Linear Transformation Assignment Report

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.