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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
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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