Hence, any linear combination of the rows of is in the row space of N. Therefore, the row space of is contained in the row space of N. From this, it follows that the dimension of the row space of is less than or equal to the dimension of the row space of N that is,. I already have one algorithm for testing whether a set of vectors in is independent. That algorithm involves constructing a matrix with the vectors as the columns , then row reducing.
The algorithm will also produce a linear combination of the vectors which adds up to the zero vector if the set is dependent. If all you care about is whether or not a set of vectors in is independent i.
In this approach, you construct a matrix with the given vectors as the rows. The object is to determine whether the set is independent. Let R be a row reduced echelon matrix which is row equivalent to M. If R has m nonzero rows, then is independent.
Otherwise, the set is dependent. If R has p nonzero rows, then R and M have rank p. They have the same rank, because they have the same row space. Since spans, some subset of is a basis. However, a basis must contain elements. Therefore, must be independent. Any independent subset of the row space must contain elements. The vectors , , and in are dependent:. The row reduced echelon matrix has only two nonzero rows. I already know that every matrix can be row reduced to a row reduced echelon matrix.
The next result completes the discussion by showing that the row reduced echelon form is unique. Every matrix can be row reduced to a unique row reduced echelon matrix. Suppose M row reduces to R, a row reduced echelon matrix with nonzero rows. In component form,. Claim: The first nonzero component of v must occur in column , for. Suppose is the first which is nonzero.
The sum looks like. The first nonzero element of is a 1 at. The first nonzero element in lies to the right of column. Thus, for , and. Evidently, this is the first nonzero component of v. This proves the claim. This establishes that if a row reduced echelon matrix is row equivalent to M, its leading coefficients must lie in the same columns as those of R.
For the rows of are elements of W, and the claim applies. Next, I'll show that the nonzero rows of are the same as the nonzero row of R. Consider, for instance, the first nonzero rows of R and. Their first nonzero components are 1's lying in column. Moreover, both and have zeros in columns , , Then is a nonzero vector in W whose first nonzero component is not in column , , The same argument applies to show that for all k.
I showed earlier that you can add vectors to an independent set to make a basis. Here's how it works in a particular case. Add vectors to the following set to make a basis for :.
If I make a matrix with these vectors as rows and row reduce, the row reduced echelon form will have the same row space i. Since there are four nonzero rows, and since these rows are clearly independent vectors, the original set of vectors is independent.
By examining the row reduced echelon form, I see that the vector will not be a linear combination of the others. I'm choosing a standard basis vector with a "1" in a position not occupied by a leading coefficient. Therefore, I can add it to the set and get a new independent set:. There are 5 vectors in this set, so it is a basis for. I also showed earlier that you can remove vectors from a spanning set to get a basis. You can do this using the same algorithm that gives a basis for the column space of a matrix.
First, here's a reminder about matrix multiplication. If A is an matrix and , then the product is a linear combination of the columns of A. In fact, if is the i-th column of A and ,. Let A be a matrix, and let R be the row reduced echelon matrix which is row equivalent to A.
Suppose the leading coefficients of R occur in columns , where , and let denote the i-th column of A. Then is independent. Form the vector , where. The equation above implies that. It follows that v is in the solution space of the system. Since has the same solution space,. Let denote the i-th column of R. However, since R is in row reduced echelon form, is a vector with 1 in the k-th row and 0's elsewhere.
Hence, is independent, and. The leading coefficients occur in the first three columns. Hence, the first three columns of A are independent:. In fact, they form a basis for the column space of A. Find a subset of the following set of vectors which forms a basis for. The leading coefficients occur in columns 1, 2, and 4.
Therefore, the corresponding columns of the original matrix are independent, and form a basis for :. Let R be the row reduced echelon matrix which is row equivalent to A. By the preceding lemma, is independent. There is one vector in this set for each leading coefficient, and the number of leading coefficients equals the row rank.
Now consider. This is A with the rows and columns swapped, so. Applying the first part of the proof to ,. The proof provides an algorithm for finding a basis for the column space of a matrix. Specifically, row reduce the matrix A to a row reduced echelon matrix R. If the leading coefficients of R occur in columns , then consider the columns of A. These columns form a basis for the column space of A. The leading coefficients occur in columns 1 and 2.
Therefore, and form a basis for the column space of A. If A and B are row equivalent, they don't necessarily have the same column space. For example,. However, all the elements of the column space of the second matrix have their second component equal to 0; this is obviously not true of elements of the column space of the first matrix.
I showed earlier that. This was row rank; a similar proof shows that. Since row rank and column rank are the same,. Create a free Team What is Teams? Learn more. Can a matrix have a null space that is equal to its column space? Ask Question. Asked 8 years, 3 months ago. Active 8 months ago. Viewed 32k times. Mirrana Mirrana 8, 33 33 gold badges 78 78 silver badges bronze badges. Add a comment. Active Oldest Votes. Gerry Myerson Gerry Myerson k 12 12 gold badges silver badges bronze badges.
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