Range null space theorem

These are, apparently, two sides of the same coin, and this symmetry can be made precise via the following theorem. Theorem 6. Let be a linear map. Then is finite-dimensional and. Proof sketch. Let be a basis for , and extend this list of vectors to a basis of of the form. Show that spans , and also show that is linearly independent.

Conclude that. Possible Answers:. Correct answer:. Explanation : The first step is to create an augmented matrix having a column of zeros. The next step is to get this into RREF. We can simplify to This tells us the following. Now we need to write this as a linear combination. The null space is then. Report an Error. Correct answer: None of the other answers.

Explanation : We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form. Using a calculator or row reduction, we obtain for the reduced row echelon form. Explanation : The null space of the matrix is the set of solutions to the equation.

After that, our system becomes Hence a basis for the null space is just the zero vector;. Explanation : The null space of the operator is the set of solutions to the equation. After that, our system becomes Hence the null space consists of only the zero vector. Possible Answers: None of the other answers. After that, our system becomes Multiplying this vector by gets rid of the fraction, and does not affect our answer, since there is an arbitrary constant behind it.

After that, our system becomes Hence the null space consists of all vectors spanned by ;. Explanation : The rank is equal to the dimension of the row space and the column space both spaces always have the same dimension.

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Do not fill in this field. By definition, is invertible if is one-one and onto. But we have shown that is one-one if and only if is onto. Thus, we have the last equivalent condition. If either is one-one or is onto, then is invertible. The following are some of the consequences of the rank-nullity theorem. The proof is left as an exercise for the reader. There exist exactly columns of that are linearly independent. There is a submatrix of with non-zero determinant and every submatrix of has zero determinant.

The dimension of the range space of is There is a subset of consisting of exactly linearly independent vectors such that the system for is consistent. The dimension of the null space of. If is finite dimensional then show that the null space and the range space of are also finite dimensional.

If and are both finite dimensional then show that if then is onto. Let be an real matrix.