← Back to GATE Mathematics

Linear Algebra

Master vector spaces, linear transformations, eigenvalue theory, inner product spaces, and canonical forms for GATE Mathematics.

Vector Spaces Linear Transformations Eigenvalues & Diagonalization Inner Product Spaces Canonical Forms
0% complete5 units
01
Vector Spaces

The foundation of linear algebra: understanding vector spaces, subspaces, bases, and dimension over arbitrary fields.

Definition and Axioms

A vector space is the central object of study in linear algebra. It provides the abstract framework for understanding systems of linear equations, transformations, and geometric intuition in higher dimensions.

Definition — Vector Space A vector space $V$ over a field $F$ is a set equipped with two operations — vector addition $(+)$ and scalar multiplication $(\cdot)$ — satisfying 8 axioms: closure under addition and scalar multiplication, commutativity and associativity of addition, existence of additive identity and inverses, compatibility of scalar multiplication with field multiplication, and distributivity.

Common examples include $\mathbb{R}^n$ over $\mathbb{R}$, the space of $m \times n$ matrices $M_{m \times n}(\mathbb{R})$, and the space of polynomials $P_n(\mathbb{R})$ of degree at most $n$.

  • A subspace $W \subseteq V$ is a non-empty subset closed under addition and scalar multiplication
  • The span of a set $S$ is the set of all finite linear combinations of elements in $S$
  • A set is linearly independent if no vector can be written as a linear combination of the others
Basis and Dimension

A basis of a vector space $V$ is a linearly independent set that spans $V$. The number of elements in any basis is the dimension of $V$.

Theorem — Invariance of Dimension Every basis of a finite-dimensional vector space has the same number of elements. This common cardinality is called $\dim(V)$.
★ Example
Show that $\beta = \{(1,0,0),\;(0,1,0),\;(0,0,1)\}$ forms a basis for $\mathbb{R}^3$.

Linear independence: Suppose $a_1(1,0,0) + a_2(0,1,0) + a_3(0,0,1) = (0,0,0)$. Then $(a_1, a_2, a_3) = (0,0,0)$, so $a_1 = a_2 = a_3 = 0$. The set is linearly independent.

Spanning: Any vector $(x,y,z) \in \mathbb{R}^3$ can be written as $x(1,0,0) + y(0,1,0) + z(0,0,1)$.

Since $\beta$ is linearly independent and spans $\mathbb{R}^3$, it is a basis. Hence $\dim(\mathbb{R}^3) = 3$.

Change of basis: If $\beta$ and $\beta'$ are two bases for $V$, the change of basis matrix $P$ satisfies $[\mathbf{v}]_{\beta'} = P^{-1}[\mathbf{v}]_{\beta}$, where $P$ is the matrix whose columns are the $\beta$-coordinates of the vectors in $\beta'$.

★ Key Takeaways — Vector Spaces
  • Always verify both closure properties when checking if a subset is a subspace
  • $\dim(\mathbb{R}^n) = n$, $\dim(M_{m \times n}) = mn$, $\dim(P_n) = n+1$
  • A set of $k$ vectors in $\mathbb{R}^n$ is linearly dependent if $k > n$
02
Linear Transformations

Functions between vector spaces that preserve the linear structure, along with their fundamental properties and matrix representations.

Definition, Kernel, and Image

A function $T: V \to W$ between vector spaces is a linear transformation if $T(\alpha u + \beta v) = \alpha T(u) + \beta T(v)$ for all $u, v \in V$ and scalars $\alpha, \beta$.

  • Kernel: $\ker(T) = \{v \in V : T(v) = \mathbf{0}\}$ — a subspace of $V$
  • Image: $\text{Im}(T) = \{T(v) : v \in V\}$ — a subspace of $W$
  • $T$ is injective (one-to-one) if and only if $\ker(T) = \{\mathbf{0}\}$
Theorem — Rank-Nullity Theorem If $T: V \to W$ is a linear transformation and $V$ is finite-dimensional, then $$\dim(\ker T) + \dim(\text{Im}\, T) = \dim V$$ Equivalently, $\text{nullity}(T) + \text{rank}(T) = \dim(V)$.
Matrix Representation

Every linear transformation between finite-dimensional vector spaces can be represented by a matrix. If $T: V \to W$ with ordered bases $\beta$ for $V$ and $\gamma$ for $W$, then $[T]_{\beta}^{\gamma}$ is the matrix whose $j$-th column is $[T(v_j)]_{\gamma}$.

★ Example
Find the kernel of $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x,y,z) = (x+y,\; y+z)$.

Step 1: Set $T(x,y,z) = (0,0)$, giving the system $x + y = 0$ and $y + z = 0$.

Step 2: From the first equation, $x = -y$. From the second, $z = -y$.

Step 3: Let $y = t$ (free variable). Then $(x,y,z) = (-t, t, -t) = t(-1, 1, -1)$.

Therefore $\ker(T) = \text{span}\{(-1, 1, -1)\}$ and $\dim(\ker T) = 1$.

Verification: By rank-nullity, $\text{rank}(T) = 3 - 1 = 2 = \dim(\mathbb{R}^2)$, so $T$ is surjective.

★ Key Takeaways — Linear Transformations
  • The rank-nullity theorem is one of the most frequently tested results in GATE
  • For a square matrix $A$: $\text{rank}(A) + \text{nullity}(A) = n$
  • Changing bases changes the matrix representation: $[T]_{\beta'} = P^{-1}[T]_{\beta}P$
03
Eigenvalues and Diagonalization

Eigenvalue theory is central to matrix analysis and appears heavily in GATE. Topics include characteristic polynomials, diagonalizability criteria, and the Cayley-Hamilton theorem.

Eigenvalues and Eigenvectors

A scalar $\lambda$ is an eigenvalue of a linear operator $T$ (or matrix $A$) if there exists a nonzero vector $v$ such that $Av = \lambda v$. The vector $v$ is the corresponding eigenvector.

Definition — Characteristic Polynomial The characteristic polynomial of an $n \times n$ matrix $A$ is $p(\lambda) = \det(A - \lambda I)$. The eigenvalues of $A$ are the roots of $p(\lambda) = 0$.
  • Algebraic multiplicity: the multiplicity of $\lambda$ as a root of $p(\lambda)$
  • Geometric multiplicity: $\dim(\ker(A - \lambda I))$ — the dimension of the eigenspace
  • For every eigenvalue: $1 \leq \text{geometric mult.} \leq \text{algebraic mult.}$
Diagonalization and Cayley-Hamilton

A matrix $A$ is diagonalizable if there exists an invertible matrix $P$ such that $P^{-1}AP = D$ is diagonal. This happens if and only if the geometric multiplicity equals the algebraic multiplicity for every eigenvalue.

Theorem — Cayley-Hamilton Every square matrix satisfies its own characteristic polynomial. If $p(\lambda) = \det(A - \lambda I)$, then $p(A) = O$ (the zero matrix).
★ Example
Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.

Step 1: Characteristic polynomial: $\det(A - \lambda I) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3)$.

Step 2: Eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 3$.

Step 3 ($\lambda = 1$): Solve $(A - I)v = 0$: $\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$. So $x = -y$, giving eigenvector $v_1 = \begin{pmatrix}-1\\1\end{pmatrix}$.

Step 4 ($\lambda = 3$): Solve $(A - 3I)v = 0$: $\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$. So $x = y$, giving eigenvector $v_2 = \begin{pmatrix}1\\1\end{pmatrix}$.

Since $A$ has 2 distinct eigenvalues in a $2 \times 2$ matrix, $A$ is diagonalizable: $P^{-1}AP = \begin{pmatrix}1&0\\0&3\end{pmatrix}$ where $P = \begin{pmatrix}-1&1\\1&1\end{pmatrix}$.

★ Key Takeaways — Eigenvalues
  • $\text{trace}(A) = \sum \lambda_i$ and $\det(A) = \prod \lambda_i$
  • Real symmetric matrices are always diagonalizable with real eigenvalues
  • Cayley-Hamilton is useful for computing $A^{-1}$ and powers of $A$
04
Inner Product Spaces

Adding geometric structure to vector spaces: inner products, orthogonality, and the Gram-Schmidt process.

Inner Products and Norms
Definition — Inner Product An inner product on a real vector space $V$ is a function $\langle \cdot, \cdot \rangle : V \times V \to \mathbb{R}$ that is (i) linear in the first argument, (ii) symmetric: $\langle u,v \rangle = \langle v,u \rangle$, and (iii) positive definite: $\langle v,v \rangle > 0$ for $v \neq 0$.

The norm induced by an inner product is $\|v\| = \sqrt{\langle v, v \rangle}$.

Theorem — Cauchy-Schwarz Inequality For all $u, v$ in an inner product space: $|\langle u, v \rangle| \leq \|u\| \cdot \|v\|$. Equality holds if and only if $u$ and $v$ are linearly dependent.
Orthogonality and Gram-Schmidt

Vectors $u, v$ are orthogonal if $\langle u, v \rangle = 0$. An orthonormal basis is a basis of mutually orthogonal unit vectors. The Gram-Schmidt process converts any basis into an orthogonal one.

★ Example
Apply the Gram-Schmidt process to $\{v_1 = (1,1,0),\; v_2 = (1,0,1)\}$ in $\mathbb{R}^3$ with the standard inner product.

Step 1: Set $u_1 = v_1 = (1, 1, 0)$.

Step 2: Compute $u_2 = v_2 - \dfrac{\langle v_2, u_1 \rangle}{\langle u_1, u_1 \rangle} u_1$.

$\langle v_2, u_1 \rangle = (1)(1) + (0)(1) + (1)(0) = 1$ and $\langle u_1, u_1 \rangle = 1 + 1 + 0 = 2$.

$u_2 = (1,0,1) - \frac{1}{2}(1,1,0) = \left(\frac{1}{2}, -\frac{1}{2}, 1\right)$.

Step 3 (Normalize): $e_1 = \frac{u_1}{\|u_1\|} = \frac{1}{\sqrt{2}}(1,1,0)$ and $e_2 = \frac{u_2}{\|u_2\|} = \frac{1}{\sqrt{3/2}}\left(\frac{1}{2}, -\frac{1}{2}, 1\right) = \frac{1}{\sqrt{6}}(1, -1, 2)$.

The orthonormal set is $\left\{\frac{1}{\sqrt{2}}(1,1,0),\; \frac{1}{\sqrt{6}}(1,-1,2)\right\}$.

Orthogonal projection: The projection of $v$ onto subspace $W$ is $\text{proj}_W(v) = \sum_{i} \langle v, e_i \rangle e_i$ where $\{e_i\}$ is an orthonormal basis for $W$. This is the foundation of the least squares method.

★ Key Takeaways — Inner Product Spaces
  • Gram-Schmidt is a standard procedure tested in GATE — practice the computation
  • Orthogonal matrices satisfy $Q^T Q = I$ and preserve norms and inner products
  • The least squares solution to $Ax = b$ is $x = (A^T A)^{-1} A^T b$
05
Canonical Forms

The structure theorem for linear operators: Jordan and rational canonical forms provide the deepest insight into the behaviour of matrices.

Jordan Canonical Form

When a matrix is not diagonalizable, the next best structure is the Jordan canonical form. Every complex matrix is similar to a block-diagonal matrix of Jordan blocks.

Definition — Jordan Block A Jordan block $J_k(\lambda)$ is a $k \times k$ upper triangular matrix with $\lambda$ on the diagonal and $1$'s on the superdiagonal: $J_k(\lambda) = \begin{pmatrix} \lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ & & & \lambda \end{pmatrix}$.
  • The minimal polynomial is the monic polynomial of least degree such that $m(A) = O$
  • The minimal polynomial divides the characteristic polynomial, and they share the same roots
  • $A$ is diagonalizable if and only if the minimal polynomial has no repeated roots
Rational Canonical Form and Applications

The rational canonical form (companion matrix form) works over any field and depends only on the invariant factors of $A - \lambda I$. It is unique and does not require knowledge of eigenvalues.

Applications to differential equations: For a system $\mathbf{x}' = A\mathbf{x}$, the Jordan form of $A$ determines the solution structure. Each Jordan block $J_k(\lambda)$ contributes terms of the form $t^j e^{\lambda t}$ for $j = 0, 1, \ldots, k-1$.

★ Key Takeaways — Canonical Forms
  • Jordan form exists for every complex matrix; it is unique up to ordering of blocks
  • The minimal polynomial determines the size of the largest Jordan block for each eigenvalue
  • Rational canonical form is field-independent and uniquely determined by invariant factors
✎ Practice Problems
Problem 1
Let $V$ be the vector space of $2 \times 2$ real symmetric matrices. Find $\dim(V)$.
Show Solution ▼
A $2 \times 2$ symmetric matrix has the form $\begin{pmatrix}a & b \\ b & c\end{pmatrix}$. A basis is $\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&1\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}\right\}$. Therefore $\dim(V) = 3$.
Problem 2
If $A$ is a $5 \times 3$ matrix with $\text{rank}(A) = 2$, find the dimension of the null space of $A$.
Show Solution ▼
By the rank-nullity theorem: $\text{nullity}(A) = 3 - \text{rank}(A) = 3 - 2 = 1$. (Note: the number of columns determines the domain dimension.)
Problem 3
Find the eigenvalues of $A = \begin{pmatrix}0&1&0\\0&0&1\\6&-11&6\end{pmatrix}$.
Show Solution ▼
The characteristic polynomial is $-\lambda^3 + 6\lambda^2 - 11\lambda + 6 = -(\lambda-1)(\lambda-2)(\lambda-3)$. The eigenvalues are $\lambda = 1, 2, 3$.
Problem 4
Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the orthogonal projection onto the plane $x + y + z = 0$. Find $\text{rank}(T)$ and the eigenvalues of $T$.
Show Solution ▼
The plane $x+y+z=0$ is a 2-dimensional subspace, so $\text{rank}(T) = 2$. Since $T$ is a projection, $T^2 = T$, so eigenvalues satisfy $\lambda^2 = \lambda$, giving $\lambda \in \{0, 1\}$. The eigenvalue $1$ has multiplicity 2 (the plane) and $0$ has multiplicity 1 (the normal direction $(1,1,1)$).
Problem 5
If $A$ is a $3 \times 3$ matrix with characteristic polynomial $(\lambda - 2)^3$ and minimal polynomial $(\lambda - 2)^2$, determine the Jordan canonical form of $A$.
Show Solution ▼
The only eigenvalue is $\lambda = 2$ with algebraic multiplicity 3. The minimal polynomial $(\lambda-2)^2$ means the largest Jordan block has size 2. The Jordan form must be $\begin{pmatrix}2&1&0\\0&2&0\\0&0&2\end{pmatrix}$ (one $2\times 2$ block and one $1 \times 1$ block).
⚙ Interactive Quiz
1. What is the dimension of the vector space of all $3 \times 3$ skew-symmetric matrices over $\mathbb{R}$?
A 6
B 3
C 9
D 5
2. If $A$ is a $4 \times 4$ matrix with $\text{rank}(A) = 3$, what is $\dim(\ker A)$?
A 1
B 2
C 3
D 0
3. Which of the following matrices is NOT diagonalizable?
A $I_2$ (identity matrix)
B $\begin{pmatrix}1&1\\0&1\end{pmatrix}$
C $\begin{pmatrix}2&1\\1&2\end{pmatrix}$
D $\begin{pmatrix}3&0\\0&5\end{pmatrix}$
4. The Cayley-Hamilton theorem states that every square matrix:
A Is diagonalizable
B Has minimal polynomial equal to characteristic polynomial
C Satisfies its own characteristic equation
D Has only real eigenvalues
5. If $\{u, v, w\}$ is orthonormal in $\mathbb{R}^3$, what is $\|2u - 3v + w\|$?
A 6
B $\sqrt{14}$
C $\sqrt{12}$
D 14