Diagonalization — conditions, procedure

4.5.31 · Maths › Linear Algebra (Full)

"Diagonalizable" ka matlab actually kya hai?

$P$ kya hai? $D$ kya hai? Yahi toh asli baat hai. Chaliye derive karte hain ki woh kya hone chahiye.

Diagonalizability ki condition

KYUN yahi exact condition? Kyunki $P$ invertible honi chahiye, aur ek $n\times n$ matrix invertible hoti hai $⟺$ uske columns linearly independent hote hain. Columns eigenvectors hain. Toh humein $n$ independent eigenvectors chahiye. Spanning set nahi $\Rightarrow$ invertible $P$ nahi $\Rightarrow$ diagonalization nahi.

Do useful sufficient/necessary refinements

Procedure (step by step)

Worked Example 1 — distinct eigenvalues

Maano $A=\left(\begin{array}{cc}4& 1\\ 2& 3\end{array}\right)$.

Step 1 — characteristic equation. $\det (A-\lambda I)=\det \left(\begin{array}{cc}4-\lambda & 1\\ 2& 3-\lambda \end{array}\right)=(4-\lambda )(3-\lambda )-2={\lambda }^2-7\lambda +10=(\lambda -5)(\lambda -2).$ Yeh step kyun? Eigenvalues wahan hote hain jahan $A-\lambda I$ singular ho jaata hai (uska nonzero null vector hota hai). ${\lambda }_1=5,{\lambda }_2=2$ — do distinct values, toh already diagonalizable guaranteed hai.

Step 2 — eigenvectors. $\lambda =5$ ke liye: $A-5I=\left(\begin{array}{cc}-1& 1\\ 2& -2\end{array}\right)$. Row reduce: $-x_1+x_2=0\Rightarrow x_1=x_2$. Eigenvector ${\mathbf{p}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$. $\lambda =2$ ke liye: $A-2I=\left(\begin{array}{cc}2& 1\\ 2& 1\end{array}\right)\Rightarrow 2x_1+x_2=0\Rightarrow x_2=-2x_1$. Eigenvector ${\mathbf{p}}_2=\left(\begin{array}{c}1\\ -2\end{array}\right)$. Yeh step kyun? Har eigenvector $A-\lambda I$ ke null space ko span karta hai; yeh woh direction hai jo sirf stretch hoti hai.

Step 3–5 — assemble karo. $P=\left(\begin{array}{cc}1& 1\\ 1& -2\end{array}\right),D=\left(\begin{array}{cc}5& 0\\ 0& 2\end{array}\right).$ Yeh step kyun? Column 1 ($\lambda =5$ wala vector) $D_{11}=5$ ke saath pair karta hai; column 2, $D_{22}=2$ ke saath — same order.

Check: $\det P=-3\ne 0$, toh $P$ invertible hai. ✓ Aur verify kiya ja sakta hai ki $AP=PD$.

Worked Example 2 — repeated eigenvalue, NOT diagonalizable

Maano $A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$ (ek shear).

Step 1: $\det (A-\lambda I)=(2-\lambda {)}^2=0\Rightarrow \lambda =2$, algebraic multiplicity $a=2$.

Step 2: $A-2I=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Solve karo $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\mathbf{x}=0\Rightarrow x_2=0$, $x_1$ free. Sirf ek independent eigenvector $\left(\begin{array}{c}1\\ 0\end{array}\right)$, toh geometric multiplicity $g=1$.

Step 3: $g(2)=1<a(2)=2$. Hamare paas ek $2\times 2$ matrix ke liye sirf 1 eigenvector hai → invertible $P$ nahi ban sakta → NOT diagonalizable.

Yeh kyun matter karta hai: Yeh prototype "defective" matrix hai. Ek shear ki ek hi fixed direction hoti hai; koi doosri independent direction nahi hai jo woh sirf stretch kare, toh koi diagonalizing basis exist nahi karti.

Worked Example 3 — repeated eigenvalue jo IS diagonalizable hai

Maano $A=\left(\begin{array}{cc}3& 0\\ 0& 3\end{array}\right)=3I$.

$\det (A-\lambda I)=(3-\lambda {)}^2$, toh $\lambda =3$ with $a=2$. Ab $A-3I=0$, jiska null space poora ${\mathbb{R}}^2$ hai, toh $g=2$. Kyunki $g=a$, yeh diagonalizable hai — asliye mein yeh already diagonal hai. Seekh: repeated eigenvalues theek hain jab tak woh enough independent eigenvectors provide karte hain.

Common mistakes (Steel-man)

Active recall

Connections

• Eigenvalues and Eigenvectors — raw ingredients.
• Characteristic Polynomial — algebraic multiplicities ka source.
• Similar Matrices — diagonalization = diagonal matrix ke saath similarity.
• Matrix Powers and Exponentials — diagonalization ka main payoff.
• Spectral Theorem — symmetric matrices hamesha (orthogonally) diagonalizable hoti hain.
• Jordan Normal Form — jab matrix not diagonalizable ho toh kya karo.
• Change of Basis — $P$ ek change-of-basis matrix hai eigenbasis mein.

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