Science:Math Exam Resources/Courses/MATH221/April 2013/Question 02 (c)/Solution 1

The column space is the same as the range of the matrix, which we know has has dimension 3, thus we are looking for a basis made up of 3 elements.

While row operations (or left multiplication by an invertible matrix) changes the column space, it still preserves the relations among the columns. The reason for this being is that left matrix multiplication acts on the columns of a matrix as if they were vectors, thus, for example, if $c_{1}$ is the first column of a matrix B, then $Mc_{1}$ is the first column of the matrix $MB$, furthermore if we had $c_{1}=c_{2}+c_{3}$, a relationship among the columns of A, then $Mc_{1}=Mc_{2}+Mc_{3}$ will be a relationship in the columns of $MA$. Since M is invertible you can reverse this step, so that linearly independent columns in B correspond to linearly independent columns in A.

Due to this fact, we can read the relations that the columns of the matrix A by looking at those off of the matrix B, in particular we can simply read which column of A are linear independent by looking at the columns of B that contain a pivot (i.e. a set of columns that is linearly independent in B).

The column space of A has a basis equal to $\left[{\begin{array}{c}1\\3\\2\\2\\\end{array}}\right],\left[{\begin{array}{c}2\\4\\3\\2\\\end{array}}\right],\left[{\begin{array}{c}3\\0\\1\\-3\\\end{array}}\right]$.