Day 9:
Rank Nullity
Theorem. The following quantities are equal:
- The rank of \(A\)
- The number of pivots in \(\text{rref}(A)\)
- The number of pivots in \(\text{rref}(A^{\top})\)
- The rank of \(A^{\top}\)
Proof. The only thing we need to show is that \(\text{rref}(A)\) and \(\text{rref}(A^{\top})\) have the same number of pivots. But we'll actually show that \(\dim C(A^{\top})\) equals the number of pivots in \(\operatorname{rref}(A)\).
Theorem. If \(A\) is any matrix, then \(C(A^{\top}) = C(\operatorname{rref}(A)^{\top})\).
Let's recall the following useful theorem:
Example. \[A = \left[\begin{array}{rrrr} 2 & \phantom{-}4 & -2 & 8\\ 1 & 2 & 2 & -5\\ 1 & 2 & 0 & 1\end{array}\right]\]
\[\operatorname{rref}(A) = \left[\begin{array}{rrrr} 1 & \phantom{-}2 & \phantom{-}0 & 1\\ 0 & 0 & 1 & -3\\ 0 & 0 & 0 & 0\end{array}\right]\]
Since \(\left\{\begin{bmatrix} 2\\ 1\\ 1\end{bmatrix},\begin{bmatrix}-2\\ 2\\ 0\end{bmatrix}\right\}\) is a basis for \(C(A)\) we see that
\[\#\text{ pivots in }\operatorname{rref}(A) = \dim C(A)\]
Since \(C(A^{\top}) = C(\operatorname{rref}(A)^{\top})\), and hence \(\left\{\begin{bmatrix} 1\\ 2\\ 0\\ 1\end{bmatrix},\begin{bmatrix} 0\\ 0\\ 1\\ -3\end{bmatrix}\right\}\) spans \(C(A^{\top})\).
This set is also clearly independent, and hence is a basis for \(C(A^{\top})\). Hence
\[\dim(C(A^{\top}) = \#\text{ pivots in }\operatorname{rref}(A).\]
Proof. Let \(A\) be a matrix, and let \(w_{1}^{\top},w_{2}^{\top},\ldots,w_{\ell}^{\top}\) denote the nonzero rows of \(\operatorname{rref}(A)\), that is,
\[\operatorname{rref}(A) = \begin{bmatrix} - & w_{1}^{\top} & -\\ - & w_{2}^{\top} & -\\ & \vdots & \\ - & w_{\ell}^{\top} & -\\ - & 0 & -\\ & \vdots & \\ - & 0 & -\end{bmatrix}.\]
Each nonzero row contains a unique pivot, hence
\[\ell = \#\text{ pivots in }\operatorname{rref}(A).\]
Since \(C(A^{\top}) = C(\operatorname{rref}(A)^{\top})\), we deduce that \(\{w_{1},w_{2},\ldots,w_{\ell}\}\) spans \(C(A^{\top})\). However, looking at the pivots we see that is \(\{w_{1},w_{2},\ldots,w_{\ell}\}\) independent, and hence a basis for \(C(A^{\top}\), that is, \(\ell = \dim C(A^{\top})\). \(\Box\)
Theorem. The following quantities are equal:
- The rank of \(A\)
- The number of pivots in \(\text{rref}(A)\)
- The number of pivots in \(\text{rref}(A^{\top})\)
- The rank of \(A^{\top}\)
Corollary. The subspaces \(C(A)\) and \(C(A^{\top})\) have the same dimension.
Caution: \(C(A)\) and \(C(A^{\top})\) are almost never the same subspace. Indeed, if \(A\) is \(m\times n\), then \(C(A)\) is a subspace of \(\R^{m}\) and \(C(A^{\top})\) is a subspace of \(\R^{n}\).
Definition. Given a matrix \(A\in\mathbb{R}^{m\times n}\), the dimension of the null space of \(A\) is called the nullity of \(A\), and is denoted \(\operatorname{nullity}(A).\)
Hence, we now have at least three symbols for the same quantity:
\[\dim N(A) = \dim\operatorname{ker}(A) = \operatorname{nullity}(A).\]
We have already claimed that the nullity of \(A\) is equal to the number of non-pivot columns in \(\operatorname{rref}(A)\). Then next theorem shows that this is true.
Lemma. If \(A\in\mathbb{R}^{m\times n}\), then \(C(A^{\top}) \cap N(A) = \{0\}.\)
Proof. Suppose \(x\in C(A^{\top})\) and \(x\in N(A)\). By the definition of the row space there is some \(y\in\mathbb{R}^{m}\) such that \(x=A^{\top}y\). Multiplying by \(A\) on both sides we have
\[0 = Ax = AA^{\top}y.\]
Multiplying on the left by \(y^{\top}\) we obtain
\[ 0 = y^{\top} 0 = y^{\top}AA^{\top}y = (A^{\top}y)^{\top}(A^{\top}y) = x^{\top}x.\]
Now, suppose \(x = [x_{1}\ \ x_{2}\ \ \cdots\ \ x_{n}]^{\top}\). This last equality shows that
\[x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2} = 0,\]
which clearly implies \(x=0\). \(\Box\)
Rank Nullity
Theorem (The Rank-nullity theorem). If \(A\) is an \(m\times n\) matrix, then \[\text{rank}(A) + \operatorname{nullity}(A) = n.\]
Proof.
- Let W:=\(\{w_{1},w_{2},\ldots,w_{\ell}\}\) be the nonzero columns of \(\operatorname{rref}(A)^{\top}\).
- We have already seen that \(W\) is a basis for \(C(A^{\top})\), and hence \(\ell = \dim C(A^{\top}) = \dim C(A) = \operatorname{rank}(A)\).
- Let \(\{v_{1},\ldots,v_{k}\}\) be a basis for \(N(A)\) (hence \(k=\operatorname{nullity}(A)\))
- Note that \(\operatorname{rref}(A)v_{i} = 0\) for all \(i\in\{1,2,\ldots,k\}\).
- I claim that \(\{w_{1},w_{2},\ldots,w_{\ell},v_{1},v_{2},\ldots,v_{k}\}\) is an independent set in \(\R^{n}\).
- Suppose there are scalars \(a_{1},\ldots,a_{\ell}\) and \(b_{1},\ldots,b_{k}\) such that \[\sum_{i=1}^{\ell}a_{i}w_{i} + \sum_{j=1}^{k}b_{j}v_{j} = \boldsymbol{0}\]
Rank Nullity
Proof continued.
- This implies that \[\sum_{i=1}^{\ell}a_{i}w_{i} = - \sum_{j=1}^{k}b_{j}v_{j} = \begin{bmatrix}c_{1}\\ c_{2}\\ \vdots\\ c_{n}\end{bmatrix} = \boldsymbol{c}\]
- Hence \[ \boldsymbol{c} = \sum_{i=1}^{\ell}a_{i}w_{i} \in C(A^{\top}) \quad\text{and}\quad \boldsymbol{c} = - \sum_{j=1}^{k}b_{j}v_{j} \in N(A).\]
- By the lemma, this implies \[\mathbf{c} = \sum_{i=1}^{\ell}a_{i}w_{i} = - \sum_{j=1}^{k}b_{j}v_{j} = \boldsymbol{0}\]
Theorem (The Rank-nullity theorem). If \(A\) is an \(m\times n\) matrix, then \[\text{rank}(A) + \operatorname{nullity}(A) = n.\]
Rank Nullity
Proof continued.
- Since both sets \(\{w_{1},\ldots,w_{\ell}\}\) and \(\{v_{1},\ldots,v_{k}\}\) are independent, we conclude that \(a_{1}=a_{2}=\cdots =a_{\ell} = 0 = b_{1} = b_{2} = \cdots = b_{k}\)
- \(\{w_{1},w_{2},\ldots,w_{\ell},v_{1},v_{2},\ldots,v_{k}\}\) is an independent set in \(\R^{n}\).
- Therefore, \( k+\ell \leq n\)
- We have already seen how to construct an independent set \(\{u_{1},\ldots,u_{r}\}\subset N(A)\) where \[r = (\#\text{ of non-pivot columns in }\text{rref}(A)) = n-\ell.\]
- This implies \(\ k = \operatorname{nullity}(A) \geq r = n-\ell\).
- Thus \(n\leq k+\ell\leq n\), that is \(k+\ell = n\). \(\Box\)
Theorem (The Rank-nullity theorem). If \(A\) is an \(m\times n\) matrix, then \[\text{rank}(A) + \operatorname{nullity}(A) = n.\]
Example 2. For matrices \(A\in\mathbb{R}^{m\times n}\) such that \(\operatorname{rank}(A) = n\), it _____________ holds that \(\operatorname{nullity}(A) = 0\).
Fill in the blank with always, sometimes, or never:
always
Example 3. For matrices \(A\in\mathbb{R}^{m\times n}\) such that \(\operatorname{rank}(A) = m\) and \(n\leq m\) it _____________ holds that \(\operatorname{nullity}(A) = 0\).
always
Example 1. For matrices \(A\in\mathbb{R}^{3\times 4}\) such that \(\operatorname{rank}(A) = 3\), it _____________ holds that \(\operatorname{nullity}(A) = 1\).
always
Example 4. For matrices \(A\in\mathbb{R}^{m\times n}\) such that \(m<n\), it _____________ holds that \(\operatorname{nullity}(A) = \operatorname{nullity}(A^{\top})\).
never
Theorem 1. \(\text{rank}(AB)\leq \min\{\text{rank}(A),\text{rank}(B)\}\).
Rank theorems
Proof.
- Note that the columns of \(AB\) are all in \(C(A)\), hence the dimension of \(C(AB)\) is at most the dimension of \(C(A)\), that is, \[\text{rank}(AB)\leq \text{rank}(A).\]
- \(\text{rank}(AB)=\text{rank}((AB)^{\top}) = \text{rank}(B^{\top}A^{\top})\)
- By the first bullet \(\text{rank}(B^{\top}A^{\top})\leq \text{rank}(B^{\top}).\)
- Since \(\text{rank}(B^{\top})=\text{rank}(B)\) we have \[\text{rank}(AB) = \text{rank}(B^{\top}A^{\top})\leq \text{rank}(B^{\top}) = \text{rank}(B).\]
- Hence, \[\text{rank}(AB)\leq \text{rank}(A)\quad\text{ and }\quad\text{rank}(AB)\leq \text{rank}(B).\]
- This is the same as \[\text{rank}(AB)\leq \min\{\text{rank}(A),\text{rank}(B)\}.\]
Theorem 2. \(\text{rank}(A+B)\leq \text{rank}(A) + \text{rank}(B)\).
Rank theorems
Proof.
- Let \(v_{1},\ldots,v_{k}\) be a basis for \(C(A)\), where \(k=\text{rank}(A)\)
- Let \(w_{1},\ldots,w_{\ell}\) be a basis for \(C(B)\), where \(\ell=\text{rank}(B)\)
- The columns of \(A+B\) are live in \(\text{span}\{v_{1},\ldots,v_{k},w_{1},\ldots,w_{\ell}\}\)
- So, \(C(A+B)\subset\text{span}\{v_{1},\ldots,v_{k},w_{1},\ldots,w_{\ell}\}\)
- Hence \[\text{rank}(A+B)\leq k+\ell = \text{rank}(A)+\text{rank}(B)\]
Lemma. \(N(A)=N(A^{\top}A).\)
Rank theorems
Proof. First we show that \(N(A)\subset N(A^{\top}A)\).
- \(x\in N(A)\) \(\Rightarrow\) \(Ax=0\) \(\Rightarrow\) \(A^{\top}Ax=0\) \(\Rightarrow\) \(x\in N(A^{\top}A)\).
Next, we show the reverse inclusion \(N(A)\supset N(A^{\top}A).\)
- Let \(x\in N(A^{\top}A)\)
- Then \(A^{\top}Ax=0\).
- Either \(x\in N(A)\) or \(Ax\in N(A^{\top})\).
- But \(Ax\in C(A)\).
- We have already shown that \(C(A)\) and \(N(A^{\top})\) are orthogonal.
- If \(Ax\) is also in \(N(A^{\top})\), then \((Ax)\cdot(Ax) = 0\)
- This only happens if \(Ax=0\).
Theorem 3. \(\text{rank}(A^{\top}A)=\text{rank}(AA^{\top}) = \text{rank}(A) = \text{rank}(A^{\top})\).
Rank theorems
Proof.
- By the lemma \(N(A) = N(A^{\top}A)\).
- Let \(v_{1},\ldots,v_{k}\) be a basis for \(C(A)\).
- Then, \(A^{\top}v_{1},\ldots,A^{\top}v_{k}\) spans \(C(A^{\top}A)\).
- We claim that this set of vectors is a basis.
- Assume we have scalars \(\alpha_{1},\ldots,\alpha_{k}\) so that \[0 = \alpha_{1}A^{\top}v_{1} + \cdots + \alpha_{k}A^{\top}v_{k} = A^{\top}(\alpha_{1}v_{1} + \cdots + \alpha_{k}v_{k})\]
- Set \(x = \alpha_{1}v_{1} + \cdots + \alpha_{k}v_{k}\) and note that \(x\in C(A)\).
- There is some vector \(y\) so that \(y=Ax\).
- \(A^{\top}Ay=0\)
- \(A^{\top}Ay=0\) \(\Rightarrow\) \(Ay=0\)
- \(Ay=0\) \(\Rightarrow\) \(x=0\)
- \(v_{1},\ldots,v_{k}\) is independent \(\Rightarrow\) the \(\alpha_{i}\)'s are all zero.
Linear Algebra Day 9
By John Jasper
Linear Algebra Day 9
- 506