7. Linear Algebra

Linear sum of subspaces

The linear sum of subspaces is a fundamental algebraic operation in vector spaces that combines two or more subspaces into a larger vector space containing all their vector additions.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Modular Identity for Subspaces (Grassmann's Dimension Theorem)

Let $W_1$ and $W_2$ be finite-dimensional subspaces of a vector space $V$. Then $W_1 + W_2$ is a subspace of $V$, and \begin{align*} \dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) \end{align*}

📐 Geometric Interpretation

Geometrically in $\mathbb{R}^3$, if $W_1$ and $W_2$ are distinct planes passing through the origin, their sum $W_1 + W_2$ spans the entire space $\mathbb{R}^3$ (dimension 3), their intersection $W_1 \cap W_2$ is a line (dimension 1), and their individual dimensions are both 2. Thus, $3 = 2 + 2 - 1$.

⚙️ Algebraic & Structural Meaning

It establishes the exact relationship between the generators of sums and intersections of algebraic structures, providing the foundational bridge to direct sum decompositions $V = W_1 \oplus W_2$ where $\dim(W_1 \cap W_2) = 0$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Finite-Dimensionality of Subspaces	`Algebraic finiteness condition`	Strictly Mandatory / Necessary	If $W_1$ or $W_2$ is infinite-dimensional, the dimension formula $\dim(W_1+W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$ becomes indeterminate (involving infinite cardinals).
Subspace Closure under Vector Addition and Scalar Multiplication	`Subspace axioms`	Strictly Mandatory	If subsets are arbitrary and not closed under vector addition, $W_1 + W_2$ fails to form a vector subspace, rendering dimension formulas inapplicable.
Field Consistency	`Scalar field $\mathbb{F}$`	Essential for basis expansion	Mismatched underlying scalar fields (e.g., treating a complex vector space over $\mathbb{R}$ vs $\mathbb{C}$) invalidates linear independence counts.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Grassmann Dimension Calculation

$$\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$$

Rule: If any three dimensions are given, instantly calculate the fourth without constructing bases.

Application: Frequently asked in UP PGT where dimensions of two planes or polynomial spaces are given.

Direct Sum Condition Test

$$W_1 \cap W_2 = \{0\} \implies \dim(W_1 \oplus W_2) = \dim(W_1) + \dim(W_2)$$

Rule: Check if intersection contains only the zero vector. If true, the sum is direct and dimensions simply add up.

Application: Directly tested in DSSSB PGT and KVS PGT matrix/subspace questions.

Codmention Rule for Hyperplanes

$$\text{codim}(W_1 \cap W_2) \le \text{codim}(W_1) + \text{codim}(W_2)$$

Rule: Use codimension arithmetic when dealing with solution spaces of homogeneous linear equations.

Symmetric and Skew-Symmetric Matrix Spaces

$$M_n(\mathbb{R}) = \text{Sym}_n(\mathbb{R}) \oplus \text{Skew}_n(\mathbb{R})$$

Rule: The space of all $n \times n$ matrices is the direct sum of symmetric and skew-symmetric matrices with dimensions $\frac{n(n+1)}{2} + \frac{n(n-1)}{2} = n^2$.

Application: Direct numerical questions in UP PGT Mathematics.

Union of Subspaces Rule

$$W_1 \cup W_2 \text{ is a subspace } \iff W_1 \subseteq W_2 \text{ or } W_2 \subseteq W_1$$

Rule: Never confuse linear sum ($W_1 + W_2$) with union ($W_1 \cup W_2$). Union is rarely a subspace unless one contains the other.

Application: Classic conceptual trap question in UP TGT/PGT.

Solution Space Dimension (Rank-Nullity Link)

$$\dim(\text{Null}(A)) = n - \text{rank}(A)$$

Rule: Combine rank-nullity with intersection of null spaces corresponding to row-combined matrices.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming Union of Subspaces is a Subspace

Common Pitfall: Candidates often assume that the union of two subspaces $W_1 \cup W_2$ is also a subspace.

Counterexample: Let $W_1 = \{(x, 0) : x \in \mathbb{R}\}$ and $W_2 = \{(0, y) : y \in \mathbb{R}\}$ in $\mathbb{R}^2$.

Why it fails: Their union contains $(1, 0)$ and $(0, 1)$, but their sum $(1, 1)$ does not lie in $W_1 \cup W_2$. Thus $W_1 \cup W_2$ is not closed under addition.

🛡️ Exam Defense: Remember: The union of two subspaces is a subspace if and only if one is contained in the other.

⚠️ COMMON ERROR #02

Ignoring Intersection Dimension in Grassmann's Formula

Common Pitfall: Applying $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2)$ without checking whether $W_1 \cap W_2 = \{0\}$.

Counterexample: Two identical planes $W_1 = W_2$ in $\mathbb{R}^3$.

Why it fails: If $W_1 = W_2$, $W_1 + W_2 = W_1$, so dimension is 2, not $2 + 2 = 4$.

🛡️ Exam Defense: Always subtract $\dim(W_1 \cap W_2)$ unless explicitly stated that the sum is a direct sum.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Linear Sum	Any two subspaces $W_1, W_2$	$\dim(W_1+W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$	Span of combined geometric objects	Keywords: 'sum of subspaces', 'span'
Direct Sum	$W_1 \cap W_2 = \{0\}$	$V = W_1 \oplus W_2 \implies \dim(V) = \dim(W_1) + \dim(W_2)$	Independent orthogonal or oblique components	Keywords: 'direct sum', 'unique representation'
Union of Subspaces	Subspaces $W_1, W_2$	$W_1 \cup W_2$ is a subspace $\iff W_1 \subseteq W_2$ or $W_2 \subseteq W_1$	Set-theoretic union of lines/planes	Keywords: 'union', 'when is union a subspace'
Orthogonal Complement	Inner product space $V$, Subspace $W$	$V = W \oplus W^\perp$, $\dim(W) + \dim(W^\perp) = \dim(V)$	Perpendicular geometric complement	Keywords: 'inner product', 'orthogonal complement'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $W_1$ and $W_2$ are two subspaces of a vector space $V$ such.that $\dim(W_1) = 3$, $\dim(W_2) = 4$, and $\dim(W_1 \cap W_2) = 1$, then the dimension of $W_1 + W_2$ is:

(A) 7(B) 6(C) 5(D) 8

📚 Method A: Formal Textbook Derivation

Apply Grassmann's dimension formula: $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) = 3 + 4 - 1 = 6$.

⚡ Method B: 25-Second Exam Shortcut

Mental arithmetic: $3 + 4 - 1 = 6$. Done in 5 seconds.

💡 Tip: Direct formula substitution. Never forget to subtract the intersection.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $W_1$ be the space of symmetric matrices of order $n$ and $W_2$ be the space of skew-symmetric matrices of order $n$. Then $\dim(W_1 + W_2)$ is equal to:

(A) $\frac{n(n-1)}{2}$(B) $\frac{n(n+1)}{2}$(C) $n^2$(D) $n^2 - 1$

📚 Method A: Formal Textbook Derivation

Every $n \times n$ matrix can be uniquely written as the sum of a symmetric and a skew-symmetric matrix. Thus $M_n(\mathbb{R}) = W_1 \oplus W_2$. The dimension is $\frac{n(n+1)}{2} + \frac{n(n-1)}{2} = n^2$.

⚡ Method B: 25-Second Exam Shortcut

Total number of independent entries in an $n \times n$ matrix is $n^2$. Since $M_n(\mathbb{R}) = W_1 \oplus W_2$, the sum dimension is $n^2$.

💡 Tip: Recognize standard matrix decompositions instantly.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $W_1$ and $W_2$ are subspaces of $\mathbb{R}^5$ with $\dim(W_1) = 3$ and $\dim(W_2) = 3$, what is the minimum possible dimension of $W_1 \cap W_2$?

A 0

B 1

C 2

D 3

Derivation & Analysis:

Since $\dim(W_1 + W_2) \le \dim(\mathbb{R}^5) = 5$, by Grassmann's formula: $5 \ge \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) = 3 + 3 - \dim(W_1 \cap W_2) \implies \dim(W_1 \cap W_2) \ge 1$.

MCQ 2: Let $W_1, W_2$ be distinct 2-dimensional subspaces of a 3-dimensional vector space $V$. What is $\dim(W_1 + W_2)$?

A 1

B 2

C 3

D 4

Derivation & Analysis:

Since $W_1 \neq W_2$ are distinct 2-dimensional subspaces in $\mathbb{R}^3$, their intersection is a 1-dimensional line. Thus $\dim(W_1 + W_2) = 2 + 2 - 1 = 3 = \dim(V)$.

MCQ 3: The union of two subspaces $W_1$ and $W_2$ of a vector space $V$ is a subspace if and only if:

A $W_1 \cap W_2 = \{0\}$

B $W_1 \subseteq W_2 \text{ or } W_2 \subseteq W_1$

C $W_1 + W_2 = V$

D $W_1$ and $W_2$ are orthogonal

Derivation & Analysis:

Standard theorem: The union of two subspaces is a subspace if and only if one is contained in the other.

MCQ 4: If $V$ is a finite-dimensional vector space and $W_1, W_2$ are subspaces such that $V = W_1 \oplus W_2$, then:

A $W_1 \cap W_2 = V$

B $W_1 \cap W_2 = \{0\} \text{ and } W_1 + W_2 = V$

C $W_1 = W_2$

D $W_1$ and $W_2$ must be orthogonal

Derivation & Analysis:

By definition of a direct sum, $V = W_1 \oplus W_2$ implies $W_1 \cap W_2 = \{0\}$ and $W_1 + W_2 = V$.

MCQ 5: Let $\dim(W_1) = 4$, $\dim(W_2) = 5$, and $\dim(W_1 + W_2) = 7$. Then $\dim(W_1 \cap W_2)$ is:

A 1

B 2

C 3

D 0

Derivation & Analysis:

Using Grassmann's identity: $7 = 4 + 5 - \dim(W_1 \cap W_2) \implies \dim(W_1 \cap W_2) = 9 - 7 = 2$.

MCQ 6: If $W_1, W_2, W_3$ are subspaces of a finite-dimensional vector space, which of the following is always a subspace?

A $W_1 \cup W_2$

B $W_1 \setminus W_2$

C $W_1 + W_2$

D All of the above

Derivation & Analysis:

The sum of subspaces $W_1 + W_2$ is always a subspace, whereas the union and set-difference in general are not.

MCQ 7: In the vector space of polynomials $\mathbb{R}_n[x]$, let $W_1$ be polynomials of degree $\le 2$ and $W_2$ be polynomials of degree $\le 3$. What is $W_1 + W_2$?

A $W_1$

B $W_2$

C $W_1 \cap W_2$

D $\mathbb{R}_3[x]$

Derivation & Analysis:

Since $W_1 \subseteq W_2$, their sum $W_1 + W_2 = W_2$, which is $\mathbb{R}_3[x]$.

MCQ 8: If $\dim(W_1) = p$ and $\dim(W_2) = q$, what is the maximum possible dimension of $W_1 + W_2$ in an $n$-dimensional space $V$?

A $\min(p+q, n)$

B $p+q$

C $n$

D $|p-q|$

Derivation & Analysis:

The dimension of $W_1 + W_2$ cannot exceed the dimension of the ambient space $n$, nor can it exceed $p+q$. Thus $\min(p+q, n)$.

MCQ 9: Let $W_1$ and $W_2$ be subspaces of $V$ such that $\dim(V) = 8$, $\dim(W_1) = 5$, and $\dim(W_2) = 4$. What is the minimum possible dimension of $W_1 \cap W_2$?

A 0

B 1

C 2

D 3

Derivation & Analysis:

$\dim(W_1 + W_2) \le 8$. By Grassmann: $8 \ge 5 + 4 - \dim(W_1 \cap W_2) \implies \dim(W_1 \cap W_2) \ge 1$.

Q10

MCQ 10: If $W_1$ and $W_2$ are distinct hyperplanes in an $n$-dimensional vector space $V$ ($n \ge 2$), then $\dim(W_1 + W_2)$ is:

A $n-1$

B $n$

C $2n-2$

D $2n-1$

Derivation & Analysis:

A hyperplane has dimension $n-1$. Two distinct hyperplanes in $V$ span the entire space $V$ unless they are identical. Thus $\dim(W_1 + W_2) = n$.

Common questions

No. Since $W_1 + W_2$ is a subspace of $V$, its dimension is bounded above by $\dim(V)$. In fact, $\dim(W_1 + W_2) \le \min(\dim(V), \dim(W_1) + \dim(W_2))$.-.

Forgetting to subtract $\dim(W_1 \cap W_2)$ when the subspaces are not known to be in direct sum, or incorrectly assuming $\dim(W_1 \cap W_2) = 0$ by default.

Memorize standard dimensions (e.g., symmetric/skew-symmetric matrices, diagonal matrices, polynomial spaces) and apply Grassmann's formula directly using mental arithmetic.