Linear Algebra Assignment: Basis, Transformations, Diagonalization

This assignment delves into fundamental concepts of linear algebra. The first part focuses on linear transformations, requiring the determination of bases for the kernel and image of a given transformation, along with finding the matrix representation with respect to a different basis. Furthermore, it involves calculating the transformed vector under this new basis. The second part of the assignment explores matrix diagonalization, requiring the calculation of eigenvalues and eigenvectors, and determining an invertible matrix to diagonalize a given matrix. The solution provides a step-by-step approach to solve these problems. This document is contributed to Desklib, a platform offering AI-based study tools for students to access past papers and solved assignments, providing valuable resources for learning and exam preparation.

Solution 1: Given is a linear transformation with respect to the standard basis of
defined by the matrix:
. Let’s first find the row reduced echelon form of matrix A. Perform the
elementary row operations to find row reduced echelon form.
, we get
1. To find bases for ker T, solve using ,where , we get
So,
So, basis for ker T is
Now, since the rref(A) contain pivot elements in only 1st column, so the corresponding
column in original matrix form the basis for ImT. So, basis for Im T is .
2. Let
Now,

So
, matrix of T with respect to the basis B is
3. Let and since , so
Now,
T
herefore,
Solution 2: Given .

1. Eigenvalues:
Solve , we get
Since, eigenvalues are distinct and hence corresponding eigenvectors are linearly
independent. And we know that if eigenvalues are linearly independent, then matrix is
diagonalizable. Therefore matrix A is diagonalizable.
2. To find invertible matrix P and diagonal matrix we have to find eigenvectors of matrix A
corresponding to the eigenvalues.
Now, for , solve
. Using elementary row operations we get
. Let , we get . So eigenvector
Now, for , solve
. Using elementary row operations we get

. Let , we get . So eigenvector
Now, for , solve
. Using elementary row operations we get
. Let , we get . So eigenvector
Therefore, invertible matrix P is
Such that , a diagonal matrix.