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## How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix  By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. What does the "zeroing-out" of two rows tell us? How can we use what non-zero rows remain to construct a basis for span S? Notice that these are five -dimensional vectors, so we are already starting out "short a coordinate variable", making it "free". Taking the hint from Omnomnomnom or the above, the subspace spanned by your set of four vectors only has dimension 3. So we need to set up three linearly independent vectors, using the columns of the row-reduced matrix. With the matrix fully "reduced", we need to pick out three five-dimensional column vectors which are linearly independent. The third column is a linear combination of the first two, so we can toss that one out. A suitable basis for span S is then. I think colormegone's procedure to find basis is correct in terms of row reducing the matrix. These 3 vectors correspond to the first, second and fourth column in the original matrix, so a basis should be the set of corresponding column vectors in the original matrix, i. Sign up to join this community. The best answers are voted up and rise to the top. Home Questions Tags Users Unanswered. Finding the dimension of subspace span S Ask Question. Asked 5 years, 11 months ago. Active 1 year, 6 months ago. Viewed 66k times. I know that dimension is the maximum number of linearly independent vectors in a subspace. So is the dimension in this case 4? Since there are 4 vectors? How do I solve this? CunningTF 3 3 silver badges 10 10 bronze badges. Con Con 31 2 2 gold badges 2 2 silver badges 6 6 bronze badges. Okay so dimension could be for. How do I determine if one or more vectors is lin combo of the others? Guess and check? This will give you homogeneous system of linear equations. You can then row reduce the matrix to find out the rank of the matrix, and the dimension of the subspace will be equal to this rank. Showing that the equation provided by Hayden has only the trivial solution is equivalent to showing that the set is linearly independent. Active Oldest Votes. Also, which set of vectors is a basis for Span S? 