Subspaces

At a Glance

• Frequency: 8 sub-parts across 7 of 14 years (2014, 2016, 2017, 2020, 2021, 2022, 2026)
• Priority tier: T2
• Marks (count): 10 (4), 15 (1), 3 (1)
• Average solve time: ~6 min
• Difficulty mix: easy 4, medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Subspaces appear in Section A every few years — 6 of 13 years, always for 3–15 marks — and the questions divide cleanly into two archetypes that each have a mechanical procedure. The first archetype is entirely driven by one formula: $\dim(U+W) = \dim U + \dim W - \dim(U\cap W)$; once you know it, dimension questions become arithmetic. The second archetype is a three-step proof that is identical every time: show $\mathbf{0}\in S$, closure under addition, closure under scalar multiplication. Master these two tools and subspace questions are free marks.

Minimum Theory

Subspace definition and test. A non-empty subset $S$ of a vector space $V$ over $\mathbb{R}$ is a subspace if and only if for all $\mathbf{u},\mathbf{v}\in S$ and all $\alpha,\beta\in\mathbb{R}$, $\alpha\mathbf{u}+\beta\mathbf{v}\in S.$ Equivalently (and most useful for exam proofs): (i) $\mathbf{0}\in S$, (ii) $S$ is closed under addition ($\mathbf{u}+\mathbf{v}\in S$), and (iii) $S$ is closed under scalar multiplication ($c\mathbf{u}\in S$). A subset defined by a system of homogeneous linear equations always passes this test — the zero vector satisfies every homogeneous equation, and linearity ensures closure.

Kernels are subspaces. The solution set of $A\mathbf{x}=\mathbf{0}$ (the null space / kernel of $A$) is always a subspace of $\mathbb{R}^n$. This is the key structural fact: any set of constraints of the form $L(\mathbf{x})=\mathbf{0}$ for a linear map $L$ is a subspace. UPSC regularly disguises subspaces as solution sets of systems.

Intersection and sum. If $U,W$ are subspaces of $V$ then $U\cap W$ (vectors in both) is a subspace, and $U+W = \{\mathbf{u}+\mathbf{w}: \mathbf{u}\in U, \mathbf{w}\in W\}$ (all sums) is a subspace. Their dimensions satisfy the dimension formula: $\dim(U+W) = \dim U + \dim W - \dim(U\cap W).$ Rearranging gives bounds on the intersection: $\dim U + \dim W - \dim V \;\le\; \dim(U\cap W) \;\le\; \min(\dim U,\dim W).$

Question Archetypes

intersection-sum-dimension (3 question(s); 2014, 2016, 2017)

Compute or bound dim of intersection and sum of subspaces via the dimension formula

Recognition Cues

• The question explicitly mentions intersection ($U\cap W$) or sum ($U+W$) of two or more named subspaces.
• It asks to find or determine the possible values of a dimension.
• Subspaces are given as spans of vectors, as solution sets of linear systems, or described abstractly (e.g. “a 4-dimensional subspace of a 6-dimensional space”).

Solution Template

1. Identify $\dim U$ and $\dim W$ directly or by row-reduction.
2. Find $\dim(U+W)$ by checking whether $U+W$ fills $V$, or by row-reducing the combined spanning set.
3. Apply the formula: $\dim(U\cap W) = \dim U + \dim W - \dim(U+W)$.
4. For abstract problems: use the bounds $\max(\dim U,\dim W)\le\dim(U+W)\le\dim V$ to enumerate possible values, then eliminate values ruled out by extra conditions (e.g. distinctness).
5. Exhibit an example for each remaining value to confirm it is actually attained.

Worked Example 1

2014 Paper 1, 2014-P1-Q1a (10 marks)

$V$ is the $xy$-plane in $\mathbb{R}^3$ and $W = \operatorname{span}\{(1,2,3),(1,-1,1)\}$. Find a vector that generates $V\cap W$.

Step 1 — Describe $W$ explicitly. A general element of $W$ is $\mathbf{w} = a(1,2,3)+b(1,-1,1) = (a+b,\;2a-b,\;3a+b).$

Step 2 — Impose membership in $V$. $V$ is the $xy$-plane, so membership requires the $z$-coordinate to be zero: $3a+b=0 \;\Longrightarrow\; b=-3a.$

Step 3 — Substitute back. With $b=-3a$: $\mathbf{w} = (a-3a,\;2a+3a,\;0) = a(-2,5,0).$

Conclusion. $V\cap W = \operatorname{span}\{(-2,5,0)\}$ is one-dimensional, generated by $(-2,5,0)$.

Dimension check: $\dim V = 2$, $\dim W = 2$, $\dim(V+W)=3$ (since $W$ has a non-zero $z$-component while $V$ does not, so $V+W=\mathbb{R}^3$). The formula gives $\dim(V\cap W)=2+2-3=1$. Consistent.

Worked Example 2

2016 Paper 1, 2016-P1-Q1b-ii (3 marks)

$W_1:\,x+y-z=0$, $W_2:\,3x+y-2z=0$, $W_3:\,x-7y+3z=0$. Find $\dim(W_1\cap W_2\cap W_3)$ and $\dim(W_1+W_2)$.

Form the matrix of normal vectors and row-reduce: $M = \begin{pmatrix}1&1&-1\\3&1&-2\\1&-7&3\end{pmatrix} \xrightarrow{R_2-3R_1,\,R_3-R_1} \begin{pmatrix}1&1&-1\\0&-2&1\\0&-8&4\end{pmatrix} \xrightarrow{R_3-4R_2} \begin{pmatrix}1&1&-1\\0&-2&1\\0&0&0\end{pmatrix}.$

$\operatorname{rank}(M)=2$, so the solution space of all three equations simultaneously has $\dim(W_1\cap W_2\cap W_3) = 3-2 = \boxed{1}$.

For $W_1+W_2$: their normal vectors $(1,1,-1)$ and $(3,1,-2)$ are non-parallel, so $W_1\ne W_2$; both are 2-dimensional planes in $\mathbb{R}^3$ that share a line, hence $\dim(W_1+W_2) = \boxed{3}$.

Worked Example 3

2017 Paper 1, 2017-P1-Q2d (10 marks)

$U$ and $W$ are distinct 4-dimensional subspaces of a 6-dimensional vector space $V$. Find all possible values of $\dim(U\cap W)$.

Step 1 — Apply the dimension formula. $\dim(U\cap W) = \dim U + \dim W - \dim(U+W) = 4+4-\dim(U+W) = 8-\dim(U+W).$

Step 2 — Bound $\dim(U+W)$. Since $U+W\subseteq V$ and $\dim V=6$: $4 \;\le\; \dim(U+W) \;\le\; 6 \qquad\Longrightarrow\qquad 2\;\le\;\dim(U\cap W)\;\le\; 4.$

Step 3 — Eliminate the value 4. If $\dim(U\cap W)=4$, then $U\cap W$ is a 4-dimensional subspace of $U$; but $\dim U=4$, so $U\cap W = U$, forcing $U\subseteq W$. Similarly $W\subseteq U$, giving $U=W$. This contradicts the hypothesis that $U$ and $W$ are distinct. So $\dim(U\cap W)\ne 4$.

Step 4 — Exhibit subspaces for the remaining values.

• $\dim(U\cap W)=3$: Take $V=\mathbb{R}^6$; $U=\operatorname{span}\{e_1,e_2,e_3,e_4\}$, $W=\operatorname{span}\{e_1,e_2,e_3,e_5\}$. Then $U\cap W=\operatorname{span}\{e_1,e_2,e_3\}$, dimension 3.

• $\dim(U\cap W)=2$: Take $W=\operatorname{span}\{e_1,e_2,e_5,e_6\}$. Then $U\cap W=\operatorname{span}\{e_1,e_2\}$, dimension 2.

Conclusion. $\dim(U\cap W)\in\{2,3\}$.

Common Traps

• Off-by-one in the formula. The dimension formula is $\dim U + \dim W - \dim(U\cap W)$, not a sum. Write it out before plugging in numbers.
• Confusing rank with nullity. When subspaces are given as solution sets of $A\mathbf{x}=\mathbf{0}$, use the nullity $= n - \operatorname{rank}(A)$, not the rank itself.
• Forgetting to exhibit examples. For an “all possible values” question, stating the range is not sufficient — you must show each value is actually attained.
• Distinctness eliminates equality. $U\ne W$ rules out $U\cap W$ having the same dimension as $U$ (or $W$) whenever they have equal dimension, because that would force $U=W$.

subspace-verification (3 question(s); 2020, 2021, 2022)

Apply the subspace test (zero vector + closure); find bases and dimension

Recognition Cues

• The question says “show $S$ is a subspace”, “prove $S$ is a vector space over $\mathbb{R}$”, or “verify the set is a subspace”.
• A set $S$ is described explicitly (by a parametric formula or by conditions on coordinates).
• Additional parts ask to “find a basis” and the “dimension”.

Solution Template

1. Identify the ambient space (usually $\mathbb{R}^n$ or $M_n(\mathbb{R})$).
2. Check $\mathbf{0}\in S$: substitute the zero element and verify all conditions are satisfied.
3. Check closure under addition: take arbitrary $\mathbf{u},\mathbf{v}\in S$ (satisfying the defining condition) and verify $\mathbf{u}+\mathbf{v}$ satisfies the same condition. Use linearity.
4. Check closure under scalar multiplication: take $\mathbf{u}\in S$ and $c\in\mathbb{R}$; verify $c\mathbf{u}\in S$.
5. Conclude $S$ is a subspace by the subspace criterion.
6. Find a basis: write a general element of $S$ as a linear combination of a minimal spanning set; identify the vectors in the combination.
7. State the dimension as the number of vectors in the basis.

Worked Example 1

2020 Paper 1, 2020-P1-Q1a (10 marks)

Show that the set $M$ of all $n\times n$ real magic squares is a vector space over $\mathbb{R}$. Give two distinct $2\times 2$ magic squares.

A magic square (in the linear-algebra sense used here) is an $n\times n$ real matrix in which all row sums and all column sums are equal. The set of all such matrices is a subset of $M_n(\mathbb{R})$; it suffices to show it is a subspace.

Subspace proof. Let $A,B\in M$ (both are magic with equal line sum $s_A$ and $s_B$ respectively) and $c\in\mathbb{R}$.

• Zero matrix: The $n\times n$ zero matrix has every line sum equal to $0$, so $\mathbf{0}\in M$.
• Closure under addition: Each row (and column) of $A+B$ has sum $s_A + s_B$ — the same for every row and column, so $A+B\in M$.
• Closure under scalar multiplication: Each line sum of $cA$ equals $c\,s_A$, the same for every row and column, so $cA\in M$.

Since $M$ is a non-empty subset of the vector space $M_n(\mathbb{R})$ closed under addition and scalar multiplication, $M$ is a subspace — hence a vector space over $\mathbb{R}$.

Two distinct $2\times 2$ magic squares. For $n=2$, a magic square requires all four row and column sums to be equal. Let the common sum be $k$; then every entry can be taken equal to $k/2$, giving the constant matrix $\frac{k}{2}J$ where $J$ is the all-ones matrix. Thus every $2\times 2$ magic square is a constant matrix:

$A = \begin{pmatrix}1&1\\1&1\end{pmatrix}, \qquad B = \begin{pmatrix}3&3\\3&3\end{pmatrix}.$

Both have equal line sums ($2$ and $6$ respectively), and $A\ne B$.

Worked Example 2

2021 Paper 1, 2021-P1-Q2c (15 marks)

Let $S = \{(x,2y,3x) : x,y\in\mathbb{R}\}$. Show that $S$ is a subspace of $\mathbb{R}^3$, find two distinct bases for $S$, and determine its dimension.

Part (i) — $S$ is a subspace.

• Zero vector: $(0,0,0) = (0,2\cdot 0,3\cdot 0)\in S$ (take $x=y=0$). $\checkmark$
• Closure under addition: Let $\mathbf{u}=(x_1,2y_1,3x_1)$ and $\mathbf{v}=(x_2,2y_2,3x_2)$ be in $S$. Then $\mathbf{u}+\mathbf{v} = (x_1+x_2,\; 2(y_1+y_2),\; 3(x_1+x_2)) \in S \quad\text{(with }x = x_1+x_2,\; y = y_1+y_2\text{)}.\; \checkmark$
• Closure under scalar multiplication: For $c\in\mathbb{R}$, $c\mathbf{u} = (cx_1,\; 2(cy_1),\; 3(cx_1)) \in S \quad\text{(with }x = cx_1,\; y = cy_1\text{)}.\; \checkmark$

Hence $S$ is a subspace of $\mathbb{R}^3$.

Part (ii) — Spanning set. Every element of $S$ has the form $(x,2y,3x) = x(1,0,3) + y(0,2,0).$ The two vectors $(1,0,3)$ and $(0,2,0)$ span $S$ and are clearly linearly independent (neither is a scalar multiple of the other). Hence

$\mathbf{B}_1 = \{(1,0,3),\;(0,2,0)\}$

is a basis for $S$.

Part (iii) — A second basis. Any two linearly independent vectors in $S$ form a basis. Take $x=2,y=0$: gives $(2,0,6)$; take $x=0,y=1$: gives $(0,2,0)$. These are independent (not scalar multiples), so

$\mathbf{B}_2 = \{(2,0,6),\;(0,1,0)\}$

is a second basis for $S$. (Note $(0,1,0) = \frac{1}{2}(0,2,0)\in S$ since $x=0,y=\frac{1}{2}$.)

Part (iv) — Dimension. Both bases contain exactly 2 vectors, so $\dim S = 2$.

Worked Example 3

2022 Paper 1, 2022-P1-Q3a-i (10 marks)

Let $P = \{\mathbf{x}\in\mathbb{R}^3 : x-y-z=0,\; 2x-y+z=0\}$. Prove that $P$ is a subspace of $\mathbb{R}^3$.

$P$ is the solution set of the homogeneous linear system $A\mathbf{x}=\mathbf{0}$ with $A = \begin{pmatrix}1&-1&-1\\2&-1&1\end{pmatrix}.$ We apply the three-part subspace test directly.

Zero vector. $(0,0,0)$ satisfies both equations ($0-0-0=0$ and $2\cdot 0-0+0=0$), so $\mathbf{0}\in P$. $\checkmark$

Closure under addition. Let $\mathbf{u}=(x_1,y_1,z_1)\in P$ and $\mathbf{v}=(x_2,y_2,z_2)\in P$, so $x_1-y_1-z_1=0,\quad 2x_1-y_1+z_1=0,\quad x_2-y_2-z_2=0,\quad 2x_2-y_2+z_2=0.$ Consider $\mathbf{u}+\mathbf{v}=(x_1+x_2,\,y_1+y_2,\,z_1+z_2)$: $(x_1+x_2)-(y_1+y_2)-(z_1+z_2) = (x_1-y_1-z_1)+(x_2-y_2-z_2) = 0+0=0,$ $2(x_1+x_2)-(y_1+y_2)+(z_1+z_2) = (2x_1-y_1+z_1)+(2x_2-y_2+z_2) = 0+0=0.$ So $\mathbf{u}+\mathbf{v}\in P$. $\checkmark$

Closure under scalar multiplication. Let $\mathbf{u}=(x_1,y_1,z_1)\in P$ and $c\in\mathbb{R}$; consider $c\mathbf{u}=(cx_1,cy_1,cz_1)$: $cx_1-cy_1-cz_1 = c(x_1-y_1-z_1) = c\cdot 0 = 0,$ $2cx_1-cy_1+cz_1 = c(2x_1-y_1+z_1) = c\cdot 0 = 0.$ So $c\mathbf{u}\in P$. $\checkmark$

Since $P$ satisfies all three conditions, $P$ is a subspace of $\mathbb{R}^3$. $\blacksquare$

Common Traps

• Using the recreational definition of magic squares. The recreational definition (distinct integers 1 through $n^2$) does not form a vector space — two such squares do not add to another square with distinct integer entries. The linear-algebra definition (equal line sums, any real entries) is required here.
• Forgetting the $z=3x$ constraint. For $S=\{(x,2y,3x)\}$ the third coordinate is tied to the first, not free. Students who write general $(x,y,z)$ miss the constraint and obtain $\dim S=3$ incorrectly.
• Testing only two conditions. UPSC expects all three: zero vector, addition, scalar multiplication. Showing only two earns partial credit.
• Not verifying $\mathbf{0}\in S$. For solution sets of homogeneous systems this is obvious, but you must still state it explicitly for full marks.

direct-sum-complement (1 question(s); 2026)

Recognition Cues

• “Find a subspace $K$ such that $V=H\oplus K$” — a construction, not a verification.
• $H$ is usually the kernel of a single linear functional (an integral, a trace, a coefficient sum), so $\dim H=\dim V-1$ and $K$ can be one-dimensional.

Solution Template

1. Find $\dim H$ — for a single non-trivial linear condition it is $\dim V-1$.
2. Pick any vector $v\notin H$; set $K=\operatorname{span}\{v\}$. Choose the simplest such $v$.
3. Prove $H\cap K=\{0\}$: a general element of $K$ is $cv$; the condition defining $H$ forces $c=0$.
4. Prove $H+K=V$ by dimension: $\dim H+\dim K=\dim V$ together with $H\cap K=\{0\}$ gives the direct sum.
5. Say explicitly that $K$ is not unique.

Worked Example(s)

2026 Paper 1, 2026-P1-Q3a-ii (7 marks)

$H=\left\{p\in P_3[x]\ :\ \int_{-1}^{1}p(x)\,dx=0\right\}$. Find $K$ with $P_3[x]=H\oplus K$.

For $p=a_0+a_1x+a_2x^2+a_3x^3$, the odd powers integrate to zero over $[-1,1]$, so $\int_{-1}^{1}p\,dx=2a_0+\tfrac23a_2 .$ The condition $2a_0+\tfrac23a_2=0$ is one non-trivial linear equation on a 4-dimensional space, so $\dim H=3$, with basis $\{\,x,\ x^3,\ x^2-\tfrac13\,\}$.

Take $K=\operatorname{span}\{1\}$. Then for $p=c\in K$, $\int_{-1}^{1}c\,dx=2c$, which vanishes only if $c=0$; hence $H\cap K=\{0\}$. Since $\dim H+\dim K=3+1=4=\dim P_3[x]$, $\boxed{\;P_3[x]=H\oplus K,\qquad K=\operatorname{span}\{1\}.\;}$

Any polynomial with non-zero integral would serve equally — $K=\operatorname{span}\{x^2\}$ works too. Say so.

Common Traps

• Trying to “solve” for $K$ as if it were determined. It is not: any complement works, and stating that earns the mark rather than losing it.
• Forgetting the odd powers integrate to zero and carrying $a_1,a_3$ into the condition.
• Asserting $H\oplus K$ without checking $H\cap K=\{0\}$. Dimensions alone are not enough — you need both.

Marks-Aware Writing

For a 10-mark subspace proof: Show all three parts of the subspace test — zero vector, closure under addition, closure under scalar multiplication — as separate, labelled steps. Each step should display the algebra explicitly (do not say “by linearity” and skip the calculation). A correct conclusion with no intermediate algebra earns at most 3–4 marks.

For finding a basis and dimension (part of a 10- or 15-mark question): Write the general element in the form $x\mathbf{v}_1 + y\mathbf{v}_2 + \ldots$, identify $\mathbf{v}_1,\mathbf{v}_2,\ldots$ as your spanning set, then state why they are linearly independent. Dimension = number of basis vectors — state this explicitly.

For a 15-mark question: The extra marks reward completeness. Write all parts in full: the subspace proof, both bases with explicit independence arguments, and the dimension statement. A one-sentence basis identification without the independence check will lose marks.

For a 3-mark dimension question: Be concise. Row-reduce the matrix, read off the rank, apply nullity $= n - \operatorname{rank}$. A single well-annotated row-reduction with the final answer is sufficient.

Practice Set

• 2014-P1-Q2a (15 m) — — Hint: write a general element as a linear combination, identify spanning vectors, prove independence to extract a basis.
• 2025-P1-Q4c-ii (8 m) — — Hint: apply the dimension formula to bound $\dim(U\cap W)$, then check which values are ruled out by any extra constraints given.