09. 기본행렬연산Elementary Matrix Operation

Solution.

1. Columns 1 and 3 of $A$ agree with the corresponding columns of $B$, and only column 2 differs. Subtracting twice column 1 from column 2 of $A$ gives $$\begin{pmatrix}2\\0\\-1\end{pmatrix}-2\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}0\\-2\\-3\end{pmatrix},$$ which matches column 2 of $B$. Hence the elementary operation transforming $A$ into $B$ is the column operation $C_2 \to C_2-2C_1$.
2. Applying $R_2 \to R_2-R_1$, $R_3 \to R_3-R_1$ to $B$ gives $$\begin{pmatrix}1&0&3\\1&-2&1\\1&-3&1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&0&3\\0&-2&-2\\0&-3&-2\end{pmatrix}.$$ Applying $R_2 \to -\frac12R_2$, $R_3 \to R_3+3R_2$ then gives $$\begin{pmatrix}1&0&3\\0&-2&-2\\0&-3&-2\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&0&3\\0&1&1\\0&0&1\end{pmatrix},$$ and finally applying $R_1 \to R_1-3R_3$, $R_2 \to R_2-R_3$ gives $$\begin{pmatrix}1&0&3\\0&1&1\\0&0&1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I_3.$$ Thus the six elementary row operations $R_2\to R_2-R_1,\ R_3\to R_3-R_1,\ R_2\to-\frac12R_2,\ R_3\to R_3+3R_2,\ R_1\to R_1-3R_3,\ R_2\to R_2-R_3$, applied in order, transform $B$ into $I_3$.

Solution.

1. $\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}$ is the elementary matrix of type 1 obtained from $I_3$ by interchanging rows 1 and 3. Performing the same interchange twice returns the original matrix, so this matrix is its own inverse: $$\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}^{-1}=\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\end{pmatrix}.$$ Indeed, multiplying the two matrices immediately gives $I_3$.
2. $\begin{pmatrix}1&0&0\\0&3&0\\0&0&1\end{pmatrix}$ is the elementary matrix of type 2 obtained from $I_3$ by scaling row 2 by $3$. The inverse operation scales row 2 by $\frac13$, so $$\begin{pmatrix}1&0&0\\0&3&0\\0&0&1\end{pmatrix}^{-1}=\begin{pmatrix}1&0&0\\0&\frac13&0\\0&0&1\end{pmatrix}.$$ Indeed, multiplying the two diagonal matrices gives $\operatorname{diag}(1,\,3\cdot\tfrac13,\,1)=I_3$.
3. $\begin{pmatrix}1&0&0\\0&1&0\\-2&0&1\end{pmatrix}$ is the elementary matrix of type 3 obtained from $I_3$ by applying $R_3 \to R_3-2R_1$. The inverse operation is $R_3 \to R_3+2R_1$, so $$\begin{pmatrix}1&0&0\\0&1&0\\-2&0&1\end{pmatrix}^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\2&0&1\end{pmatrix}.$$ Indeed, multiplying the two matrices gives row 3 equal to $(-2+2,\,0,\,1)=(0,0,1)$, so the product is $I_3$.

Proof. For an $m\times n$ matrix $A=(a_{ij})$, call $A$ upper triangular (for a rectangular matrix) if the $(i,j)$ entry is $0$ whenever $i>j$. We induct on the number of rows $m$.

If $m=1$, there are no rows below $A$'s single row, so $A$ is trivially upper triangular.

Suppose $m>1$ and the statement holds for every matrix with fewer than $m$ rows.

Case 1: column 1 of $A$ is the zero vector. Then column 1 already satisfies the required condition (the $(i,1)$ entry is $0$ for every $i>1$). Applying the induction hypothesis to the $(m-1)\times(n-1)$ submatrix formed by rows $2,\dots,m$ and columns $2,\dots,n$, there is a sequence of type 1 and type 3 elementary row operations (applied only to rows $2,\dots,m$, leaving row 1 untouched) that transforms this submatrix into upper triangular form. These operations do not affect the entries of column 1 (which are all $0$), so the entire matrix $A$ becomes upper triangular.

Case 2: column 1 of $A$ is not the zero vector. Then there is some $i_0$ with $a_{i_01}\neq0$. If $i_0\neq1$, a type 1 operation interchanging rows $1$ and $i_0$ makes the $(1,1)$ entry nonzero. Now, for $j=2,\dots,m$, applying the type 3 operations $R_j \to R_j - \dfrac{a_{j1}}{a_{11}}R_1$ in succession makes every entry of column 1 below row 1 equal to $0$. Applying the induction hypothesis to the $(m-1)\times(n-1)$ submatrix formed by rows $2,\dots,m$ and columns $2,\dots,n$, a sequence of type 1 and type 3 operations (applied only to rows $2,\dots,m$) transforms this submatrix into upper triangular form. These operations do not disturb the already-zero entries of column 1, so the entire matrix $A$ ends up with $(i,j)$ entry equal to $0$ whenever $i>j$, i.e. $A$ is upper triangular.

In both cases only type 1 and type 3 operations were used, so by induction every $m\times n$ matrix can be transformed into upper triangular form by a sequence of elementary row operations of type 1 and 3.

Proof. Fix distinct row indices $i\neq j$. We show that the type 1 operation interchanging $R_i$ and $R_j$ is obtained by the following succession of type 2 and type 3 operations (all other rows are left unchanged). Let $a,b$ denote the initial values of $R_i,R_j$.

1. Apply $R_i \to R_i+R_j$ (type 3): $(R_i,R_j)=(a+b,\ b)$.
2. Apply $R_j \to R_j-R_i$ (type 3): $(R_i,R_j)=(a+b,\ b-(a+b))=(a+b,\ -a)$.
3. Apply $R_i \to R_i+R_j$ (type 3): $(R_i,R_j)=((a+b)+(-a),\ -a)=(b,\ -a)$.
4. Apply $R_j \to -R_j$ (type 2): $(R_i,R_j)=(b,\ a)$.

After these four steps $(R_i,R_j)$ has changed from $(a,b)$ to $(b,a)$, so the two rows have been exactly interchanged. Each step affects only rows $i$ and $j$, so every other row remains unchanged throughout. Hence the type 1 operation (interchanging two rows) is obtained by composing four elementary row operations in the order type 3, type 3, type 3, type 2.

Solution.

1. Applying $R_3 \to R_3-R_1$ gives $$\begin{pmatrix}1&1&0\\0&1&1\\1&1&0\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&1&0\\0&1&1\\0&0&0\end{pmatrix}.$$ This is in row echelon form with $2$ nonzero rows, so the rank is $2$.
2. Applying $R_2 \to R_2-R_1$ gives $$\begin{pmatrix}1&0&2\\1&1&4\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&0&2\\0&1&2\end{pmatrix}.$$ Both rows are nonzero and the matrix is in row echelon form, so the rank is $2$ (the maximum possible, since there are only two rows).
3. Applying $R_2 \to R_2-R_1$, $R_4 \to R_4-R_1$ gives $$\begin{pmatrix}1&2&3&1&1\\1&4&0&1&2\\0&2&-3&0&1\\1&0&0&0&0\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&2&3&1&1\\0&2&-3&0&1\\0&2&-3&0&1\\0&-2&-3&-1&-1\end{pmatrix}.$$ Applying $R_3 \to R_3-R_2$, $R_4 \to R_4+R_2$ then gives $$\begin{pmatrix}1&2&3&1&1\\0&2&-3&0&1\\0&2&-3&0&1\\0&-2&-3&-1&-1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&2&3&1&1\\0&2&-3&0&1\\0&0&0&0&0\\0&0&-6&-1&0\end{pmatrix}.$$ Interchanging rows 3 and 4 ($R_3 \leftrightarrow R_4$) gives the row echelon form $$\begin{pmatrix}1&2&3&1&1\\0&2&-3&0&1\\0&0&-6&-1&0\\0&0&0&0&0\end{pmatrix},$$ which has $3$ nonzero rows, so the rank is $3$.

Solution. We first find the rank of $A=\begin{pmatrix}2&1\\-1&2\\2&1\end{pmatrix}$. Applying $R_3 \to R_3-R_1$ makes row 3 the zero vector, and the remaining two rows $(2,1),(-1,2)$ are linearly independent since $2\cdot2-1\cdot(-1)=5\neq0$. Hence $\operatorname{rank}(A)=2=\min(3,2)$, so $D=\begin{pmatrix}I_2\\O\end{pmatrix}=\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}$ and no column operations are needed ($Q=I_2$).

Applying $R_3 \to R_3-R_1$ to $A$ gives $$\begin{pmatrix}2&1\\-1&2\\2&1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}2&1\\-1&2\\0&0\end{pmatrix},$$ and applying $R_2 \to R_2+\frac12R_1$ gives $$\begin{pmatrix}2&1\\-1&2\\0&0\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}2&1\\0&\frac52\\0&0\end{pmatrix}.$$ Applying $R_1 \to \frac12R_1$, $R_2 \to \frac25R_2$ gives $$\begin{pmatrix}2&1\\0&\frac52\\0&0\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&\frac12\\0&1\\0&0\end{pmatrix},$$ and finally applying $R_1 \to R_1-\frac12R_2$ gives $$\begin{pmatrix}1&\frac12\\0&1\\0&0\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}=D.$$

Applying the same five row operations, in the same order, to $I_3$ directly produces the product $E$ of the corresponding elementary matrices, which transforms $A$ into $D$: $$E=\begin{pmatrix}\frac25&-\frac15&0\\\frac15&\frac25&0\\-1&0&1\end{pmatrix},\qquad EA=D$$ (verified directly by multiplying $E$ and $A$). Computing the inverse of $E$ gives $$P:=E^{-1}=\begin{pmatrix}2&1&0\\-1&2&0\\2&1&1\end{pmatrix},$$ and expanding $\det(P)$ along column 3 gives $\det(P)=1\cdot\det\begin{pmatrix}2&1\\-1&2\end{pmatrix}=5\neq0$, so $P$ is invertible.

Taking $Q=I_2$, $$PDQ=PD=\begin{pmatrix}2&1&0\\-1&2&0\\2&1&1\end{pmatrix}\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}=\begin{pmatrix}2\cdot1+1\cdot0+0\cdot0&2\cdot0+1\cdot1+0\cdot0\\-1\cdot1+2\cdot0+0\cdot0&-1\cdot0+2\cdot1+0\cdot0\\2\cdot1+1\cdot0+1\cdot0&2\cdot0+1\cdot1+1\cdot0\end{pmatrix}=\begin{pmatrix}2&1\\-1&2\\2&1\end{pmatrix}=A,$$ so $$A=PDQ,\qquad P=\begin{pmatrix}2&1&0\\-1&2&0\\2&1&1\end{pmatrix},\quad D=\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix},\quad Q=I_2$$ is the desired decomposition.

Proof. Let $L_A:F^n\to F^m$ be the left-multiplication transformation $L_A(x)=Ax$, so that the rank of a matrix is defined by $\operatorname{rank}(A):=\dim(\operatorname{Im}(L_A))$. For a nonzero scalar $c$, $L_{cA}(x)=(cA)x=c(Ax)=cL_A(x)$, so $L_{cA}=cL_A$ and hence $$\operatorname{Im}(L_{cA})=\{cL_A(x):x\in F^n\}=\{cy:y\in\operatorname{Im}(L_A)\}.$$

Since $\operatorname{Im}(L_A)$ is a subspace, it is closed under scalar multiplication, so $\{cy:y\in\operatorname{Im}(L_A)\}\subseteq\operatorname{Im}(L_A)$. Conversely, for any $w\in\operatorname{Im}(L_A)$, since $c\neq0$ we have $w=c(c^{-1}w)$ with $c^{-1}w\in\operatorname{Im}(L_A)$, so $w\in\{cy:y\in\operatorname{Im}(L_A)\}$. Therefore $$\operatorname{Im}(L_{cA})=\{cy:y\in\operatorname{Im}(L_A)\}=\operatorname{Im}(L_A),$$ and consequently $$\operatorname{rank}(cA)=\dim(\operatorname{Im}(L_{cA}))=\dim(\operatorname{Im}(L_A))=\operatorname{rank}(A).$$

Solution.

1. Starting from $(A\mid I)$, $$\left(\begin{array}{cc|cc}1&2&1&0\\1&1&0&1\end{array}\right)$$ applying $R_2 \to R_2-R_1$ gives $$\left(\begin{array}{cc|cc}1&2&1&0\\0&-1&-1&1\end{array}\right),$$ applying $R_2 \to -R_2$ gives $$\left(\begin{array}{cc|cc}1&2&1&0\\0&1&1&-1\end{array}\right),$$ and finally applying $R_1 \to R_1-2R_2$ gives $$\left(\begin{array}{cc|cc}1&0&-1&2\\0&1&1&-1\end{array}\right).$$ Hence $$\begin{pmatrix}1&2\\1&1\end{pmatrix}^{-1}=\begin{pmatrix}-1&2\\1&-1\end{pmatrix}.$$ Indeed $$\begin{pmatrix}1&2\\1&1\end{pmatrix}\begin{pmatrix}-1&2\\1&-1\end{pmatrix}=\begin{pmatrix}-1+2&2-2\\-1+1&2-1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix},$$ confirming this.
2. Starting from $(A\mid I)$, $$\left(\begin{array}{ccc|ccc}1&2&1&1&0&0\\1&3&4&0&1&0\\2&3&-1&0&0&1\end{array}\right)$$ applying $R_2 \to R_2-R_1$, $R_3 \to R_3-2R_1$ gives $$\left(\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&1&3&-1&1&0\\0&-1&-3&-2&0&1\end{array}\right).$$ Applying $R_3 \to R_3+R_2$ then gives $$\left(\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&1&3&-1&1&0\\0&0&0&-3&1&1\end{array}\right),$$ so row 3 of the left block becomes the zero vector. Indeed, the rows of the original matrix $A$ satisfy $$-3(1,2,1)+(1,3,4)+(2,3,-1)=(0,0,0),\qquad\text{i.e.}\quad R_3=3R_1-R_2,$$ a linear dependence relation. Hence $\operatorname{rank}(A)\leq2$, which is less than $3$. Since a square matrix whose rank is less than its number of rows is not invertible, $A$ has no inverse.
3. Starting from $(A\mid I)$, $$\left(\begin{array}{cccc|cccc}1&2&1&0&1&0&0&0\\2&5&5&1&0&1&0&0\\-2&-3&0&3&0&0&1&0\\3&4&-2&-3&0&0&0&1\end{array}\right)$$ applying $R_2 \to R_2-2R_1$, $R_3 \to R_3+2R_1$, $R_4 \to R_4-3R_1$ gives $$\left(\begin{array}{cccc|cccc}1&2&1&0&1&0&0&0\\0&1&3&1&-2&1&0&0\\0&1&2&3&2&0&1&0\\0&-2&-5&-3&-3&0&0&1\end{array}\right).$$ Applying $R_1 \to R_1-2R_2$, $R_3 \to R_3-R_2$, $R_4 \to R_4+2R_2$ then gives $$\left(\begin{array}{cccc|cccc}1&0&-5&-2&5&-2&0&0\\0&1&3&1&-2&1&0&0\\0&0&-1&2&4&-1&1&0\\0&0&1&-1&-7&2&0&1\end{array}\right),$$ and applying $R_3 \to -R_3$ to make the pivot $1$ gives $$\left(\begin{array}{cccc|cccc}1&0&-5&-2&5&-2&0&0\\0&1&3&1&-2&1&0&0\\0&0&1&-2&-4&1&-1&0\\0&0&1&-1&-7&2&0&1\end{array}\right).$$ Applying $R_1 \to R_1+5R_3$, $R_2 \to R_2-3R_3$, $R_4 \to R_4-R_3$ gives $$\left(\begin{array}{cccc|cccc}1&0&0&-12&-15&3&-5&0\\0&1&0&7&10&-2&3&0\\0&0&1&-2&-4&1&-1&0\\0&0&0&1&-3&1&1&1\end{array}\right),$$ and finally applying $R_1 \to R_1+12R_4$, $R_2 \to R_2-7R_4$, $R_3 \to R_3+2R_4$ gives $$\left(\begin{array}{cccc|cccc}1&0&0&0&-51&15&7&12\\0&1&0&0&31&-9&-4&-7\\0&0&1&0&-10&3&1&2\\0&0&0&1&-3&1&1&1\end{array}\right).$$ Hence $$A^{-1}=\begin{pmatrix}-51&15&7&12\\31&-9&-4&-7\\-10&3&1&2\\-3&1&1&1\end{pmatrix}.$$

As a check, we verify $AA^{-1}=I_4$ on columns 1 and 4 of $A^{-1}$. Taking the dot product of each row of $A$ with column 1 of $A^{-1}$, namely $(-51,31,-10,-3)^t$, gives $$(1,2,1,0)\cdot(-51,31,-10,-3)=-51+62-10+0=1,$$ $$(2,5,5,1)\cdot(-51,31,-10,-3)=-102+155-50-3=0,$$ $$(-2,-3,0,3)\cdot(-51,31,-10,-3)=102-93+0-9=0,$$ $$(3,4,-2,-3)\cdot(-51,31,-10,-3)=-153+124+20+9=0,$$ which is exactly $e_1=(1,0,0,0)^t$. Likewise, for column 4 of $A^{-1}$, namely $(12,-7,2,1)^t$, $$(1,2,1,0)\cdot(12,-7,2,1)=12-14+2+0=0,\quad(2,5,5,1)\cdot(12,-7,2,1)=24-35+10+1=0,$$ $$(-2,-3,0,3)\cdot(12,-7,2,1)=-24+21+0+3=0,\quad(3,4,-2,-3)\cdot(12,-7,2,1)=36-28-4-3=1,$$ which is exactly $e_4=(0,0,0,1)^t$. Thus $AA^{-1}=I_4$ is confirmed on both columns.

Solution.

1. With respect to the standard basis $\{1,x,x^2\}$, $T(1)=-1$, $T(x)=2-x$, $T(x^2)=2+4x-x^2$, so the matrix representation of $T$ is $$M:=[T]_\beta=\begin{pmatrix}-1&2&2\\0&-1&4\\0&0&-1\end{pmatrix}.$$ $M$ is upper triangular with all diagonal entries equal to $-1\neq0$, so $\det(M)=(-1)^3=-1\neq0$ and $T$ is invertible. Applying $R_1 \to -R_1$, $R_2 \to -R_2$, $R_3 \to -R_3$ to $(M\mid I)$ gives $$\left(\begin{array}{ccc|ccc}1&-2&-2&-1&0&0\\0&1&-4&0&-1&0\\0&0&1&0&0&-1\end{array}\right),$$ applying $R_2 \to R_2+4R_3$, $R_1 \to R_1+2R_3$ gives $$\left(\begin{array}{ccc|ccc}1&-2&0&-1&0&-2\\0&1&0&0&-1&-4\\0&0&1&0&0&-1\end{array}\right),$$ and finally applying $R_1 \to R_1+2R_2$ gives $$\left(\begin{array}{ccc|ccc}1&0&0&-1&-2&-10\\0&1&0&0&-1&-4\\0&0&1&0&0&-1\end{array}\right).$$ Hence $$M^{-1}=\begin{pmatrix}-1&-2&-10\\0&-1&-4\\0&0&-1\end{pmatrix},$$ so $T^{-1}(1)=-1$, $T^{-1}(x)=-2-x$, $T^{-1}(x^2)=-10-4x-x^2$. That is, for $f(x)=a+bx+cx^2$, $$T^{-1}(f)=-(a+2b+10c)-(b+4c)x-cx^2.$$ As a check, $$T(-10-4x-x^2)=(-2)+2(-4-2x)+(10+4x+x^2)=x^2,$$ confirming the computation.
2. Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Then $\operatorname{tr}(A)=a+d$, and since $A^t=\begin{pmatrix}a&c\\b&d\end{pmatrix}$, $\operatorname{tr}(A^t)=a+d=\operatorname{tr}(A)$. Also $$EA=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\a&b\end{pmatrix},\qquad AE=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}b&a\\d&c\end{pmatrix},$$ so $\operatorname{tr}(EA)=b+c=\operatorname{tr}(AE)$. Hence $$T(A)=(a+d,\ a+d,\ b+c,\ b+c),$$ and for the standard basis matrices $E_{11},E_{12},E_{21},E_{22}$ we get $T(E_{11})=T(E_{22})=(1,1,0,0)$ and $T(E_{12})=T(E_{21})=(0,0,1,1)$, so the range of $T$ is exactly the $2$-dimensional subspace $\{(s,s,t,t):s,t\in\mathbb{R}\}$. Thus $\operatorname{rank}(T)=2\neq4=\dim\mathbb{R}^4$, so $T$ is not onto. Since the domain $M_{2\times2}(\mathbb{R})$ is $4$-dimensional, the rank-nullity theorem gives $\operatorname{nullity}(T)=4-2=2>0$, so $T$ is not one-to-one either. Hence $T$ is not bijective, and no inverse transformation exists.

Solution. We transform $A=\begin{pmatrix}1&2&1\\1&0&1\\1&1&2\end{pmatrix}$ into $I_3$ by the following sequence of elementary row operations.

Applying $R_2 \to R_2-R_1$, $R_3 \to R_3-R_1$ gives $$\begin{pmatrix}1&2&1\\1&0&1\\1&1&2\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&2&1\\0&-2&0\\0&-1&1\end{pmatrix},$$ and applying $R_2 \to -\frac12R_2$ gives $$\begin{pmatrix}1&2&1\\0&-2&0\\0&-1&1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&2&1\\0&1&0\\0&-1&1\end{pmatrix}.$$ Applying $R_3 \to R_3+R_2$ gives $$\begin{pmatrix}1&2&1\\0&1&0\\0&-1&1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&2&1\\0&1&0\\0&0&1\end{pmatrix},$$ and finally applying $R_1 \to R_1-2R_2$, $R_1 \to R_1-R_3$ in succession gives $$\begin{pmatrix}1&2&1\\0&1&0\\0&0&1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I_3.$$

Let $E_1,\dots,E_6$ denote the elementary matrices corresponding to these six operations, in the order applied, so that $E_6E_5E_4E_3E_2E_1A=I_3$ and hence $$A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}E_5^{-1}E_6^{-1}.$$ The elementary matrices for each operation and its inverse operation are as follows.

$$E_1^{-1}=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}\ (R_2\to R_2+R_1),\qquad E_2^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\1&0&1\end{pmatrix}\ (R_3\to R_3+R_1),$$ $$E_3^{-1}=\begin{pmatrix}1&0&0\\0&-2&0\\0&0&1\end{pmatrix}\ (R_2\to -2R_2),\qquad E_4^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\0&-1&1\end{pmatrix}\ (R_3\to R_3-R_2),$$ $$E_5^{-1}=\begin{pmatrix}1&2&0\\0&1&0\\0&0&1\end{pmatrix}\ (R_1\to R_1+2R_2),\qquad E_6^{-1}=\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}\ (R_1\to R_1+R_3).$$

Hence $$A=E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}E_5^{-1}E_6^{-1}=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\1&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&-2&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\0&-1&1\end{pmatrix}\begin{pmatrix}1&2&0\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}.$$

As a check, we multiply from the right. $$\begin{pmatrix}1&2&0\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}1&2&1\\0&1&0\\0&0&1\end{pmatrix},\qquad\begin{pmatrix}1&0&0\\0&1&0\\0&-1&1\end{pmatrix}\begin{pmatrix}1&2&1\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}1&2&1\\0&1&0\\0&-1&1\end{pmatrix},$$ $$\begin{pmatrix}1&0&0\\0&-2&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&2&1\\0&1&0\\0&-1&1\end{pmatrix}=\begin{pmatrix}1&2&1\\0&-2&0\\0&-1&1\end{pmatrix},\qquad\begin{pmatrix}1&0&0\\0&1&0\\1&0&1\end{pmatrix}\begin{pmatrix}1&2&1\\0&-2&0\\0&-1&1\end{pmatrix}=\begin{pmatrix}1&2&1\\0&-2&0\\1&1&2\end{pmatrix},$$ $$\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&2&1\\0&-2&0\\1&1&2\end{pmatrix}=\begin{pmatrix}1&2&1\\1&0&1\\1&1&2\end{pmatrix}=A,$$ confirming that the decomposition above is correct.

Proof.

1. Let $A=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ ($3\times1$, $A\neq O$) and $B=(b_1,b_2,b_3)$ ($1\times3$, $B\neq O$). Then $AB$ is a $3\times3$ matrix with $(AB)_{ij}=a_ib_j$, so row $i$ of $AB$ is $$(\,a_ib_1,\ a_ib_2,\ a_ib_3\,)=a_i\,B,$$ a scalar multiple of $B$. Since all three rows of $AB$ are scalar multiples of the single vector $B$, the row space of $AB$ is spanned by $\{B\}$ and hence has dimension at most $1$. Therefore $\operatorname{rank}(AB)\leq1$.
2. Since $\operatorname{rank}(C)=1$, $C\neq O$ and the row space of $C$ is $1$-dimensional. Let $B$ be a nonzero row of $C$; then $B\neq O$, so $B$ is a $1\times3$ matrix of rank $1$. Since the row space of $C$ is $1$-dimensional and $B$ is a nonzero vector in that row space, $B$ spans the row space, so every row of $C$ is a scalar multiple of $B$: there exist scalars $a_1,a_2,a_3$ such that row $i$ of $C$ equals $a_iB$ ($i=1,2,3$). Let $A:=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$. Since $C\neq O$, at least one of $a_1,a_2,a_3$ is nonzero, so $A\neq O$ and hence $A$ is a $3\times1$ matrix of rank $1$. Since $(AB)_{ij}=a_ib_j$ equals exactly the $(i,j)$ entry of $C$, we get $C=AB$.

Proof. Let $\vec a_k$ denote column $k$ of $A$ ($m\times1$, $k=1,\dots,n$) and let $\vec b_k^t$ denote row $k$ of $B$ ($1\times p$, $k=1,\dots,n$). The $(i,j)$ entry of $AB$ is $$(AB)_{ij}=\sum_{k=1}^nA_{ik}B_{kj}=\sum_{k=1}^n(\vec a_k)_i(\vec b_k^t)_j=\sum_{k=1}^n(\vec a_k\vec b_k^t)_{ij},$$ so $$AB=\sum_{k=1}^n\vec a_k\,\vec b_k^t.$$

For each $k$, $\vec a_k\vec b_k^t$ is an $m\times p$ matrix. If $\vec a_k=O$ or $\vec b_k=O$, then $\vec a_k\vec b_k^t=O$, which has rank $0$. Otherwise, row $i$ of $\vec a_k\vec b_k^t$ is $(\vec a_k)_i\cdot\vec b_k^t$, a scalar multiple of $\vec b_k^t$, so every row is spanned by the single vector $\vec b_k^t$ and the row space has dimension $1$. In either case, $\vec a_k\vec b_k^t$ has rank at most $1$. Hence $AB=\sum_{k=1}^n\vec a_k\vec b_k^t$ is written as a sum of $n$ matrices, each of rank at most one.

Solution. Let $L_A:F^n\to F^m$ and $L_B:F^p\to F^n$ be the left-multiplication transformations corresponding to $A$ and $B$, so that $\operatorname{rank}(A)=\dim(\operatorname{Im}(L_A))$ and $\operatorname{rank}(B)=\dim(\operatorname{Im}(L_B))$.

Since $\operatorname{rank}(B)=n=\dim F^n$ and $\operatorname{Im}(L_B)$ is a subspace of $F^n$, a subspace whose dimension equals the dimension of the whole space must equal the whole space, so $\operatorname{Im}(L_B)=F^n$; that is, $L_B$ is onto. Likewise, since $\operatorname{rank}(A)=m=\dim F^m$, we get $\operatorname{Im}(L_A)=F^m$, so $L_A$ is also onto.

The left-multiplication transformation corresponding to $AB$ is the composition $L_A\circ L_B:F^p\to F^m$. Since $L_B$ is onto, $\operatorname{Im}(L_B)=F^n$, so $$\operatorname{Im}(L_A\circ L_B)=L_A(\operatorname{Im}(L_B))=L_A(F^n)=\operatorname{Im}(L_A)=F^m.$$ Hence $L_A\circ L_B$ is also onto, so $$\operatorname{rank}(AB)=\dim\big(\operatorname{Im}(L_A\circ L_B)\big)=\dim F^m=m.$$ Therefore the rank of $AB$ is $m$.