Exercises — Rank of a matrix — definition, row rank = column rank theorem

What to notice: row 1 is exactly $2\times$ row 2. So the two rows point in the same direction — only one independent row. Why that fixes the rank: row rank = dimension of the span of the rows. One direction ⇒ dimension $1$. $\mathrm{rank}(A)=1.$ Check via columns: column 2 $=2\times$ column 1, so column rank $=1$ too. The theorem holds. ✔️

What to notice: each column already sits on its own axis — $(1,0,0),(0,1,0),(0,0,1)$ point along three perpendicular directions, so none is a mix of the others. Why: three independent columns span a $3$-dimensional column space. $\mathrm{rank}(I_3)=3.$ This is full rank for a $3\times 3$ matrix — see Invertible Matrix Theorem (full rank ⇔ invertible).

What to notice: every row and every column is the zero vector. The span of a bunch of zero vectors is just the single point $\{0\}$, which has dimension $0$. $\mathrm{rank}(0_{2\times 4})=0.$ This is the degenerate floor of the bound $0\le \mathrm{rank}(A)$.

Step 1 — clear column 1. $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$. Why? Row operations replace a row by a combination of rows, which never changes the row space (see Gaussian Elimination & Echelon Form), so rank is preserved. $\to \left(\begin{array}{ccc}1& 2& 3\\ 0& 0& 0\\ 0& -1& -2\end{array}\right)$ Step 2 — tidy the staircase. Swap $R_2\leftrightarrow R_3$ so the zero row falls to the bottom. $\to \left(\begin{array}{ccc}1& 2& 3\\ 0& -1& -2\\ 0& 0& 0\end{array}\right)$ Step 3 — count pivots: positions $(1,1)$ and $(2,2)$. Two pivots. $\mathrm{rank}(A)=2.$

Step 1: $R_3\to R_3-R_1$. $\to \left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 1& 1\end{array}\right)$ Step 2: $R_3\to R_3-R_2$ (kill the duplicate row). $\to \left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)$ Step 3: two pivots at $(1,1),(2,2)$. $\mathrm{rank}(B)=2.$ Notice column 3 $=2\cdot (\text{col }1)+1\cdot (\text{col }2)$ — the missing pivot column is exactly the dependent column.

Idea: rank drops below full precisely when the rows become multiples of each other. Row 2 $=3\times$ Row 1 means $(3,k)=(3,6)$, i.e. $k=6$. Elimination check: $R_2\to R_2-3R_1$ gives $(0,k-6)$. That row is zero ⇔ $k=6$. $\boxed{k=6}\Rightarrow \mathrm{rank}=1,\text{else }\mathrm{rank}=2.$ Equivalently the determinant $1\cdot k-2\cdot 3=k-6$ vanishes at $k=6$.

Tool — Rank–Nullity: $\mathrm{rank}(A)+\dim \ker (A)=n$, where $n$ is the number of columns. See Rank–Nullity Theorem. Why $n$ and not $m$? Because $x$ lives in ${\mathbb{R}}^n$ (it must have one entry per column to be multipliable), so the input space we split has dimension $n=6$. $\dim \ker (A)=6-3=3.$

Shape first: $AB$ is $3\times 5$, so $\mathrm{rank}(AB)\le \min (3,5)=3$. Deeper bound: every column of $AB$ is a combination of the columns of $A$ (see Matrix Multiplication as Linear Combination), so $\mathrm{rank}(AB)\le \mathrm{rank}(A)\le 2$. Rows of $AB$ are combinations of rows of $B$, so also $\le \mathrm{rank}(B)\le 2$. $0\le \mathrm{rank}(AB)\le 2.$ The inner dimension $2$ throttles the product — a genuinely tighter cap than the shape alone gave.

Yes. For a square $n\times n$ matrix, $\mathrm{rank}=n$ means all $n$ columns are independent, so they form a basis of ${\mathbb{R}}^5$; by the Invertible Matrix Theorem full rank $\iff$ invertible. $\mathrm{rank}(A)=5=n\Rightarrow A^{-1}\text{ exists.}$

Step 1 — find the pivot columns. Reducing (parent note) gives pivots in columns $1$ and $3$. So take those original columns for $C$: $C=\left(\begin{array}{cc}1& 1\\ 2& 3\\ 3& 4\end{array}\right)(3\times 2).$ Step 2 — express every column of $A$ in the basis $\{c_1,c_3\}$. These coefficient vectors become the columns of $R$:

• col $1=1\cdot c_1+0\cdot c_3\Rightarrow (1,0{)}^T$
• col $2=2\cdot c_1+0\cdot c_3\Rightarrow (2,0{)}^T$ (since col $2=2\times$col $1$)
• col $3=0\cdot c_1+1\cdot c_3\Rightarrow (0,1{)}^T$ $R=\left(\begin{array}{ccc}1& 2& 0\\ 0& 0& 1\end{array}\right)(2\times 3).$ Step 3 — verify $CR=A$: $CR=\left(\begin{array}{cc}1& 1\\ 2& 3\\ 3& 4\end{array}\right)\left(\begin{array}{ccc}1& 2& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1& 2& 1\\ 2& 4& 3\\ 3& 6& 4\end{array}\right)=A.✔️$ The shared inner dimension $2$ is $\mathrm{rank}(A)$.

Strategy: make rows $3$ a combination of rows $1$ and $2$, keeping rows $1,2$ independent, and avoid any zero entries. Take $M=\left(\begin{array}{ccc}1& 1& 2\\ 1& 2& 1\\ 2& 3& 3\end{array}\right),R_3=R_1+R_2.$ Why rank $2$: rows $1,2$ are independent (not multiples — compare $1:1$ vs $1:2$ in the first two entries), so rank $\ge 2$; row $3$ adds nothing since $R_3=R_1+R_2$, so rank $\le 2$. Hence exactly $2$. All nine entries are nonzero. ✔️

Setup — what transpose does: $A^T$ turns the rows of $A$ into columns and columns into rows. Therefore the columns of $A^T$ are literally the rows of $A$, so $\text{column rank}(A^T)=\text{row rank}(A).$ Now apply the central theorem to $A^T$: for any matrix, row rank = column rank, hence $\mathrm{rank}(A^T)=\text{column rank}(A^T)=\text{row rank}(A).$ But the theorem applied to $A$ says $\text{row rank}(A)=\mathrm{rank}(A)$. Chaining: $\mathrm{rank}(A^T)=\mathrm{rank}(A).■$

Rank $0$: let $B=-A$ with $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$. Then $A+B=0$, rank $0$. (Overlap perfectly cancels.) Rank $1$: let $A=B=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$. Then $A+B=2A$, still rank $1$. Rank $2$: let $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$, $B=\left(\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$. Then $A+B=\mathrm{diag}(1,1,0)$, rank $2$. Conclusion: all three of $0,1,2$ occur, so $\mathrm{rank}(A+B)=\mathrm{rank}A+\mathrm{rank}B$ is false in general. Only the inequality $\mathrm{rank}(A+B)\le \mathrm{rank}A+\mathrm{rank}B=2$ survives, and it is tight (rank $2$ case). ✔️

Column argument: the column space is spanned by the $n$ columns of $A$, each living in ${\mathbb{R}}^m$. A span of $n$ vectors has dimension $\le n$; a subspace of ${\mathbb{R}}^m$ has dimension $\le m$. So $r\le n$ and $r\le m$, giving $r\le \min (m,n)$ (uses Linear Independence and Basis). Limiting cases:

• If $m\le n$ and $r=m$: the rows are independent, $A$ has full row rank — the map $A$ is onto ${\mathbb{R}}^m$.
• If $n\le m$ and $r=n$: the columns are independent, $A$ has full column rank — the map is one-to-one, $\ker (A)=\{0\}$ (nullity $0$ by Rank–Nullity).
• Degenerate floor $r=0$ ⇔ $A=0$. $0\le \mathrm{rank}(A)\le \min (m,n).■$