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Linear Algebra

Vector spaces, linear maps, canonical forms, bilinear and quadratic forms, and inner product spaces at the CSIR NET level with advanced topics like Jordan form computation.

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01 Vector Spaces & Subspaces

Axioms of vector spaces, subspaces, bases, dimension, direct sums, and quotient spaces.

◆ Bases & Dimension
Definition A basis of a vector space $V$ over a field $F$ is a linearly independent spanning set. The dimension $\dim V$ is the cardinality of any basis (well-defined by the Steinitz exchange lemma).
Dimension Formula For subspaces $W_1, W_2$ of a finite-dimensional space $V$: $\dim(W_1 + W_2) = \dim W_1 + \dim W_2 - \dim(W_1 \cap W_2)$.
Direct Sum Decomposition $V = W_1 \oplus W_2$ if and only if $V = W_1 + W_2$ and $W_1 \cap W_2 = \{0\}$. Equivalently, every $v \in V$ has a unique expression $v = w_1 + w_2$ with $w_i \in W_i$.
📝 Example
Let $V = \mathbb{R}^3$, $W_1 = \{(x,y,0): x,y \in \mathbb{R}\}$, $W_2 = \{(0,0,z): z \in \mathbb{R}\}$. Show $V = W_1 \oplus W_2$.
Any $(a,b,c) = (a,b,0) + (0,0,c) \in W_1 + W_2$, so $V = W_1 + W_2$. If $v \in W_1 \cap W_2$, then $v = (x,y,0) = (0,0,z)$, giving $x=y=z=0$. So $W_1 \cap W_2 = \{0\}$, confirming $V = W_1 \oplus W_2$.
02 Linear Transformations & Matrix Representations

Kernel, image, rank-nullity theorem, change of basis, and similarity of matrices.

◆ Rank-Nullity Theorem
Rank-Nullity Theorem If $T: V \to W$ is a linear map between finite-dimensional spaces, then $\dim V = \dim(\ker T) + \dim(\text{im}\, T) = \text{nullity}(T) + \text{rank}(T)$.

Two matrices $A, B \in M_n(F)$ are similar ($A = P^{-1}BP$) if and only if they represent the same linear transformation under different bases. Similar matrices have the same characteristic polynomial, eigenvalues, rank, determinant, and trace.

📝 Example
Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ with $\text{rank}(T) = 2$. Find $\dim(\ker T)$.
By rank-nullity: $\dim(\ker T) = \dim \mathbb{R}^4 - \text{rank}(T) = 4 - 2 = 2$.
◆ Cayley-Hamilton Theorem
Cayley-Hamilton Theorem Every square matrix satisfies its own characteristic polynomial. If $p(\lambda) = \det(\lambda I - A)$, then $p(A) = 0$.

This is a key tool: it shows that $A^{-1}$ (when it exists) can be expressed as a polynomial in $A$. It also implies that the minimal polynomial divides the characteristic polynomial.

03 Eigenvalues & Diagonalizability

Algebraic and geometric multiplicity, criteria for diagonalizability, and simultaneous diagonalization.

◆ Eigenvalue Theory
Definition $\lambda$ is an eigenvalue of $T$ if $Tv = \lambda v$ for some nonzero $v$. The algebraic multiplicity $a_\lambda$ is the multiplicity of $\lambda$ as a root of the characteristic polynomial. The geometric multiplicity $g_\lambda = \dim \ker(T - \lambda I)$.
Diagonalizability Criterion $T$ is diagonalizable if and only if $g_\lambda = a_\lambda$ for every eigenvalue $\lambda$. Equivalently, the minimal polynomial of $T$ splits into distinct linear factors.
Simultaneous Diagonalization Two diagonalizable operators $S, T$ on a finite-dimensional space are simultaneously diagonalizable if and only if $ST = TS$.
📝 Example
Is $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ diagonalizable?
The characteristic polynomial is $(1-\lambda)^2$, so $\lambda = 1$ with $a_1 = 2$. But $\ker(A - I) = \ker \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \text{span}\{(1,0)\}$, giving $g_1 = 1 \neq 2 = a_1$. So $A$ is not diagonalizable.
04 Jordan Canonical Form & Rational Canonical Form

Structure theorem for finitely generated modules over a PID applied to linear operators, Jordan blocks, and computation techniques.

◆ Jordan Canonical Form
Jordan Block A Jordan block $J_k(\lambda)$ is the $k \times k$ matrix with $\lambda$ on the diagonal, $1$'s on the superdiagonal, and $0$'s elsewhere: $J_k(\lambda) = \lambda I_k + N_k$, where $N_k$ is the nilpotent shift.
Jordan Normal Form Theorem Over an algebraically closed field, every $n \times n$ matrix is similar to a block diagonal matrix $J = \text{diag}(J_{k_1}(\lambda_1), \ldots, J_{k_r}(\lambda_r))$. This form is unique up to permutation of blocks.

To find the Jordan form: (1) Find eigenvalues from the characteristic polynomial. (2) For each eigenvalue $\lambda$, compute $\dim \ker (A - \lambda I)^k$ for $k = 1, 2, \ldots$ to determine the sizes of Jordan blocks. The number of blocks for $\lambda$ of size $\geq k$ equals $\dim \ker(A-\lambda I)^k - \dim \ker(A-\lambda I)^{k-1}$.

📝 Example
Find the Jordan form of $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$.
The only eigenvalue is $\lambda = 2$ with $a_2 = 3$. $\text{rank}(A - 2I) = 2$, so $g_2 = 1$ (one Jordan block). The Jordan form is $J_3(2)$, a single $3 \times 3$ Jordan block. In fact, $A$ is already in Jordan form.
◆ Rational Canonical Form
Rational Canonical Form Every matrix over any field is similar to a block diagonal matrix of companion matrices $C(p_1), \ldots, C(p_r)$ where $p_1 | p_2 | \cdots | p_r$ are the invariant factors and $p_r$ is the minimal polynomial.

The rational canonical form exists over any field (unlike Jordan form which requires algebraic closure). The invariant factors $p_1 | p_2 | \cdots | p_r$ satisfy $p_1 p_2 \cdots p_r = $ characteristic polynomial.

05 Bilinear & Quadratic Forms

Bilinear forms, symmetric and skew-symmetric forms, Sylvester's law of inertia, and classification of quadratic forms.

◆ Quadratic Forms & Classification
Definition A bilinear form on $V$ is a map $B: V \times V \to F$ that is linear in each argument. If $B(u,v) = B(v,u)$, it is symmetric. The associated quadratic form is $Q(v) = B(v,v)$.
Sylvester's Law of Inertia Over $\mathbb{R}$, any symmetric bilinear form can be diagonalized. The number of positive, negative, and zero diagonal entries (the signature $(p, q, z)$) is an invariant independent of the diagonalizing basis. The form is positive definite iff $q = z = 0$.
📝 Example
Classify the quadratic form $Q(x,y,z) = x^2 + 4xy + 3y^2 + 2z^2$.
The matrix is $A = \begin{pmatrix} 1 & 2 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 2 \end{pmatrix}$. Eigenvalues of the upper-left $2\times 2$ block are $2 \pm \sqrt{5}$, and the third eigenvalue is $2$. Since $2 - \sqrt{5} < 0$, the signature is $(2, 1, 0)$: indefinite.
06 Inner Product Spaces & Orthogonality

Inner products, Gram-Schmidt, orthogonal complements, self-adjoint operators, spectral theorem, and normal operators.

◆ Orthogonality & Projections
Gram-Schmidt Process Given linearly independent $\{v_1, \ldots, v_k\}$ in an inner product space, one can construct an orthonormal set $\{e_1, \ldots, e_k\}$ with $\text{span}\{e_1,\ldots,e_j\} = \text{span}\{v_1,\ldots,v_j\}$ for each $j$.
Spectral Theorem (Finite-Dimensional) A linear operator $T$ on a finite-dimensional inner product space over $\mathbb{R}$ is self-adjoint ($T = T^*$) if and only if there exists an orthonormal basis of eigenvectors. Over $\mathbb{C}$, $T$ is normal ($TT^* = T^*T$) if and only if it is unitarily diagonalizable.
Best Approximation If $W$ is a finite-dimensional subspace of an inner product space $V$ and $v \in V$, the closest point in $W$ to $v$ is the orthogonal projection $\text{proj}_W(v)$. The error $v - \text{proj}_W(v)$ is orthogonal to $W$.
📝 Example
Find an orthonormal basis for the column space of $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \\ 0 & 1 \end{pmatrix}$.
$v_1 = (1,1,0)^T$, $v_2 = (1,0,1)^T$. Apply Gram-Schmidt: $e_1 = \frac{v_1}{\|v_1\|} = \frac{1}{\sqrt{2}}(1,1,0)^T$. $u_2 = v_2 - \langle v_2, e_1 \rangle e_1 = (1,0,1)^T - \frac{1}{2}(1,1,0)^T = (\frac{1}{2}, -\frac{1}{2}, 1)^T$. $e_2 = \frac{u_2}{\|u_2\|} = \frac{1}{\sqrt{3/2}}(\frac{1}{2}, -\frac{1}{2}, 1)^T = \frac{1}{\sqrt{6}}(1,-1,2)^T$.
★ Key Takeaways
✍ Practice Problems
Problem 1
Let $T: V \to V$ be a linear operator on a $5$-dimensional space with $\text{rank}(T^2) = 2$. What are the possible values of $\text{rank}(T)$?
Show Solution ▼
Since $\text{im}(T^2) \subseteq \text{im}(T)$, we have $\text{rank}(T) \geq \text{rank}(T^2) = 2$. Also, $\text{rank}(T^2) \leq \text{rank}(T)$ and applying rank-nullity to $T$ restricted to $\text{im}(T)$: $\text{rank}(T^2) \geq 2\text{rank}(T) - 5$, giving $\text{rank}(T) \leq \lfloor(5+2)/2\rfloor$. Possible values: $\text{rank}(T) \in \{2, 3\}$ (since rank 4 or 5 would force rank($T^2$) $\geq 3$). Actually rank(T) can be 2, 3, or 4. For rank 4: nullity(T)=1, so rank($T^2$) can be as low as rank(T)-nullity(T)=3, which exceeds 2. So rank(T) $\in \{2, 3\}$.
Problem 2
Find all possible Jordan forms for a $4 \times 4$ matrix with characteristic polynomial $(x-2)^2(x-3)^2$.
Show Solution ▼
For $\lambda = 2$ (algebraic mult. 2): either $J_2(2)$ (one block) or $J_1(2) \oplus J_1(2)$ (two blocks). Similarly for $\lambda = 3$. The possible Jordan forms are: (1) $J_2(2) \oplus J_2(3)$, (2) $J_2(2) \oplus J_1(3) \oplus J_1(3)$, (3) $J_1(2) \oplus J_1(2) \oplus J_2(3)$, (4) $J_1(2) \oplus J_1(2) \oplus J_1(3) \oplus J_1(3)$ (diagonal case).
Problem 3
Prove that if $A$ and $B$ are $n \times n$ matrices, then $AB$ and $BA$ have the same eigenvalues.
Show Solution ▼
If $\lambda \neq 0$ is an eigenvalue of $AB$ with $ABv = \lambda v$, let $w = Bv$. Then $BAw = B(ABv) = B(\lambda v) = \lambda Bv = \lambda w$ and $w \neq 0$ (since $ABv = \lambda v \neq 0$). So $\lambda$ is an eigenvalue of $BA$. For $\lambda = 0$: $\det(AB) = \det(A)\det(B) = \det(BA)$, so $0$ is an eigenvalue of $AB$ iff it is of $BA$. In fact, $AB$ and $BA$ have the same characteristic polynomial (non-trivially proved via $\det(\lambda I - AB) = \det(\lambda I - BA)$).
Problem 4
Let $A$ be a real symmetric $3 \times 3$ matrix with eigenvalues $1, 2, 3$. Compute $\text{tr}(A^2)$ and $\det(e^A)$.
Show Solution ▼
$\text{tr}(A^2) = 1^2 + 2^2 + 3^2 = 14$. For $\det(e^A)$: since $A$ is diagonalizable with eigenvalues $1,2,3$, the eigenvalues of $e^A$ are $e^1, e^2, e^3$. So $\det(e^A) = e^1 \cdot e^2 \cdot e^3 = e^{1+2+3} = e^6$. In general, $\det(e^A) = e^{\text{tr}(A)}$.
Problem 5
Show that a nilpotent matrix $N$ ($N^k = 0$ for some $k$) has all eigenvalues equal to $0$.
Show Solution ▼
If $Nv = \lambda v$ for $v \neq 0$, then $N^k v = \lambda^k v$. Since $N^k = 0$, we get $\lambda^k v = 0$, and since $v \neq 0$, $\lambda^k = 0$, so $\lambda = 0$. Alternatively, by Cayley-Hamilton, $N$ satisfies its characteristic polynomial $p(\lambda) = \lambda^n$ (since all eigenvalues are $0$), and the minimal polynomial divides $\lambda^n$, which is $\lambda^m$ for some $m \leq n$.
🎯 Test Your Understanding
1. Let $T: \mathbb{R}^4 \to \mathbb{R}^4$ have minimal polynomial $(x-1)(x-2)$. Then $T$ is:
A Diagonalizable
B Not diagonalizable but triangularizable
C A scalar multiple of $I$
D Nilpotent
2. The number of similarity classes of $3 \times 3$ matrices over $\mathbb{C}$ with characteristic polynomial $(x-1)^3$ is:
A 1
B 2
C 3
D 4
3. If $A$ is a real $n \times n$ skew-symmetric matrix ($A^T = -A$), then:
A All eigenvalues are real
B All eigenvalues are purely imaginary or zero
C $A$ is diagonalizable over $\mathbb{R}$
D $\det(A) = 0$ always
4. The rank of the matrix $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix}$ is:
A 1
B 2
C 3
D 0
5. Let $V$ be a $6$-dimensional vector space and $W_1, W_2$ subspaces with $\dim W_1 = 4$, $\dim W_2 = 3$. The minimum possible dimension of $W_1 \cap W_2$ is:
A 0
B 1
C 2
D 3