LINEAR ALGEBRA/Student Engineering Notebook
Page 03·notes
VECTOR SPACES, SUBSPACES & LINEAR TRANSFORMATIONS—~30 MIN READ

Linear Maps, Matrix Representations & Change of Basis

“Linearity tests, standard matrix representations [T], kernel, range, and similarity transformations A' = P^{-1} A P.”

1.Linear Transformations & Coordinate Representations

A mapping T: V -> W is a Linear Transformation if it preserves vector addition and scalar multiplication.

Kernel and Range:

Kernel(T) = { v in V : T(v) = 0 } (corresponds to null space of representation matrix).

Range(T) = { T(v) : v in V } (corresponds to column space of representation matrix).

Change of Basis Matrix:

Let B = {b1, ..., bn} and B' = {b1', ..., bn'} be two bases for V. The transition matrix P = P_{B -> B'} transforms coordinates: [x]_{B'} = P^{-1} [x]_B.

The matrix of transformation in the new basis is given by the Similarity Transformation:

[T]_{B'} = P^{-1} [T]_B P.

WORKED EXAM ANALYSIS · STEP-BY-STEP DERIVATION2 Marks (Original Practice Problem · Modeled on GATE DA Pattern)

2. Worked Practice Problem: Change of Basis Transformation Matrix

⚠️ Pitfall Warning:

Remember P^{-1} A P ordering: columns of P are the new basis vectors expressed in standard coordinates.

Problem: Let T: R^2 -> R^2 be defined by T(x, y) = (2x + 3y, x - y). Find the standard matrix A = [T]_E, and compute the representation matrix B = [T]_B in the basis B = { b1 = (1, 1)^T, b2 = (1, -1)^T }.

Step 1: Construct standard matrix A:

T(e1) = T(1, 0) = (2, 1)^T

T(e2) = T(0, 1) = (3, -1)^T

A = [ [2, 3], [1, -1] ].

Step 2: Construct transition matrix P with columns b1 and b2:

P = [ [1, 1], [1, -1] ]

det(P) = (1)(-1) - (1)(1) = -2.

P^{-1} = (1 / -2) * [ [-1, -1], [-1, 1] ] = [ [1/2, 1/2], [1/2, -1/2] ].

Step 3: Compute product A * P:

A P = [ [2, 3], [1, -1] ] [ [1, 1], [1, -1] ]

Col 1: [ 21 + 31, 11 + (-1)1 ]^T = [ 5, 0 ]^T

Col 2: [ 21 + 3(-1), 11 + (-1)(-1) ]^T = [ -1, 2 ]^T

A * P = [ [5, -1], [0, 2] ].

Step 4: Compute B = P^{-1} (A P):

B = [ [1/2, 1/2], [1/2, -1/2] ] * [ [5, -1], [0, 2] ]

Row 1: [ (1/2)5 + (1/2)0, (1/2)(-1) + (1/2)2 ] = [ 5/2, 1/2 ]

Row 2: [ (1/2)5 - (1/2)0, (1/2)(-1) - (1/2)2 ] = [ 5/2, -3/2 ]

B = [ [2.5, 0.5], [2.5, -1.5] ].

Conclusion: Under basis B, the transformation matrix is B = [ [2.5, 0.5], [2.5, -1.5] ].

SOURCE TRANSPARENCY & CITATIONS

Primary Academic References & Exam Sources

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Curated University Lecture Series
Linear Algebra Done Right (3rd Edition) ·
LECTURE SERIES