Exercises — Matrices — review, operations, types

WHAT we do: count rows, then columns. WHY: order is always rows-first. $P$ has $2$ rows and $3$ columns, so its order is $2\times 3$. The entry $p_{23}$ sits in row $2$, column $3$. Walking to row 2 gives $[5,3,7]$; its 3rd number is $7$. $\boxed{\text{order }2\times 3,p_{23}=7}$

Rule used: for a product the inner dimensions must agree $(m\times \mathbf{p})(\mathbf{p}\times n)$; for a sum the orders must be identical.

• $AB$: $(2\times 3)(3\times 4)$ → inner $3=3$ ✅ → order $2\times 4$.
• $BA$: $(3\times 4)(2\times 3)$ → inner $4\ne 2$ ❌ does not exist.
• $AC$: $(2\times 3)(2\times 3)$ → inner $3\ne 2$ ❌ does not exist.
• $A+C$: both $2\times 3$ ✅ → order $2\times 3$.

Compare each with its transpose (and recall diagonal = off-diagonal entries all zero, from the reminders above).

• ${\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]}^{\top }=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]=-A$ → skew-symmetric (note the zero diagonal, as forced by $a_{ii}=-a_{ii}$).
• $\left[\begin{array}{cc}1& 0\\ 0& 5\end{array}\right]$: both off-diagonal entries are $0$, so nonzero numbers sit only where $i=j$ → diagonal (and since $A^{\top }=A$, also symmetric).
• ${\left[\begin{array}{cc}3& 1\\ 1& 3\end{array}\right]}^{\top }=$ itself → symmetric.

WHAT: each output entry $(i,j)$ = row $i$ of $A$ · column $j$ of $B$. WHY: that is the definition forced by composition, $(AB)\mathbf{x}=A(B\mathbf{x})$. $AB=\left[\begin{array}{cc}1\cdot 0+2\cdot (-1)& 1\cdot 1+2\cdot 2\\ 3\cdot 0+4\cdot (-1)& 3\cdot 1+4\cdot 2\end{array}\right]=\left[\begin{array}{cc}-2& 5\\ -4& 11\end{array}\right]$

$BA=\left[\begin{array}{cc}0\cdot 1+1\cdot 3& 0\cdot 2+1\cdot 4\\ -1\cdot 1+2\cdot 3& -1\cdot 2+2\cdot 4\end{array}\right]=\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$ Since $\left[\begin{array}{cc}-2& 5\\ -4& 11\end{array}\right]\ne \left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$, indeed $AB\ne BA$. Order of actions matters.

WHY $S$ is guaranteed symmetric. Transpose it and use two facts — transpose distributes over addition, $(X+Y{)}^{\top }=X^{\top }+Y^{\top }$, and $(A^{\top }{)}^{\top }=A$: $S^{\top }={\left(\frac{1}{2}(A+A^{\top })\right)}^{\top }=\frac{1}{2}\left(A^{\top }+(A^{\top }{)}^{\top }\right)=\frac{1}{2}(A^{\top }+A)=S.$ Since $S^{\top }=S$, $S$ is symmetric — no matter what $A$ was. WHY $K$ is guaranteed skew. Same move, watching the sign: $K^{\top }={\left(\frac{1}{2}(A-A^{\top })\right)}^{\top }=\frac{1}{2}\left(A^{\top }-A\right)=-\frac{1}{2}(A-A^{\top })=-K.$ Since $K^{\top }=-K$, $K$ is skew-symmetric (and so its diagonal must be zero, since $k_{ii}=-k_{ii}$). Now the numbers. $A^{\top }=\left[\begin{array}{cc}4& 6\\ 2& 8\end{array}\right]$. $S=\frac{1}{2}\left[\begin{array}{cc}8& 8\\ 8& 16\end{array}\right]=\left[\begin{array}{cc}4& 4\\ 4& 8\end{array}\right],K=\frac{1}{2}\left[\begin{array}{cc}0& -4\\ 4& 0\end{array}\right]=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]$ Check: $S+K=\left[\begin{array}{cc}4& 2\\ 6& 8\end{array}\right]=A$ ✅, and $K$ has zero diagonal ✅ (exactly as the proof predicted).

$(AB{)}^{\top }=\left[\begin{array}{cc}-2& -4\\ 5& 11\end{array}\right]$. $B^{\top }=\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right],A^{\top }=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$, so $B^{\top }{A}^{\top }=\left[\begin{array}{cc}0\cdot 1+(-1)\cdot 2& 0\cdot 3+(-1)\cdot 4\\ 1\cdot 1+2\cdot 2& 1\cdot 3+2\cdot 4\end{array}\right]=\left[\begin{array}{cc}-2& -4\\ 5& 11\end{array}\right]$ Matches $(AB{)}^{\top }$. The order reversed — socks-and-shoes.

Part 1. Take $A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],B=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$. Using $(AB{)}_{ij}={\sum }_k{a}_{ik}{b}_{kj}$ (row $i$ of $A$ dotted with column $j$ of $B$, legal since inner dims $2=2$): $AB=\left[\begin{array}{cc}1\cdot 0+0\cdot 0& 1\cdot 0+0\cdot 1\\ 0\cdot 0+0\cdot 0& 0\cdot 0+0\cdot 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ Yet neither $A$ nor $B$ is the zero matrix — zero divisors exist. (Here $BA$ also happens to be $0$, so this pair does not prove $BA$ can be nonzero.)

Part 2 — the edge case. Take $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$. WHY these: $A$ kills the first column and shifts the second up; $B$ keeps only the first column. Computing entry by entry with the same summation rule $(AB{)}_{ij}={\sum }_k{a}_{ik}{b}_{kj}$: $AB=\left[\begin{array}{cc}0\cdot 1+1\cdot 0& 0\cdot 0+1\cdot 0\\ 0\cdot 1+0\cdot 0& 0\cdot 0+0\cdot 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ But swapping the order, again by $(BA{)}_{ij}={\sum }_k{b}_{ik}{a}_{kj}$: $BA=\left[\begin{array}{cc}1\cdot 0+0\cdot 0& 1\cdot 1+0\cdot 0\\ 0\cdot 0+0\cdot 0& 0\cdot 1+0\cdot 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\ne 0$ So $AB=0$ definitely does not force $BA=0$. The single fact you may conclude from $AB=0$ is: nothing about $A$ or $B$ individually being zero.

$N^2=\left[\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ As a linear map, $N$ sends the $y$-axis onto the $x$-axis and the $x$-axis to $0$. Apply it twice: everything first lands on the $x$-axis, then the $x$-axis is squashed to $0$ — that is exactly nilpotency with $k=2$. Reading the figure below (caption: "$N=[[0,1],[0,0]]$ is nilpotent: $N^2=0$"). The horizontal line is the $x$-axis, the vertical line is the $y$-axis (both drawn in navy). The orange arrow is $e_1=\left[\begin{array}{c}1\\ 0\end{array}\right]$, the unit vector pointing along the $x$-axis; the violet arrow is $e_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$, the unit vector pointing up the $y$-axis. The magenta arrow shows the image $N{e}_2$: it lies flat along the $x$-axis and lands exactly on $e_1$, because $N\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$. The lone navy dot at the origin marks $N{e}_1=\left[\begin{array}{c}0\\ 0\end{array}\right]$ — the $x$-axis vector is crushed to the origin. So one application flattens the whole plane onto the $x$-axis, and a second application collapses that line to $0$: the picture is the statement $N^2=0$.

Orthogonal means $Q^{\top }Q=I$. Here $Q^{\top }=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]$, so $Q^{\top }Q=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 1& 1\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$ ✅ orthogonal. Its columns are unit length and perpendicular; it is a rotation by $45^{\circ }$ (it preserves lengths and angles). See the parent's link to Eigenvalues and Eigenvectors for why such matrices are special.

WHAT: strip variables — the matrix is the system (parent's core idea). $A=\left[\begin{array}{cc}2& 1\\ 1& 3\end{array}\right],\mathbf{b}=\left[\begin{array}{c}5\\ 10\end{array}\right].$ For a $2\times 2$, $A^{-1}=\frac{1}{\det A}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (from Determinant and Inverse of a Matrix). Here $\det A=2\cdot 3-1\cdot 1=5\ne 0$, so $A$ is non-singular. $A^{-1}=\frac{1}{5}\left[\begin{array}{cc}3& -1\\ -1& 2\end{array}\right],\mathbf{x}=\frac{1}{5}\left[\begin{array}{cc}3& -1\\ -1& 2\end{array}\right]\left[\begin{array}{c}5\\ 10\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}5\\ 15\end{array}\right]=\left[\begin{array}{c}1\\ 3\end{array}\right]$ So $x=1,y=3$. Check: $2(1)+3=5$ ✅, $1+3(3)=10$ ✅. Links to Systems of Linear Equations.

$\mathbf{v}$ is $2\times 1$, ${\mathbf{v}}^{\top }$ is $1\times 2$.

• ${\mathbf{v}}^{\top }\mathbf{v}$: $(1\times 2)(2\times 1)=1\times 1$ — a single number, the dot product $1^2+2^2=5$ (squared length).
• $\mathbf{v}{\mathbf{v}}^{\top }$: $(2\times 1)(1\times 2)=2\times 2$ — a full matrix: $\mathbf{v}{\mathbf{v}}^{\top }=\left[\begin{array}{c}1\\ 2\end{array}\right]\left[\begin{array}{cc}1& 2\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$ This is symmetric (as $\mathbf{v}{\mathbf{v}}^{\top }$ always is, since $(\mathbf{v}{\mathbf{v}}^{\top }{)}^{\top }=\mathbf{v}{\mathbf{v}}^{\top }$). Note it is not a projection here: a projector $P$ must satisfy $P^2=P$, but $(\mathbf{v}{\mathbf{v}}^{\top }{)}^2=({\mathbf{v}}^{\top }\mathbf{v})\mathbf{v}{\mathbf{v}}^{\top }=5\mathbf{v}{\mathbf{v}}^{\top }\ne \mathbf{v}{\mathbf{v}}^{\top }$. Only after dividing by ${\mathbf{v}}^{\top }\mathbf{v}$ — i.e. $\frac{\mathbf{v}{\mathbf{v}}^{\top }}{{\mathbf{v}}^{\top }\mathbf{v}}$, which uses a unit direction — do you get a genuine projection onto the line through $\mathbf{v}$.

Proof: group $ABC=(AB)C$. Then $(ABC{)}^{\top }=((AB)C{)}^{\top }=C^{\top }(AB{)}^{\top }=C^{\top }{B}^{\top }{A}^{\top }$, applying the reversal law twice. Check. $AB=\left[\begin{array}{cc}1& 1\\ 2& 3\end{array}\right]$, $ABC=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$, so $(ABC{)}^{\top }=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]$. $C^{\top }{B}^{\top }{A}^{\top }=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]$ ✅

Let $M=A^{\top }A$. Take its transpose and use the reversal law: $M^{\top }=(A^{\top }A{)}^{\top }=A^{\top }(A^{\top }{)}^{\top }=A^{\top }A=M.$ Since $M^{\top }=M$, $M$ is symmetric — no matter what $A$ was. (We used $(A^{\top }{)}^{\top }=A$.)

Step 1 — write the trace as a double sum. By definition the diagonal entry $(AB{)}_{ii}$ is row $i$ of $A$ dotted with column $i$ of $B$, i.e. $(AB{)}_{ii}={\sum }_k{a}_{ik}{b}_{ki}$. Summing over the diagonal: $\mathrm{tr}(AB)={\sum }_i(AB{)}_{ii}={\sum }_i{\sum }_k{a}_{ik}{b}_{ki}.$ Step 2 — swap the order of the two sums. Both $i$ and $k$ run over the same finite index set, and each term $a_{ik}{b}_{ki}$ is just an ordinary number; a finite sum of numbers may be added in any order (this is why the interchange is legal — no limits, no convergence worries). So ${\sum }_i{\sum }_k{a}_{ik}{b}_{ki}={\sum }_k{\sum }_i{a}_{ik}{b}_{ki}.$ Step 3 — rename and recognise $BA$. Since numbers commute, $a_{ik}{b}_{ki}=b_{ki}{a}_{ik}$. Reading the inner sum over $i$: ${\sum }_i{b}_{ki}{a}_{ik}$ is exactly row $k$ of $B$ dotted with column $k$ of $A$, i.e. $(BA{)}_{kk}$. Hence ${\sum }_k{\sum }_i{b}_{ki}{a}_{ik}={\sum }_k(BA{)}_{kk}=\mathrm{tr}(BA).$ Chaining Steps 1–3 gives $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Check. $AB=\left[\begin{array}{cc}1\cdot 0+2\cdot 1& 1\cdot 1+2\cdot 0\\ 3\cdot 0+4\cdot 1& 3\cdot 1+4\cdot 0\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$, so $\mathrm{tr}(AB)=2+3=5$. $BA=\left[\begin{array}{cc}0\cdot 1+1\cdot 3& 0\cdot 2+1\cdot 4\\ 1\cdot 1+0\cdot 3& 1\cdot 2+0\cdot 4\end{array}\right]=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]$, so $\mathrm{tr}(BA)=3+2=5$ ✅. (Note $AB\ne BA$, yet the traces agree — as proved.)

Transpose $A=S+K$: $A^{\top }=S^{\top }+K^{\top }=S-K$ (using $S^{\top }=S$, $K^{\top }=-K$). Now we have two equations: $A=S+K,A^{\top }=S-K.$ Add them: $A+A^{\top }=2S\Rightarrow S=\frac{1}{2}(A+A^{\top })$. Subtract: $A-A^{\top }=2K\Rightarrow K=\frac{1}{2}(A-A^{\top })$. The split is forced — there is exactly one symmetric+skew decomposition.

$(Q\mathbf{x}{)}^{\top }(Q\mathbf{y})={\mathbf{x}}^{\top }{Q}^{\top }Q\mathbf{y}={\mathbf{x}}^{\top }I\mathbf{y}={\mathbf{x}}^{\top }\mathbf{y}.$ Since the dot product encodes lengths and angles, $Q$ leaves both unchanged: orthogonal maps are exactly the rigid rotations/reflections. Setting $\mathbf{y}=\mathbf{x}$ shows lengths are preserved.