LAR Rank and Nullity

Section 1.3 Rank and Nullity

Subsection 1.3.1 Some fundamental subspaces

Let \(T:V\to W\) be a linear map between vector spaces over a field \(F.\) We have defined the kernel of \(T\), \(\ker(T)=\Null(T)\text{,}\) (also called the nullspace) and noted that it is a subspace of the domain \(V.\) The image of \(T\), \(\Im(T),\) is a subspace of the codomain \(W.\)

Subsection 1.3.2 The rank-nullity theorem

Given a linear map \(T:V\to W\text{,}\) with \(V\) finite dimensional, there is a fundamental theorem relating the dimension of \(V\) to the dimensions of \(\ker(T)\) and \(\Im(T).\)

Theorem 1.3.1. The Rank-Nullity Theorem (aka the dimension theorem).

Let \(T:V\to W\) be a linear map, with \(V\) a finite-dimensional vector space. Then

\begin{equation*} \dim V = \rank(T) + \nullity(T) = \dim \Im(T) + \dim \ker(T). \end{equation*}

Proof.

Let \(n = \dim V,\) and recall that if \(\{v_1, \dots, v_n\}\) is any basis for \(V\text{,}\) then \(\Im(T) = \Span(\{T(v_1), \dots, T(v_n)\}).\)

First consider the case that \(T\) is injective. This means that \(\ker(T) = \{0\}\text{,}\) so that \(\nullity(T) = 0\text{.}\) By Proposition 1.2.3, the set \(\{T(v_1), \dots, T(v_n)\}\) is linearly independent, and since this set spans \(\Im(T)\text{,}\) it is a basis for \(\Im(T)\text{,}\) so its cardinality equals the dimension of the image, i.e., \(\rank(T).\) Thus \(\rank(T)=n\text{,}\) and we see that

\begin{equation*} n = \dim V = n + 0 = \rank(T) + \nullity(T). \end{equation*}

Now consider the case where \(\ker(T) \ne \{0\}\text{.}\) Let \(\{u_1, \dots, u_k\}\) be a basis for \(\ker(T)\text{,}\) hence \(\nullity(T) = k.\) Since \(\{u_1, \dots, u_k\}\) is a linearly independent set, by [provisional cross-reference: prop-extend-independent-set-to-basis], it can be extended to a basis for \(V\text{:}\)

\begin{equation*} \{u_1, \dots, u_k, u_{k+1}, \dots, u_n\} \end{equation*}

To establish the theorem, we need only show that \(\rank(T) = n-k.\) Since \(\{u_1, \dots, u_k, u_{k+1}, \dots, u_n\}\) is a basis for \(V\text{,}\) \(\Im(T) = \Span(\{T(u_1), \dots, T(u_n)\})\text{,}\) but we recall that \(u_1, \dots, u_k \in \ker(T)\text{,}\) so that \(\Im(T) = \Span(\{T(u_{k+1}), \dots, T(u_n)\})\text{.}\) Thus we know \(\rank T \le n-k.\) To obtain an equality, we need only show that the set \(\{T(u_{k+1}), \dots, T(u_n)\}\) is linearly independent.

Suppose to the contrary, that the set is linearly independent. Then there exists scalars \(a_i \in F\text{,}\) not all zero, so that

\begin{equation*} \sum_{i=k+1}^n a_i T(u_i) = 0. \end{equation*}

By linearity, this says \(T(\sum_{i=k+1}^r a_i u_i) = 0,\) which means \(\sum_{i=k+1}^r a_i u_i \in \ker(T).\) But this in turn says that \(\sum_{i=k+1}^r a_i u_i \in \Span(\{u_1, \dots, u_k\})\) implying the full set \(\{u_1, \dots, u_n\}\) is linearly dependent, contradicting that it is a basis for \(V.\) This completes the proof.

Let’s do a simple example.

Example 1.3.2.

Consider the linear map \(T:P_3(\R) \to P_3(\R)\) given by \(T(f) = f'' + f'\text{,}\) where \(f'\) and \(f''\) are the first and second derivatives of \(f.\)

The domain has dimension 4 with standard basis \(\{1, x, x^2, x^3\}\text{,}\) so

\begin{equation*} \Im(T) = \Span\{T(1), T(x), T(x^2), T(x^3)\} \end{equation*}

One easily checks that \(\Im(T) = \Span\{0, 1, 2+2x, 6x+3x^2\}= \Span\{1,x,x^2\}\text{.}\) At the very least we know that \(\rank(T) \le 3\text{,}\) and since \(T(1)=0\text{,}\) we must have \(\nullity(T) \ge 1\text{.}\) Now since \(\{1, x, x^2\}\) is a linearly independent set, we know that \(\rank(T) =3\) which means that \(\nullity(T)=1\) by Theorem 1.3.1. It follows that \(\{1\}\) is a basis for the nullspace.

Subsection 1.3.3 Computing rank and nullity

Let \(A \in M_{m\times n}(F)\) be a matrix. Then \(T(x) = Ax\) defines a linear map \(T:F^n\to F^m\text{.}\) Indeed in Subsection 1.4.2, we shall see how to translate the action of an arbitrary linear map between finite-dimensional vectors spaces into an action of a matrix on column vectors.

Let’s recall how to extract the image and kernel of the linear map \(x\mapsto Ax\text{.}\) We know that the image of any linear map is obtained by taking the span of \(T(e_1), \dots, T(e_n)\) where \(\{e_1, \dots, e_n\}\) is any basis for \(F^n\text{,}\) the domain. Indeed if we choose the \(e_i\) to be the standard basis vectors (with a 1 in the \(i\)th coordinate and zeroes elsewhere), then \(T(e_j)\) is simply the \(j\)th column of the matrix \(A.\) Thus \(\Im(T)\) is the column space of \(A\text{.}\) However to determine the rank of \(A\text{,}\) we would need to know which columns form a basis. We’ll get to that in a moment.

The nullspace of \(T\text{,}\) is the set of solutions to the homogeneous linear system \(Ax=0.\) You may recall that a standard method to deduce the solutions is to put the matrix \(A\) in reduced echelon form. That means that all rows of zeros are at the bottom, the leading nonzero entry of each row is a one, and in every column containing a leading 1, all other entries are zero. These leading ones play several roles.

Proposition 1.3.3.

Given the variables \(x_1, \dots, x_n\) in the system \(Ax=0\text{,}\) a 1 in the \(j\)th column of the reduced row echelon form of \(A\text{,}\) called a pivot, means that the variable \(x_j\) is a constrained variable while the remaining variables are free variables. Thus if there are \(r\) pivots, there are \(n-r\) free variables, and \(n-r=\nullity(T)\text{;}\) it follows that \(r=\rank(T).\)
The pivot columns of \(A\) (the columns of \(A\) in which there is a pivot in the reduced row echelon form of \(A\)) can be taken as a basis of the column space of \(A.\)
The row rank of \(A\) (number of linearly independent rows) equals the column rank of \(A\) (number of linearly independent columns).

Subsection 1.3.4 Elementary Row and Column operations

The following are a series of facts about elementary row and column operations on an \(m\times n\) matrix \(A\text{.}\)

The matrix \(A\) is put in reduced row echelon form by a sequence of elementary row operations.
Each elementary row operation can be achieved by left multiplication of \(A\) (\(A\mapsto EA\)) by an \(m\times m\) elementary matrix.
Each elementary column operation can be achieved by right multiplication of \(A\) (\(A\mapsto AE\)) by an \(n\times n\) elementary matrix.
Every elementary matrix is invertible and its inverse in again an elementary matrix of the same type.
The rank of an \(m\times n\) matrix is unchanged by elementary row or column operations, that is \(\rank(EA) = \rank(A)\) and \(\rank(AE)=\rank(A)\) for appropriately sized elementary matrices \(E.\)

Every invertible matrix is a product of elementary matrices, and this leads to the

Algorithm 1.3.4.

To determine whether an \(n \times n\) matrix \(A\) is invertible and if so find its inverse, reduce to row-echelon form the "augmented" \(n\times 2n\) matrix

\begin{equation*} [A|I_n] \mapsto [R|A']. \end{equation*}

The matrix \(A\) is invertible if and only if \(R=I_n,\) and in that case \(A'\) is the inverse \(A^{-1}.\)

Exercises Exercises

1.

Let \(A\) be an \(m\times n\) matrix and \(E\) an elementary matrix of the appropriate size.

Are the row spaces of \(A\) and \(EA\) the same?
Are the column spaces of \(A\) and \(AE\) the same?
If \(R\) is the reduced row-echelon form of \(A\text{,}\) are the nonzero rows of \(R\) a basis for the row space of \(A\text{?}\)
If \(R\) is the reduced row-echelon form of \(A\text{,}\) is the column space of \(R\) the same as the column space of \(A\text{?}\)

Answer.

yes; yes; yes (why?); no; If \(A=\begin{bmatrix} 1\amp 1\\1\amp 1 \end{bmatrix}\text{,}\) then \(R=\begin{bmatrix}1\amp1\\0\amp0 \end{bmatrix}\text{,}\) and

\begin{equation*} \Span(\begin{bmatrix}1\\1\end{bmatrix}) \ne \Span(\begin{bmatrix}1\\0\end{bmatrix}) \end{equation*}

2.

Given an \(m\times n\) matrix \(A,\) show that there exist (appropriately sized) elementary matrices \(U,V\) so that \(UAV\) has the form

\begin{equation*} UAV = \begin{bmatrix} I_r\amp \mathbf 0\\ \mathbf 0\amp\mathbf 0 \end{bmatrix}. \end{equation*}

where \(I_r\) is an \(r\times r\) identity matrix with \(r = \rank(A)\text{,}\) and the other entries are all zeros.

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