Column space :
Since , each column of is times the corresponding column of . Thus, every column of is a linear combination of the columns of , so . Moreover, because has full row rank, its rows span , meaning the columns of (as coefficients) can generate any vector in . Applying the invertible (a bijection) to the full span of yields the full . Therefore, . The columns of form a basis:
Row space :
The row space . We can then find using where:
After computing the inner product we get which we can then use to get:
Null space :
The null space satisfies To find , we need to solve Solving the linear system we get that:
This tells us that:
Null space :
We know that the null space is . Since , and we know that we know that so .