Exercises — Matrices — review, operations, types

KYA karte hain: rows count karo, phir columns. KYUN: order hamesha rows-first hota hai. $P$ mein $2$ rows aur $3$ columns hain, isliye uska order $2\times 3$ hai. Entry $p_{23}$ row $2$, column $3$ mein hoti hai. Row 2 par jaao to milta hai $[5,3,7]$; uska 3rd number $7$ hai. $\boxed{\text{order }2\times 3,p_{23}=7}$

Rule used: product ke liye inner dimensions match karni chahiye $(m\times \mathbf{p})(\mathbf{p}\times n)$; sum ke liye orders identical hone chahiye.

• $AB$: $(2\times 3)(3\times 4)$ → inner $3=3$ ✅ → order $2\times 4$.
• $BA$: $(3\times 4)(2\times 3)$ → inner $4\ne 2$ ❌ exist nahi karta.
• $AC$: $(2\times 3)(2\times 3)$ → inner $3\ne 2$ ❌ exist nahi karta.
• $A+C$: dono $2\times 3$ ✅ → order $2\times 3$.

Har ek ko uske transpose se compare karo (aur yaad raho diagonal = off-diagonal entries sab zero, reminders se).

• ${\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]}^{\top }=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]=-A$ → skew-symmetric (note: zero diagonal, jaisa $a_{ii}=-a_{ii}$ se forced hota hai).
• $\left[\begin{array}{cc}1& 0\\ 0& 5\end{array}\right]$: dono off-diagonal entries $0$ hain, isliye nonzero numbers sirf wahan hain jahan $i=j$ → diagonal (aur kyunki $A^{\top }=A$, symmetric bhi hai).
• ${\left[\begin{array}{cc}3& 1\\ 1& 3\end{array}\right]}^{\top }=$ khud hi → symmetric.

KYA: har output entry $(i,j)$ = row $i$ of $A$ · column $j$ of $B$. KYUN: yahi definition hai jo composition se force hoti hai, $(AB)\mathbf{x}=A(B\mathbf{x})$. $AB=\left[\begin{array}{cc}1\cdot 0+2\cdot (-1)& 1\cdot 1+2\cdot 2\\ 3\cdot 0+4\cdot (-1)& 3\cdot 1+4\cdot 2\end{array}\right]=\left[\begin{array}{cc}-2& 5\\ -4& 11\end{array}\right]$

$BA=\left[\begin{array}{cc}0\cdot 1+1\cdot 3& 0\cdot 2+1\cdot 4\\ -1\cdot 1+2\cdot 3& -1\cdot 2+2\cdot 4\end{array}\right]=\left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$ Kyunki $\left[\begin{array}{cc}-2& 5\\ -4& 11\end{array}\right]\ne \left[\begin{array}{cc}3& 4\\ 5& 6\end{array}\right]$, clearly $AB\ne BA$. Actions ka order matter karta hai.

KYUN $S$ guaranteed symmetric hai. Usse transpose karo aur do facts use karo — transpose addition par distribute hota hai, $(X+Y{)}^{\top }=X^{\top }+Y^{\top }$, aur $(A^{\top }{)}^{\top }=A$: $S^{\top }={\left(\frac{1}{2}(A+A^{\top })\right)}^{\top }=\frac{1}{2}\left(A^{\top }+(A^{\top }{)}^{\top }\right)=\frac{1}{2}(A^{\top }+A)=S.$ Kyunki $S^{\top }=S$, $S$ symmetric hai — chahe $A$ kuch bhi ho. KYUN $K$ guaranteed skew hai. Same move, sign par dhyan rakhte hue: $K^{\top }={\left(\frac{1}{2}(A-A^{\top })\right)}^{\top }=\frac{1}{2}\left(A^{\top }-A\right)=-\frac{1}{2}(A-A^{\top })=-K.$ Kyunki $K^{\top }=-K$, $K$ skew-symmetric hai (aur isliye uska diagonal zero hona chahiye, kyunki $k_{ii}=-k_{ii}$). Ab 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: $S+K=\left[\begin{array}{cc}4& 2\\ 6& 8\end{array}\right]=A$ ✅, aur $K$ ka zero diagonal ✅ (bilkul jaisa proof ne predict kiya tha).

$(AB{)}^{\top }=\left[\begin{array}{cc}-2& -4\\ 5& 11\end{array}\right]$. $B^{\top }=\left[\begin{array}{cc}0& -1\\ 1& 2\end{array}\right],A^{\top }=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]$, isliye $B^{\top }{A}^{\top }=\left[\begin{array}{cc}0\cdot 1+(-1)\cdot 2& 0\cdot 3+(-1)\cdot 4\\ 1\cdot 1+2\cdot 2& 1\cdot 3+2\cdot 4\end{array}\right]=\left[\begin{array}{cc}-2& -4\\ 5& 11\end{array}\right]$ $(AB{)}^{\top }$ se match karta hai. Order reverse hua — socks-and-shoes.

Part 1. Lo $A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],B=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]$. $(AB{)}_{ij}={\sum }_k{a}_{ik}{b}_{kj}$ use karte hue (row $i$ of $A$ ko column $j$ of $B$ se dot karo, legal kyunki inner dims $2=2$): $AB=\left[\begin{array}{cc}1\cdot 0+0\cdot 0& 1\cdot 0+0\cdot 1\\ 0\cdot 0+0\cdot 0& 0\cdot 0+0\cdot 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ Phir bhi na $A$ zero matrix hai, na $B$ — zero divisors exist karte hain. (Yahan $BA$ bhi $0$ hota hai, isliye yeh pair prove nahi karta ki $BA$ nonzero ho sakta hai.)

Part 2 — edge case. Lo $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$. KYUN yeh: $A$ pehle column ko kill karta hai aur doosre ko upar shift karta hai; $B$ sirf pehle column ko rakhta hai. Wahi summation rule $(AB{)}_{ij}={\sum }_k{a}_{ik}{b}_{kj}$ se entry by entry compute karte hue: $AB=\left[\begin{array}{cc}0\cdot 1+1\cdot 0& 0\cdot 0+1\cdot 0\\ 0\cdot 1+0\cdot 0& 0\cdot 0+0\cdot 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ Lekin order swap karne par, $(BA{)}_{ij}={\sum }_k{b}_{ik}{a}_{kj}$ se: $BA=\left[\begin{array}{cc}1\cdot 0+0\cdot 0& 1\cdot 1+0\cdot 0\\ 0\cdot 0+0\cdot 0& 0\cdot 1+0\cdot 0\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\ne 0$ Toh $AB=0$ yeh definitely force nahi karta ki $BA=0$ bhi ho. $AB=0$ se sirf yeh conclude ho sakta hai: $A$ ya $B$ mein se kisi ke baare mein bhi akele kuch nahi keh sakte.

$N^2=\left[\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$ Ek linear map ke roop mein, $N$ $y$-axis ko $x$-axis par bhejta hai aur $x$-axis ko $0$ par. Isse do baar apply karo: sab kuch pehle $x$-axis par land karta hai, phir $x$-axis $0$ par squash ho jaata hai — yahi exactly $k=2$ ke saath nilpotency hai. Neeche di figure padhna (caption: "$N=[[0,1],[0,0]]$ is nilpotent: $N^2=0$"). Horizontal line $x$-axis hai, vertical line $y$-axis hai (dono navy mein drawn hain). Orange arrow hai $e_1=\left[\begin{array}{c}1\\ 0\end{array}\right]$, unit vector $x$-axis ke along; violet arrow hai $e_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$, unit vector $y$-axis ke upar. Magenta arrow image $N{e}_2$ dikhata hai: yeh $x$-axis ke along flat lie karta hai aur exactly $e_1$ par land karta hai, kyunki $N\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$. Origin par akela navy dot $N{e}_1=\left[\begin{array}{c}0\\ 0\end{array}\right]$ mark karta hai — $x$-axis vector origin par crush ho jaata hai. Toh ek application se poora plane $x$-axis par flatten ho jaata hai, aur doosra application us line ko $0$ par collapse kar deta hai: picture bilkul yahi statement hai $N^2=0$.

Orthogonal ka matlab hai $Q^{\top }Q=I$. Yahan $Q^{\top }=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]$, toh $Q^{\top }Q=\frac{1}{2}\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 1& 1\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$ ✅ orthogonal. Uske columns unit length ke hain aur perpendicular hain; yeh ek $45^{\circ }$ rotation hai (yeh lengths aur angles preserve karta hai). Dekhna Eigenvalues and Eigenvectors ka parent ka link, ki aisi matrices kyun special hain.

KYA: variables hata do — matrix hi system hai (parent ka core idea). $A=\left[\begin{array}{cc}2& 1\\ 1& 3\end{array}\right],\mathbf{b}=\left[\begin{array}{c}5\\ 10\end{array}\right].$ $2\times 2$ ke liye, $A^{-1}=\frac{1}{\det A}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ (from Determinant and Inverse of a Matrix). Yahan $\det A=2\cdot 3-1\cdot 1=5\ne 0$, toh $A$ non-singular hai. $A^{-1}=\frac{1}{5}\left[\begin{array}{cc}3& -1\\ -1& 2\end{array}\right],\mathbf{x}=\frac{1}{5}\left[\begin{array}{cc}3& -1\\ -1& 2\end{array}\right]\left[\begin{array}{c}5\\ 10\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}5\\ 15\end{array}\right]=\left[\begin{array}{c}1\\ 3\end{array}\right]$ Toh $x=1,y=3$. Check: $2(1)+3=5$ ✅, $1+3(3)=10$ ✅. Systems of Linear Equations se connect hota hai.

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

• ${\mathbf{v}}^{\top }\mathbf{v}$: $(1\times 2)(2\times 1)=1\times 1$ — ek single number, dot product $1^2+2^2=5$ (squared length).
• $\mathbf{v}{\mathbf{v}}^{\top }$: $(2\times 1)(1\times 2)=2\times 2$ — ek full matrix: $\mathbf{v}{\mathbf{v}}^{\top }=\left[\begin{array}{c}1\\ 2\end{array}\right]\left[\begin{array}{cc}1& 2\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$ Yeh symmetric hai (jaisa $\mathbf{v}{\mathbf{v}}^{\top }$ hamesha hota hai, kyunki $(\mathbf{v}{\mathbf{v}}^{\top }{)}^{\top }=\mathbf{v}{\mathbf{v}}^{\top }$). Note: yeh yahan projection nahi hai: ek projector $P$ ko $P^2=P$ satisfy karna chahiye, lekin $(\mathbf{v}{\mathbf{v}}^{\top }{)}^2=({\mathbf{v}}^{\top }\mathbf{v})\mathbf{v}{\mathbf{v}}^{\top }=5\mathbf{v}{\mathbf{v}}^{\top }\ne \mathbf{v}{\mathbf{v}}^{\top }$. Sirf ${\mathbf{v}}^{\top }\mathbf{v}$ se divide karne par — yaani $\frac{\mathbf{v}{\mathbf{v}}^{\top }}{{\mathbf{v}}^{\top }\mathbf{v}}$, jo ek unit direction use karta hai — $\mathbf{v}$ ke through line par genuine projection milti hai.

Proof: $ABC=(AB)C$ group karo. Phir $(ABC{)}^{\top }=((AB)C{)}^{\top }=C^{\top }(AB{)}^{\top }=C^{\top }{B}^{\top }{A}^{\top }$, reversal law do baar apply karke. Check. $AB=\left[\begin{array}{cc}1& 1\\ 2& 3\end{array}\right]$, $ABC=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$, toh $(ABC{)}^{\top }=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]$. $C^{\top }{B}^{\top }{A}^{\top }=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}2& 4\\ 1& 3\end{array}\right]$ ✅

Maano $M=A^{\top }A$. Uska transpose lo aur reversal law use karo: $M^{\top }=(A^{\top }A{)}^{\top }=A^{\top }(A^{\top }{)}^{\top }=A^{\top }A=M.$ Kyunki $M^{\top }=M$, $M$ symmetric hai — chahe $A$ kuch bhi ho. (Humne $(A^{\top }{)}^{\top }=A$ use kiya.)

Step 1 — trace ko double sum ke roop mein likho. Definition se diagonal entry $(AB{)}_{ii}$ hai row $i$ of $A$ dotted with column $i$ of $B$, yaani $(AB{)}_{ii}={\sum }_k{a}_{ik}{b}_{ki}$. Diagonal par sum karo: $\mathrm{tr}(AB)={\sum }_i(AB{)}_{ii}={\sum }_i{\sum }_k{a}_{ik}{b}_{ki}.$ Step 2 — dono sums ka order swap karo. $i$ aur $k$ dono same finite index set par run karte hain, aur har term $a_{ik}{b}_{ki}$ sirf ek ordinary number hai; finite numbers ka sum kisi bhi order mein add kiya ja sakta hai (isliye interchange legal hai — koi limits, koi convergence worries nahi). Toh ${\sum }_i{\sum }_k{a}_{ik}{b}_{ki}={\sum }_k{\sum }_i{a}_{ik}{b}_{ki}.$ Step 3 — rename karo aur $BA$ pehchano. Kyunki numbers commute karte hain, $a_{ik}{b}_{ki}=b_{ki}{a}_{ik}$. $i$ par inner sum padhte hue: ${\sum }_i{b}_{ki}{a}_{ik}$ exactly row $k$ of $B$ dotted with column $k$ of $A$ hai, yaani $(BA{)}_{kk}$. Isliye ${\sum }_k{\sum }_i{b}_{ki}{a}_{ik}={\sum }_k(BA{)}_{kk}=\mathrm{tr}(BA).$ Steps 1–3 chain karke milta hai $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Check. $AB=\left[\begin{array}{cc}1\cdot 0+2\cdot 1& 1\cdot 1+2\cdot 0\\ 3\cdot 0+4\cdot 1& 3\cdot 1+4\cdot 0\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 4& 3\end{array}\right]$, toh $\mathrm{tr}(AB)=2+3=5$. $BA=\left[\begin{array}{cc}0\cdot 1+1\cdot 3& 0\cdot 2+1\cdot 4\\ 1\cdot 1+0\cdot 3& 1\cdot 2+0\cdot 4\end{array}\right]=\left[\begin{array}{cc}3& 4\\ 1& 2\end{array}\right]$, toh $\mathrm{tr}(BA)=3+2=5$ ✅. (Note $AB\ne BA$, phir bhi traces agree karte hain — jaisa prove kiya.)

$A=S+K$ transpose karo: $A^{\top }=S^{\top }+K^{\top }=S-K$ ($S^{\top }=S$, $K^{\top }=-K$ use karke). Ab hamare paas do equations hain: $A=S+K,A^{\top }=S-K.$ Add karo: $A+A^{\top }=2S\Rightarrow S=\frac{1}{2}(A+A^{\top })$. Subtract karo: $A-A^{\top }=2K\Rightarrow K=\frac{1}{2}(A-A^{\top })$. Split forced hai — exactly ek hi symmetric+skew decomposition exist karta hai.

$(Q\mathbf{x}{)}^{\top }(Q\mathbf{y})={\mathbf{x}}^{\top }{Q}^{\top }Q\mathbf{y}={\mathbf{x}}^{\top }I\mathbf{y}={\mathbf{x}}^{\top }\mathbf{y}.$ Kyunki dot product lengths aur angles encode karta hai, $Q$ dono ko unchanged rakhta hai: orthogonal maps exactly rigid rotations/reflections hote hain. $\mathbf{y}=\mathbf{x}$ set karne par dikhta hai ki lengths preserve hoti hain.