Discover Linear Algebra

1. Let's apply the Test for Linear Dependence/Independence to the vectors \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_\ell\text{:}\) suppose that \(k_1,k_2,\dotsc,k_\ell\) are scalars so that
   \begin{gather} k_1\uvec{v}_1 + k_2\uvec{v}_2 + \dotsc + k_\ell\uvec{v}_\ell = \zerovec.\label{equation-col-row-null-space-dep-indep-of-images-image-indep-implies-indep-test}\tag{\(\star\)} \end{gather}
   Multiplying both sides of this equation by the matrix \(E\text{,}\) and using some matrix algebra, we get
   \begin{equation*} k_1E\uvec{v}_1 + k_2E\uvec{v}_2 + \dotsc + k_\ell E\uvec{v}_\ell = E\zerovec = \zerovec. \end{equation*}
   But we have assumed that the vectors \(E\uvec{v}_1,E\uvec{v}_2,\dotsc,E\uvec{v}_\ell\) are linearly independent, so the only way this linear combination could equal the zero vector is if all the scalars \(k_1,k_2,\dotsc,k_\ell\) are zero. Thus, the linear combination in (\(\star\)) must be the trivial one, and so the vectors \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_\ell\) are linearly independent.
2. We assume \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_\ell\) are linearly independent. Since we also assume \(E\) to be invertible, we can restate this as saying that \(\inv{E} (E \uvec{v}_1), \inv{E} (E \uvec{v}_2), \dotsc, \inv{E} (E \uvec{v}_\ell)\) are linearly independent. Now we can apply Statement 1 with \(E\) replaced by \(\inv{E}\) and \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_\ell\) replaced by \(E \uvec{v}_1, E \uvec{v}_2, \dotsc, E \uvec{v}_\ell\text{.}\)
3. Simply apply the inverse \(\inv{E}\) to both sides of
   \begin{equation*} E\uvec{w} = k_1 E\uvec{v}_1 + k_2 E\uvec{v}_2 + \dotsb + k_\ell E\uvec{v}_\ell \end{equation*}
   to obtain
   \begin{equation*} \uvec{w} = k_1\uvec{v}_1 + k_2\uvec{v}_2 + \dotsb + k_\ell\uvec{v}_\ell. \end{equation*}

1. By definition, the columns of \(A\) are a spanning set for the column space of \(A\text{.}\) By Proposition 19.5.1, this spanning set can be reduced to a basis; it's a matter of finding the largest possible linearly independent set of these spanning column vectors.

   Let \(E = E_t E_{t-1} \dotsm E_1\) be the product of elementary matrices corresponding to some sequence of row operations that reduces \(A\) to its RREF. Because of the nature of RREF, each column of \(\RREF(A)\) that contains a leading one will be a standard basis vector in \(\R^m\text{,}\) no two such leading-one columns will be the same standard basis vector, and each column that does not contain a leading one will be a linear combination of those leading-one columns that appear to its left. Therefore, the leading-one columns represent the largest set of linearly independent vectors that can be formed from the columns of \(\RREF(A)\text{.}\) Since \(E\) is invertible, the two statements of Proposition 21.5.1 tell us that the columns of \(A\) will have the same relationships: those columns in \(A\) that are in positions where the leading ones occur in \(\RREF(A)\) will be linearly independent, and that will be the largest possible collection of linearly independent columns of \(A\text{,}\) because each of the other columns will be linearly dependent with the leading-one-position columns of \(A\) to its left.

   Thus, we can reduce the spanning set made up of all columns of \(A\) to a basis for its column space by discarding the linearly dependent columns and keeping only those columns in positions corresponding to the locations of the leading ones occur in \(\RREF(A)\text{.}\)

2. Since we obtain a basis for column space by taking those columns in the matrix in positions corresponding to the leading ones in a reduced form for the matrix, the number of basis vectors is equal to the number of leading ones.

Consider an \(m \times n\) matrix \(A\) as a collection of row vectors in \(\R^n\text{:}\)

Then the row space of \(A\) is, by definition, the subspace \(\Span\{\uvec{a}_1,\uvec{a}_2,\dotsc,\uvec{a}_m\}\) of \(\R^m\text{.}\)

As we did in Discovery 21.5, we will make repeated use of Statement 2 of Proposition 17.5.6, which tells us how to determine when two spanning sets generate the same subspace.

Let's consider each type of elementary row operation in turn.

1. Suppose we swap two rows in \(A\text{:}\)
   \begin{equation*} A = \begin{bmatrix} \leftrightlinesubstitute \amp \uvec{a}_1 \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_i \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_j \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_m \amp \leftrightlinesubstitute\\ \end{bmatrix} \to A' = \begin{bmatrix} \leftrightlinesubstitute \amp \uvec{a}_1 \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_j \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_i \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_m \amp \leftrightlinesubstitute\\ \end{bmatrix}. \end{equation*}
   The row space of the new matrix, \(A'\text{,}\) is the span of its row vectors. But every row vector in \(A'\) is equal to one of the row vectors in \(A\text{,}\) and vice versa. So clearly the conditions of the above-referenced Statement 2 are satisfied, and the rowspaces of the two matrices are the same space.

2. Suppose we multiply one of the rows in \(A\) by a nonzero constant \(k\text{:}\)

   \begin{equation*} A = \begin{bmatrix} \leftrightlinesubstitute \amp \uvec{a}_1 \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_i \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_m \amp \leftrightlinesubstitute\\ \end{bmatrix} \to A'' = \begin{bmatrix} \leftrightlinesubstitute \amp \uvec{a}_1 \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp k\uvec{a}_i \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_m \amp \leftrightlinesubstitute\\ \end{bmatrix}. \end{equation*}

   Again, most of the row vectors in the new matrix \(A''\) are equal to one of the row vectors in \(A\text{,}\) and vice versa. So to fully satisfy the conditions of the above-referenced Statement 2, we need to verify that \(k\uvec{a}_i\) is somehow a linear combination of row vectors from \(A\) and that \(\uvec{a}_i\) is somehow a linear combination of row vectors from \(A''\text{.}\) But \(k\uvec{a}_i\) is already expressed as a scalar multiple of a row vector from \(A\text{,}\) and since \(k\) is nonzero we can also write

   \begin{equation*} \uvec{a}_i = \frac{1}{k} \cdot (k\uvec{a}_i), \end{equation*}

   so that \(\uvec{a}_i\) is also a scalar multiple of a row vector from \(A''\text{.}\)

   With the conditions of the above-referenced Statement 2 now fully satisfied, we can conclude that the rowspaces of the two matrices are the same space.

3. Suppose we replace one row vector in \(A\) by the sum of that row and a scalar multiple of another:

   \begin{equation*} A = \begin{bmatrix} \leftrightlinesubstitute \amp \uvec{a}_1 \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_i \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_m \amp \leftrightlinesubstitute\\ \end{bmatrix} \to A''' = \begin{bmatrix} \leftrightlinesubstitute \amp \uvec{a}_1 \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_i + k\uvec{a}_j \amp \leftrightlinesubstitute\\ \amp\vdots\\ \leftrightlinesubstitute \amp \uvec{a}_m \amp \leftrightlinesubstitute\\ \end{bmatrix}. \end{equation*}

   Once again, most of the row vectors in the new matrix \(A'''\) are equal to one of the row vectors in \(A\text{,}\) and vice versa. So to fully satisfy the conditions of the above-referenced Statement 2, we need to verify that \(\uvec{a}_i + k\uvec{a}_j\) is somehow a linear combination of row vectors from \(A\) and that \(\uvec{a}_i\) is somehow a linear combination of row vectors from \(A'''\text{.}\) But \(\uvec{a}_i + k\uvec{a}_j\) is already expressed as a linear combination of row vectors from \(A'''\text{,}\) and for \(\uvec{a}_i\) we can write

   \begin{equation*} \uvec{a}_i = 1(\uvec{a}_i + k\uvec{a}_j) + (-k)\uvec{a}_j, \end{equation*}

   a linear combination of row vectors from \(A'''\text{.}\)

   With the conditions of the above-referenced Statement 2 now fully satisfied, we can conclude that the rowspaces of the two matrices are the same space.

1. Since \(E\) is invertible, it can be expressed as a product of elementary matrices (Theorem 6.5.2), and the product \(EA\) has the same result as applying to \(A\) the sequence of row operations represented by those elementary matrices. But Proposition 21.5.3 tells us that applying those operations does not change the row space.
2. Let \(F\) be an REF for \(A\text{,}\) and let \(E_1,E_2,\dotsc,E_\ell\) be elementary matrices corresponding to some sequence of row operations that reduces \(A\) to \(F\text{.}\) Set \(E = E_\ell \dotsm E_2 E_1\text{.}\) Then \(E\) is an invertible matrix and \(F = EA\text{.}\) Therefore, \(F\) has the same row space as \(A\) by Statement 1 of this corollary.
3. Let \(F\) be an REF for \(A\text{.}\) By Statement 2 of this corollary, the rows of \(F\) are a spanning set for the row space of \(A\text{.}\) Clearly we can discard any zero rows from this spanning set, so it just remains to verify that the nonzero rows of \(F\) are linearly independent. For this, we will use Proposition 18.5.6, building up our linearly independent spanning set one vector at a time. Let \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_\ell\) represent the nonzero rows of \(F\text{,}\) from top to bottom. Start with \(\uvec{v}_\ell\text{;}\) all by itself, this one nonzero vector is linearly independent. Now, \(\uvec{v}_{\ell-1}\) cannot be in \(\Span\{\uvec{v}_\ell\}\text{,}\) because the leading one in \(\uvec{v}_{\ell-1}\) appears to the left of the leading one in \(\uvec{v}_\ell\text{,}\) and so no scalar multiple of \(\uvec{v}_\ell\) will have a nonzero entry in the component where \(\uvec{v}_{\ell-1}\) has its leading one. From this, Proposition 18.5.6 tells us that \(\{\uvec{v}_{\ell-1},\uvec{v}_\ell\}\) is linearly independent. Moving on, \(\uvec{v}_{\ell-2}\) cannot be in \(\Span\{\uvec{v}_{\ell-1},\uvec{v}_\ell\}\text{,}\) because the leading one in \(\uvec{v}_{\ell-2}\) appears to the left of both the leading one in \(\uvec{v}_{\ell-1}\) and in \(\uvec{v}_{\ell}\text{,}\) and so no linear combination of those two vectors will have a nonzero entry in the component where \(\uvec{v}_{\ell-2}\) has its leading one. From this, Proposition 18.5.6 tells us that \(\{\uvec{v}_{\ell-2},\uvec{v}_{\ell-1},\uvec{v}_\ell\}\) is linearly independent. Repeating this argument as we move up the rows of \(F\text{,}\) we see that the nonzero rows of \(F\) are linearly independent when taken altogether.
4. Applying Statement 3 of this corollary to the RREF for \(A\text{,}\) the nonzero rows of \(\RREF(A)\) form a basis for the row space of \(A\text{.}\) But the nonzero rows of \(\RREF(A)\) must all contain leading ones, so the number of vectors in a basis for the row space of \(A\) is equal to the number of leading ones in \(\RREF(A)\text{,}\) as desired.

We have previously encountered the equivalence of many of these statements, most recently in Theorem 10.5.3. So currently we only need to concern ourselves with the new statements. For each of these, if we can establish equivalence of the new statement to one of the old, then the new statement must be equivalent to all of the old, by the transitivity of logical equivalence.

The dimension of the column space of \(A\) is equal to the number of leading ones in its RREF, while the dimension of the null space of \(A\) is equal to the number of free variables, which is equal to the number of columns in the RREF that do not have a leading one. These two numbers must add up to the total number of columns in \(A\text{.}\)