to R_2 - R_1$ gives $(0 \; 1 \; 2 \mid 1)$, then $R_1 \to R_1 - R_2$ gives $(1 \; 0 \; -3 \mid 1)$. So $x = 1 + 3t$, $y = 1 - 2t$, $z = t$, that is $x = (1, 1, 0) + t(3, -2, 1)$. Not a subspace: $A0 = 0 \neq b$, so $0$ is not a solution. That witness is enough; another is that $(1, 1, 0)$ is a solution but $2(1, 1, 0) = (2, 2, 0)$ gives $A(2, 2, 0) = (4, 6) \neq b$. The solution set is the line $\text{null}(A)$ translated by the particular solution $(1, 1, 0)$: parallel to it, not through the origin, exactly like $x - 2y = 4$ beside $x - 2y = 0$ on the figure. Indeed the difference of two solutions is in the null space: $A(p - q) = b - b = 0$.

a) Augmented matrix $\begin{pmatrix} 1 & 2 & 0 \\ 2 & 5 & 1 \\ -1 & 0 & 2 \end{pmatrix}$. $R_2 \to R_2 - 2R_1$ gives $(0 \; 1 \mid 1)$; $R_3 \to R_3 + R_1$ gives $(0 \; 2 \mid 2)$; $R_3 \to R_3 - 2R_2$ gives $(0 \; 0 \mid 0)$. No row $0 = c$ with $c \neq 0$, so the system is consistent. Back substitution: $a_2 = 1$, then $a_1 + 2a_2 = 0$ gives $a_1 = -2$. Check, always: $-2(1, 2, -1) + (2, 5, 0) = (0, 1, 2)$. The check takes ten seconds and is the only protection against an arithmetic slip that would turn a yes into a wrong combination.

b) Same row operations, new last column: $R_2 \to R_2 - 2R_1$ gives $9 - 8 = 1$; $R_3 \to R_3 + R_1$ gives $-3 + 4 = 1$; $R_3 \to R_3 - 2R_2$ gives $1 - 2 = -1$. The last row reads $0a_1 + 0a_2 = -1$: inconsistent, so $b' \notin \text{span}\{v_1, v_2\}$. The trap is the argument 'neither $v_1$ nor $v_2$ is a multiple of $b'$, so it is not in the span': it is not an argument at all, since $(0, 1, 2)$ in a) is not a multiple of either and IS in the span.

c) Carry letters in the last column and use the same three operations: $R_2 \to R_2 - 2R_1$ gives $y - 2x$; $R_3 \to R_3 + R_1$ gives $z + x$; $R_3 \to R_3 - 2R_2$ gives $z + x - 2(y - 2x) = 5x - 2y + z$. The last row is $(0 \; 0 \mid 5x - 2y + z)$, so the system is consistent exactly when $5x - 2y + z = 0$. This one computation answers a) and b) at once: $0 - 2 + 2 = 0$ and $20 - 18 - 3 = -1$, the very $-1$ found in b).

d) $\text{span}\{v_1, v_2\}$ is the plane through the origin with equation $5x - 2y + z = 0$. A span is always a subspace, so it had to pass through the origin, and it does. Checks: $v_1$ gives $5 - 4 - 1 = 0$ and $v_2$ gives $10 - 10 + 0 = 0$. A second check from the previous chapter: the normal $(5, -2, 1)$ is exactly $v_1 \times v_2$. Two non-parallel vectors always span a plane; two parallel ones only a line.

e) $v_3$: $5(1) - 2(3) + 1 = 0$, so $v_3$ lies in the plane already, and indeed $v_3 = -v_1 + v_2$ (check: $(-1 + 2, -2 + 5, 1 + 0) = (1, 3, 1)$). Adding it adds nothing: $\text{span}\{v_1, v_2, v_3\}$ is still the plane $5x - 2y + z = 0$. More vectors never guarantee a bigger span. $v_4$: $0 - 0 + 1 = 1 \neq 0$, so $v_4$ is off the plane. Row reducing $(v_1 \; v_2 \; v_4 \mid (x, y, z))$ with the same operations gives the last row $(0 \; 0 \; 1 \mid 5x - 2y + z)$: there is a pivot in every row, so the system is consistent for every right-hand side, and $\text{span}\{v_1, v_2, v_4\} = \mathbb{R}^3$.

a) Put the vectors as COLUMNS and solve $Ac = 0$: $\begin{pmatrix} 1 & 2 & 0 \\ -1 & 1 & 3 \\ 2 & 1 & -3 \end{pmatrix}$. $R_2 \to R_2 + R_1$ gives $(0 \; 3 \; 3)$; $R_3 \to R_3 - 2R_1$ gives $(0 \; -3 \; -3)$; $R_3 \to R_3 + R_2$ gives a zero row. Only two pivots for three unknowns, so $c_3$ is free: $c_2 = -c_3$ and $c_1 = -2c_2 = 2c_3$. With $c_3 = 1$: $(c_1, c_2, c_3) = (2, -1, 1)$. Relation: $2(1, -1, 2) - (2, 1, 1) + (0, 3, -3) = (0, 0, 0)$, checked coordinate by coordinate: $2 - 2 + 0$, $-2 - 1 + 3$, $4 - 1 - 3$. Stopping at 'there is a free variable, so dependent' earns the decision but not the relation, and most exams ask for both.

b) $c_1(1, 0, 2) + c_2(3, 1, 4) = (0, 0, 0)$ gives three equations: $c_1 + 3c_2 = 0$, $c_2 = 0$, $2c_1 + 4c_2 = 0$. The second forces $c_2 = 0$, then the first forces $c_1 = 0$. Only the trivial solution, so the family is independent. Note the logic: the trivial solution ALWAYS exists; what matters is that it is the only one. Writing '$c_1 = c_2 = 0$ is a solution, so independent' proves nothing.

c) $(-3, 6, -9) = -\frac{3}{2}(2, -4, 6)$, so the relation $3(2, -4, 6) + 2(-3, 6, -9) = (6 - 6, -12 + 12, 18 - 18) = 0$ is non-trivial: dependent. For TWO vectors the test is exactly this: $\{u, v\}$ is dependent if and only if one is a scalar multiple of the other. Geometrically they lie on one line through the origin. The test works for two vectors only; exercise 8 shows it failing for three.

d) Look for $w = au + bv$: $a + 3b = 5$ and $2a - b = 3$. From the second, $b = 2a - 3$; substituting, $a + 6a - 9 = 5$, so $a = 2$ and $b = 1$. So $w = 2u + v$ and the relation is $2u + v - w = 0$, checked: $(2 + 3 - 5, 4 - 1 - 3) = (0, 0)$. Before computing: $c_1u + c_2v + c_3w = 0$ is a homogeneous system of $2$ equations in $3$ unknowns, which has at most $2$ pivots, so at least one free variable and a non-trivial solution. Any three vectors of $\mathbb{R}^2$ are dependent, and more generally more than $n$ vectors of $\mathbb{R}^n$ always are.

e) $0 \cdot (1, 4, 2) + 1 \cdot (0, 0, 0) + 0 \cdot (3, 0, 1) = (0, 0, 0)$, and the coefficients $(0, 1, 0)$ are not all zero: a non-trivial relation, so the family is dependent. The same line works for any family $\{0, v_2, \dots, v_k\}$: put coefficient $1$ on the zero vector and $0$ everywhere else. The trap is to think the relation needs every coefficient non-zero; 'non-trivial' only means NOT ALL zero.

a) Starting at $O$, walk $3$ steps along $v$ to $(6, 3)$, then $2$ steps along $w$ to $(6 - 2, 3 + 2) = (4, 5)$. So $b = 3v + 2w$ on the grid. Confirm: $a(2, 1) + b(-1, 1) = (4, 5)$ gives $2a - b = 4$ and $a + b = 5$; adding, $3a = 9$, so $a = 3$ and $b = 2$. The grid gives a guess, the system gives the proof: a point between grid lines would have fractional coefficients that no picture reads reliably.

b) Solve $2a - b = p$, $a + b = q$ for arbitrary $(p, q)$. Adding the equations: $3a = p + q$, so $a = \frac{p + q}{3}$, then $b = q - a = \frac{2q - p}{3}$. These exist for EVERY $(p, q)$, so every vector of $\mathbb{R}^2$ is a combination of $v$ and $w$: $\text{span}\{v, w\} = \mathbb{R}^2$. Check with a): $p = 4$, $q = 5$ gives $a = 3$, $b = 2$. Letters are compulsory here; showing that three particular vectors are reachable proves nothing about the rest.

c) $S_1$: $(-4, 2, -6) = -2(2, -1, 3)$, so both vectors lie on one line through the origin and $\text{span}\,S_1$ is that line, $\{t(2, -1, 3)\}$. Two vectors, but only a line: count what is independent, not what is listed. $S_2$: a combination is $x(1, 0, 2) + y(0, 1, -1) = (x, y, 2x - y)$, so the span is the set of $(x, y, z)$ with $z = 2x - y$, the plane $2x - y - z = 0$. $S_3$: $(1, 1, 1) = (1, 0, 2) + (0, 1, -1)$, and indeed $2 - 1 - 1 = 0$, so the third vector is already in the plane and $\text{span}\,S_3$ is the same plane.