Exercises — Diagonalization — conditions, procedure

Ek diagonal matrix pehle se hi diagonal hai, isliye trivially $A=PD{P}^{-1}$ hai jahan $P=I$ (identity) aur $D$ khud.

• WHAT we did: form ko recognize kiya. WHY it works: $I$ invertible hai aur $ID{I}^{-1}=D$.
• Standard basis vectors $\left(\begin{array}{c}1\\ 0\end{array}\right)$ aur $\left(\begin{array}{c}0\\ 1\end{array}\right)$ eigenvectors hain, eigenvalues $5$ aur $2$ ke saath.

Answer: haan; $P=I$, $D=\left(\begin{array}{cc}5& 0\\ 0& 2\end{array}\right)$.

$A$ upper triangular hai, isliye uske eigenvalues diagonal par hote hain (characteristic polynomial $(7-\lambda )(1-\lambda )$ hai, dekho Characteristic Polynomial). Eigenvalues hain $7$ aur $1$ — do distinct values.

• WHY that settles it: distinct eigenvalues linearly independent eigenvectors force karte hain, aur ${\mathbb{R}}^2$ mein $2$ independent eigenvectors use span kar lete hain.

Answer: haan, guaranteed diagonalizable (distinct eigenvalues $7,1$).

False. Teeno eigenvalues $4$ ke barabar hain (algebraic multiplicity $a(4)=3$). Diagonalizability ab poori tarah geometric multiplicity $g(4)=\dim \ker (A-4I)$ par depend karti hai.

• Agar $g(4)=3$ (yaani $A=4I$), toh diagonalizable.
• Agar $g(4)<3$ (defective), toh not diagonalizable.

Polynomial akela hume nahi bata sakta. Answer: False.

Step 1 — eigenvalues. $\det (A-\lambda I)=(1-\lambda {)}^2-4={\lambda }^2-2\lambda -3=(\lambda -3)(\lambda +1)$. Toh ${\lambda }_1=3,{\lambda }_2=-1$. Distinct → diagonalizable. Step 2 — eigenvectors. $\lambda =3$: $A-3I=\left(\begin{array}{cc}-2& 2\\ 2& -2\end{array}\right)\Rightarrow x_1=x_2\Rightarrow {\mathbf{p}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$. $\lambda =-1$: $A+I=\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right)\Rightarrow x_1=-x_2\Rightarrow {\mathbf{p}}_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$. Step 3 — assemble (order matched): $P=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right),D=\left(\begin{array}{cc}3& 0\\ 0& -1\end{array}\right).$ $\det P=-2\ne 0$ isliye $P$ invertible hai. ✓ (Yeh $A$ symmetric hai, isliye eigenvectors orthogonal aate hain — dekho Spectral Theorem.)

Eigenvalues: $(5-\lambda {)}^2=0\Rightarrow \lambda =5$, toh $a(5)=2$. Eigenvectors: $A-5I=\left(\begin{array}{cc}0& 4\\ 0& 0\end{array}\right)$. Solve karne par $x_2=0$, $x_1$ free milta hai → ek direction $\left(\begin{array}{c}1\\ 0\end{array}\right)$, isliye $g(5)=1$. Kyunki $g(5)=1<a(5)=2$, not diagonalizable (defective — yeh ek scaled shear hai).

WHY diagonalize first? $A^4=P{D}^4{P}^{-1}$ aur $D^4$ mein bas har diagonal entry ko 4th power mein uthao (yahi toh Matrix Powers and Exponentials ka pura point hai). $D^4=\left(\begin{array}{cc}{3}^4& 0\\ 0& (-1{)}^4\end{array}\right)=\left(\begin{array}{cc}81& 0\\ 0& 1\end{array}\right)$. $P^{-1}=\frac{1}{-2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)$. $A^4=P{D}^4{P}^{-1}=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{cc}81& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{cc}41& 40\\ 40& 41\end{array}\right).$

Eigenvalue: $(3-\lambda {)}^2=0\Rightarrow \lambda =3$, har $t$ ke liye $a(3)=2$. Eigenvectors: $A-3I=\left(\begin{array}{cc}0& t\\ 0& 0\end{array}\right)$.

• Agar $t\ne 0$: solve karne par $x_2=0$, $x_1$ free → $g(3)=1<2$ → not diagonalizable.
• Agar $t=0$: $A-3I=0$, pura plane eigenspace ban jaata hai → $g(3)=2$ → diagonalizable (yeh $3I$ hai).

Answer: diagonalizable sirf jab $t=0$ ho.

Characteristic polynomial: $\det (R-\lambda I)={\lambda }^2+1=0\Rightarrow \lambda =\pm i$.

• $\mathbb{R}$ par: koi real eigenvalue exist nahi karta, isliye koi real eigenvector nahi hai — geometrically ek rotation har real direction ko turn kar deta hai, isliye koi real line sirf stretch nahi hoti. $\mathbb{R}$ par diagonalizable nahi. Neeche di gayi figure yeh concretely dikhati hai: yeh kai kaale unit vectors draw karti hai aur, red mein, ek vector $v$ uski image $Rv$ ke saath jo puri $90^{\circ }$ dur turn ho gayi hai — kyunki koi kaala arrow apni line par wapas nahi aata, koi aisi direction nahi hai jo sirf stretch ho, yaani koi real eigenvector nahi.
• $\mathbb{C}$ par: do distinct eigenvalues $i,-i$ → diagonalizable, with $D=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right).$ Eigenvectors: $\lambda =i$ ke liye, $(R-iI)\mathbf{x}=0$. Pehli row deti hai $-i{x}_1-x_2=0\Rightarrow x_2=-i{x}_1$, isliye ${\mathbf{p}}_1=\left(\begin{array}{c}1\\ -i\end{array}\right)$ lo (check: $R{\mathbf{p}}_1=\left(\begin{array}{c}i\\ 1\end{array}\right)=i\left(\begin{array}{c}1\\ -i\end{array}\right)$ ✓). Isi tarah $\lambda =-i$ ke liye, ${\mathbf{p}}_2=\left(\begin{array}{c}1\\ i\end{array}\right)$.

Har eigenvalue ke liye $g=a$ check karo:

• $\lambda =2$: $a(2)=2$ aur $g(2)=2$ → matches.
• $\lambda =5$: $a(5)=1$, aur hamesha $1\le g\le a$, isliye $g(5)=1$ → matches.

Total independent eigenvectors $=2+1=3=n$. Diagonalizable. $P$ ko $\lambda =2$ eigenspace se 2 columns aur $\lambda =5$ eigenspace se 1 column milti hai.

Lower triangular → eigenvalues diagonal par: $\lambda =2,2,3$. Toh $a(2)=2,a(3)=1$. $\lambda =2$ ke liye eigenvectors: $A-2I=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 1\end{array}\right)$. Equations: $x_1=0$ aur $x_2+x_3=0$. Free variable: ek → $\mathbf{p}=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$. Isliye $g(2)=1$. Kyunki $g(2)=1<a(2)=2$, not diagonalizable (defective — $\lambda =2$ par ek Jordan block baitha hai; dekho Jordan Normal Form).

Determinant. $\det A=\det (PD{P}^{-1})=\det P\cdot \det D\cdot \det (P^{-1})$. Kyunki $\det (P^{-1})=1/\det P$, $\det P$ factors cancel ho jaate hain, bachta hai $\det A=\det D={\prod }_i{\lambda }_i$ (diagonal matrix ka determinant uske diagonal ka product hota hai). Trace. Trace cyclic hai: $\mathrm{tr}(XY)=\mathrm{tr}(YX)$. Isliye $\mathrm{tr}A=\mathrm{tr}(PD{P}^{-1})=\mathrm{tr}(D{P}^{-1}P)=\mathrm{tr}(D)={\sum }_i{\lambda }_i.$ Dono facts purely is baat se follow karte hain ki $A$, $D$ ke similar hai (dekho Similar Matrices) — waqai yeh kisi bhi matrix ke liye uske eigenvalues ke zariye hold karte hain.

WHY diagonalize: $e^A=P{e}^D{P}^{-1}$, aur ek diagonal matrix ka $e^D$ bas har diagonal entry ka $e$ hota hai (yeh exponential ko termwise define karta hai; dekho Matrix Powers and Exponentials). $e^D=\left(\begin{array}{cc}{e}^3& 0\\ 0& e^{-1}\end{array}\right)$, aur $P=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$, $P^{-1}=\left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right)$ ke saath: $e^A=\left(\begin{array}{cc}\frac{e^3+e^{-1}}{2}& \frac{e^3-e^{-1}}{2}\\ \frac{e^3-e^{-1}}{2}& \frac{e^3+e^{-1}}{2}\end{array}\right).$

Strategy: ek Jordan block lo $J=\left(\begin{array}{cc}4& 1\\ 0& 4\end{array}\right)$ (eigenvalue $4$, defective, eigenvector $e_1=\left(\begin{array}{c}1\\ 0\end{array}\right)$) aur uski eigen-direction ko $\left(\begin{array}{c}1\\ 1\end{array}\right)$ par basis change $S=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ ke zariye rotate karo (uska pehla column target eigenvector hai). $A=SJ{S}^{-1}$ set karo. $S^{-1}=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)$. $A=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)\left(\begin{array}{cc}4& 1\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}3& 1\\ -1& 5\end{array}\right).$ Eigenvalue check: $\det (A-\lambda I)=(3-\lambda )(5-\lambda )+1={\lambda }^2-8\lambda +16=(\lambda -4{)}^2$ → $a(4)=2$. ✓ Eigenvector check: $A\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}4\\ 4\end{array}\right)=4\left(\begin{array}{c}1\\ 1\end{array}\right)$. ✓ Geometric multiplicity: $A-4I=\left(\begin{array}{cc}-1& 1\\ -1& 1\end{array}\right)$ ka rank $1$ hai, isliye $g(4)=2-1=1<2$. Not diagonalizable.

Answer: $A=\left(\begin{array}{cc}3& 1\\ -1& 5\end{array}\right)$.

Eigenvalues. Agar $A\mathbf{v}=\lambda \mathbf{v}$ aur $\mathbf{v}\ne 0$, toh $A^2\mathbf{v}={\lambda }^2\mathbf{v}$. Lekin $A^2=A$ deta hai $A^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. Isliye ${\lambda }^2=\lambda \Rightarrow \lambda (\lambda -1)=0\Rightarrow \lambda \in \{0,1\}$. Diagonalizable. Kisi bhi vector $\mathbf{x}$ ke liye identity likhte hain $\mathbf{x}=\underset{\in \text{range}}{\underbrace{A\mathbf{x}}}+\underset{\in \ker A}{\underbrace{(\mathbf{x}-A\mathbf{x})}}.$ Pehla piece $\lambda =1$ eigenvector hai: $A(A\mathbf{x})=A^2\mathbf{x}=A\mathbf{x}$. Doosra $\lambda =0$ eigenvector hai: $A(\mathbf{x}-A\mathbf{x})=A\mathbf{x}-A^2\mathbf{x}=0$. Toh har vector eigenvectors mein split ho jaata hai → eigenvectors poore space ko span karte hain → $A$ diagonalizable hai aur $D=\mathrm{diag}(1,\dots ,1,0,\dots ,0)$. Yeh Spectral Theorem intuition ka ek clean special case hai: projections perfectly diagonalizable hote hain.

General reasoning. $A^k=P{D}^k{P}^{-1}$, aur $D^k=\mathrm{diag}({\lambda }_1^k,\dots ,{\lambda }_n^k)$. Har entry ka fate $|{\lambda }_i|$ se govern hota hai:

• $|{\lambda }_i|<1$: ${\lambda }_i^k\to 0$.
• ${\lambda }_i=1$: ${\lambda }_i^k=1$ wahi rahta hai.
• $|{\lambda }_i|>1$ ya ${\lambda }_i=-1$: ${\lambda }_i^k$ diverge/oscillate karta hai → convergence nahi.

Isliye $A^k$ converge karta hai (possibly nonzero limit par) exactly tab jab har eigenvalue ya toh unit circle ke andar ho ya $1$ ke barabar ho. $A=\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$ par apply karo. L2.1 se uske eigenvalues ${\lambda }_1=3$ aur ${\lambda }_2=-1$ hain.

• ${\lambda }_1=3$ ka $|3|=3>1$ hai, isliye diagonal entry $3^k\to \infty$ (yeh explode karta hai).
• ${\lambda }_2=-1$ ka $(-1{)}^k$ oscillate karta hai $1,-1,1,-1,\dots$ (yeh kabhi settle nahi karta).

Akela koi bhi ek already convergence tod deta hai. Kyunki $A^k=P{D}^k{P}^{-1}$ in divergent diagonal entries ko carry karta hai, $A^k$ converge NAHI karta — yeh blow up ho jaata hai ($3^k$ term dominate karta hai). Conclusion: koi limit exist nahi karta.