# Day 10:

More rank theorems

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 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 $$Ax\in N(A^{\top})$$.
• If $$x\in N(A)$$, then we're done, so assume $$Ax\in N(A^{\top})$$.
• But $$Ax\in C(A)$$.
• We have already shown that $$C(A)\cap N(A^{\top}) = \{0\}$$.
• Since $$Ax\in N(A^{\top})$$ and $$Ax\in C(A)$$, this implies $$Ax=0$$, and hence $$x\in N(A)$$

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 $$x=Ay$$.
• $$A^{\top}Ay=0$$
• $$A^{\top}Ay=0$$ $$\Rightarrow$$ $$Ay=0$$
• $$Ay=0$$ $$\Rightarrow$$ $$x=0$$
• $$v_{1},\ldots,v_{k}$$ is independent $$\Rightarrow$$ the $$\alpha_{i}$$'s are all zero.

### Computing $$B$$ such that $$BA=\text{rref}(A)$$.

Theorem. Let $$A$$ be an $$m\times n$$ matrix. Let $$I$$ be an $$m\times m$$ identity matrix. If

$\text{rref}([A\ |\ I]) = [D\ |\ B]$

then $$BA=\text{rref}(A)$$.

Proof. Note that $\text{rref}([A\ |\ I]) = [\text{rref}(A)\ |\ B].$ Assume $$C$$ is the matrix such that

$C\cdot [A\ \vert\ I] = \text{rref}([A\ \vert\ I])= [\text{rref}(A)\ \vert\ B].$

Finally, note that $C\cdot [A\ \vert\ I] = [CA\ \vert\ CI] = [CA\ \vert\ C],$ and hence $$C=B$$ and $$BA=\text{rref}(A)$$. $$\Box$$

### Computing $$B$$ such that $$BA=\text{rref}(A)$$.

Example.

$\text{rref}\left(\left[\begin{array}{rrr|rr} 2 & 3 & 1 & 1 & 0\\ 2 & 3 & -2 & 0 & 1 \end{array}\right]\right) = \left[\begin{array}{rrr|rr} 1 & 3/2 & 0 & 1/3 & 1/6\\ 0 & 0 & 1 & 1/3 & -1/3 \end{array}\right]$

$\left[\begin{array}{rrr} 1/3 & 1/6\\ 1/3 & -1/3 \end{array}\right]\left[\begin{array}{rrr} 2 & 3 & 1\\ 2 & 3 & -2 \end{array}\right] = \left[\begin{array}{rrr} 1 & 3/2 & 0\\ 0 & 0 & 1\end{array}\right]$

### The Inverse of a matrix

Definition. Given a square matrix $$A$$, a square matrix $$B$$ such that $$AB=BA=I$$ is called the inverse of $$A$$. The inverse of $$A$$ is denoted $$A^{-1}$$. If a matrix $$A$$ has an inverse, then we say that $$A$$ is invertible.

Examples.

• If $$A=\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}$$, then $$A^{-1} = \begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix}$$

• If $$A = \begin{bmatrix} 1 & 0 & -2\\ 3 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$, then $$A = \begin{bmatrix} 1 & 0 & 2\\ -3 & 1 & -6\\ 0 & 0 & 1\end{bmatrix}$$

Theorem. A matrix $$A$$ is invertible if and only if $$A$$ is square and $$\text{rref}(A)=I$$.

Proof. Suppose $$\text{rref}(A)=I$$. There exists a matrix $$B$$ so that $$BA=\text{rref}(A)$$. From this we deduce that $$BA=I$$. For each elementary matrix $$E$$ there is an elementary matrix $$F$$ such that $$FE=I$$. Since $$B$$ is a product of elementary matrices $B= E_{k}\cdot E_{k-1}\cdots E_{1}$

We take $$F_{i}$$ such that $$F_{i}E_{i}=I$$ for each $$i$$, and set  $C = F_{1}\cdot F_{2}\cdots F_{k}$

and we see that $$CB=I$$. Finally, we have $AB = CBAB =C I B = CB = I.$

Now, suppose $$A$$ is invertible. By definition $$A$$ is square. If $$x$$ is a vector such that $$Ax=0$$, then $$A^{-1}Ax=A^{-1}0$$. Thus, $$x=0$$ is the only solution to $$Ax=0$$. Hence, $$\{0\}=N(A) = N(\operatorname{rref}(A))$$. This implies that each row of $$\operatorname{rref}(A)$$ must have a pivot. Since it is square, every column of $$\operatorname{rref}(A)$$ contains a pivot. This implies $$\operatorname{rref}(A)=I$$. $$\Box$$

Example. Let $A = \begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 0\\ 5 & 7 & 4\end{bmatrix}$

Is $$A$$ invertible? If it is, find $$A^{-1}$$.

Note that $\text{rref}(A)\neq I$ and hence $$A$$ is not invertible.

Note that

$\text{rref}\left(\left[\begin{array}{ccc|ccc} 2 & 3 & 4 & 1 & 0 & 0\\ 3 & 4 & 0 & 0 & 1 & 0\\ 5 & 7 & 4 & 0 & 0 & 1\end{array}\right]\right) = \left[\begin{array}{ccc|ccc} 1 & 0 &-16 & 0 & 7 & -4\\ 0 & 1 & 12 & 0 & -5 & 3\\ 0 & 0 & 0 & 1 & 1 & -1\end{array}\right]$

Example.  Let $A = \begin{bmatrix} 0 & -1 & -2\\ 1 & 1 & 1\\ -1 & -1 & 0\end{bmatrix}$

Is $$A$$ invertible? If it is, find $$A^{-1}$$.

Since $$\text{rref}(A)= I$$ we see that $$A$$ is invertible. Moreover, $A^{-1} = \begin{bmatrix} 1 & 2 & 1\\ -1 & -2 & -2\\ 0 & 1 & 1\end{bmatrix}$

Note that

$\text{rref}\left(\left[\begin{array}{ccc|ccc} 0 & -1 & -2 & 1 & 0 & 0\\ 1 & 1 & 1 & 0 & 1 & 0\\ -1 & -1 & 0& 0 & 0 & 1\end{array}\right]\right) = \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 2 & 1\\ 0 & 1 & 0 & -1 & -2 & -2\\ 0 & 0 & 1 & 0 & 1 & 1\end{array}\right]$

By John Jasper

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