Diagonalization — conditions, procedure

What does "diagonalizable" actually mean?

WHAT is $P$? WHAT is $D$? This is the heart of it. Let's derive what they must be.

The condition for diagonalizability

WHY this exact condition? Because $P$ must be invertible, and an $n\times n$ matrix is invertible $\iff$ its columns are linearly independent. The columns are eigenvectors. So we need $n$ independent eigenvectors. No spanning set $\Rightarrow$ no invertible $P$ $\Rightarrow$ no diagonalization.

Two useful sufficient/necessary refinements

The procedure (step by step)

Worked Example 1 — distinct eigenvalues

Let $A=\begin{pmatrix} 4 & 1 \\ 2 & 3\end{pmatrix}$.

Step 1 — characteristic equation. $\det(A-\lambda I)=\det\begin{pmatrix}4-\lambda & 1\\ 2 & 3-\lambda\end{pmatrix}=(4-\lambda)(3-\lambda)-2 = \lambda^2-7\lambda+10=(\lambda-5)(\lambda-2).$ Why this step? Eigenvalues are where $A-\lambda I$ becomes singular (has a nonzero null vector). $\lambda_1=5,\ \lambda_2=2$ — two distinct values, so already guaranteed diagonalizable.

Step 2 — eigenvectors. For $\lambda=5$: $A-5I=\begin{pmatrix}-1 & 1\\ 2 & -2\end{pmatrix}$. Row reduce: $-x_1+x_2=0\Rightarrow x_1=x_2$. Eigenvector $\mathbf{p}_1=\begin{pmatrix}1\\1\end{pmatrix}$. For $\lambda=2$: $A-2I=\begin{pmatrix}2&1\\2&1\end{pmatrix}\Rightarrow 2x_1+x_2=0\Rightarrow x_2=-2x_1$. Eigenvector $\mathbf{p}_2=\begin{pmatrix}1\\-2\end{pmatrix}$. Why this step? Each eigenvector spans the null space of $A-\lambda I$; it's the direction that only gets stretched.

Step 3–5 — assemble. $P=\begin{pmatrix}1 & 1\\ 1 & -2\end{pmatrix},\quad D=\begin{pmatrix}5 & 0\\ 0 & 2\end{pmatrix}.$ Why this step? Column 1 ($\lambda=5$ vector) pairs with $D_{11}=5$; column 2 with $D_{22}=2$ — same order.

Check: $\det P = -3 \ne 0$, so $P$ is invertible. ✓ And one can verify $AP=PD$.

Worked Example 2 — repeated eigenvalue, NOT diagonalizable

Let $A=\begin{pmatrix}2 & 1\\ 0 & 2\end{pmatrix}$ (a shear).

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

Step 2: $A-2I=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Solve $\begin{pmatrix}0&1\\0&0\end{pmatrix}\mathbf{x}=0\Rightarrow x_2=0$, $x_1$ free. Only one independent eigenvector $\begin{pmatrix}1\\0\end{pmatrix}$, so geometric multiplicity $g=1$.

Step 3: $g(2)=1 < a(2)=2$. We have only 1 eigenvector for a $2\times2$ matrix → cannot build invertible $P$ → NOT diagonalizable.

Why this matters: This is the prototype "defective" matrix. A shear has a single fixed direction; there's no second independent direction it merely stretches, so no diagonalizing basis exists.

Worked Example 3 — repeated eigenvalue that IS diagonalizable

Let $A=\begin{pmatrix}3 & 0\\ 0 & 3\end{pmatrix}=3I$.

$\det(A-\lambda I)=(3-\lambda)^2$, so $\lambda=3$ with $a=2$. Now $A-3I=0$, whose null space is all of $\mathbb{R}^2$, so $g=2$. Since $g=a$, it's diagonalizable — indeed it's already diagonal. Lesson: repeated eigenvalues are fine as long as they supply enough independent eigenvectors.

Common mistakes (Steel-man)

Active recall

Connections

• Eigenvalues and Eigenvectors — the raw ingredients.
• Characteristic Polynomial — source of algebraic multiplicities.
• Similar Matrices — diagonalization = similarity to a diagonal matrix.
• Matrix Powers and Exponentials — main payoff of diagonalization.
• Spectral Theorem — symmetric matrices are always (orthogonally) diagonalizable.
• Jordan Normal Form — what to do when a matrix is not diagonalizable.
• Change of Basis — $P$ is a change-of-basis matrix into the eigenbasis.

Concept Map

Hinglish (regional understanding)