Quant Academy Math · Probability · Quant Finance
Phase 4 - Lesson 4.4

Diagonalization and Similarity

Rewriting a matrix in an eigenbasis so that acting, powering, and exponentiating become elementwise operations.

⏱ 50 min● Intermediate🔗 Prereqs: 4.3
↖ Phase 4 hub
Builds on: Eigenpairs from 4.3.
Leads to: The spectral theorem (4.6) guarantees orthogonal diagonalization for symmetric matrices; matrix exponentials drive linear ODEs.

Learning Objectives

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Key Vocabulary

Similar matrices
\(A\) and \(B=P^{-1}AP\) for invertible \(P\); same map in different bases.
Diagonalizable
\(A=PDP^{-1}\) with \(D\) diagonal; the columns of \(P\) are eigenvectors.
Eigenbasis
A basis of the space consisting entirely of eigenvectors of \(A\).
Algebraic multiplicity
The multiplicity of \(\lambda\) as a root of the characteristic polynomial.
Geometric multiplicity
The dimension \(\dim\ker(A-\lambda I)\) of the eigenspace.
Defective matrix
One lacking a full eigenbasis (some eigenvalue’s geometric < algebraic multiplicity); not diagonalizable.

Intuition & Motivation

Intuition
A matrix looks complicated only because you are viewing it in the wrong coordinates. If it has enough independent eigenvectors, switch to that eigenbasis and the matrix becomes diagonal - it just scales each axis. In those coordinates, applying the map a hundred times is raising each diagonal entry to the hundredth power; the exponential is the exponential of the diagonal. Similarity is the statement ‘same transformation, different coordinates’, and everything intrinsic - eigenvalues, trace, determinant, rank - is preserved by it.

Similarity

Definition - Similar matrices
\(A\) and \(B\) are similar if \(B=P^{-1}AP\) for some invertible \(P\). They represent the same linear map in different bases and share their characteristic polynomial - hence the same eigenvalues, trace, determinant, and rank.

Diagonalization

Theorem - Diagonalizability criterion
An \(n\times n\) matrix is diagonalizable iff it has \(n\) linearly independent eigenvectors. Then \(A=PDP^{-1}\) where \(D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\) and the columns of \(P\) are the corresponding eigenvectors. A sufficient condition: \(n\) distinct eigenvalues.

Diagonalization makes functions of a matrix trivial. Since \(A^k=PD^kP^{-1}\) and \(D^k\) just powers the diagonal, and more generally for any analytic \(f\), \(f(A)=P\,f(D)\,P^{-1}\) with \(f(D)\) applying \(f\) entrywise on the diagonal.

Worked Example - Compute \(A^{10}\) by diagonalizing
1
Reuse \(A=\begin{bmatrix}2&1\\1&2\end{bmatrix}\) with \(\lambda=1,3\) and eigenvectors \((1,-1),(1,1)\).
2
Form \(P=\begin{bmatrix}1&1\\-1&1\end{bmatrix},\ D=\begin{bmatrix}1&0\\0&3\end{bmatrix}\), with \(P^{-1}=\tfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\).
3
Then \(A^{10}=P D^{10} P^{-1}\) with \(D^{10}=\operatorname{diag}(1,3^{10})\) and \(3^{10}=59049\).
4
Multiply out: \(A^{10}=\tfrac12\begin{bmatrix}1+3^{10}&3^{10}-1\\3^{10}-1&1+3^{10}\end{bmatrix}=\begin{bmatrix}29525&29524\\29524&29525\end{bmatrix}\). No 10-fold matrix product needed.

When diagonalization fails

If some eigenvalue’s geometric multiplicity is strictly less than its algebraic multiplicity, there are too few eigenvectors to fill a basis and the matrix is defective. The canonical example is \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\): eigenvalue \(1\) has algebraic multiplicity 2 but only a one-dimensional eigenspace. Such matrices still admit the Jordan form, but not diagonalization.

Explore the map: when two eigen-directions exist and are distinct, the action decomposes into two independent stretches - that is exactly diagonalizability. Tune the entries toward a shear and the eigen-directions merge, previewing a defective matrix.

Common Mistakes to Avoid
  • Assuming every matrix is diagonalizable - defective matrices (repeated eigenvalues, deficient eigenspaces) are not.
  • Ordering the columns of \(P\) inconsistently with the diagonal of \(D\); the \(j\)-th column must match \(\lambda_j\).
  • Believing similarity preserves the individual entries; it preserves the eigenvalues/trace/det, not the numbers themselves.
  • Using \(A^k=PD^kP^{-1}\) but forgetting to invert \(P\) on the right.
Quant Practitioner Tips
  • Distinct eigenvalues ⇒ automatically diagonalizable - you need not even check eigenspaces.
  • To power or exponentiate a matrix, diagonalize once and act on the diagonal; this is how linear ODE solutions \(e^{At}\) are computed.
  • Symmetric matrices (covariances!) are always diagonalizable, and by an orthogonal \(P\) - the spectral theorem in 4.6.

Knowledge Check

Q1 Easy
Similar matrices \(B=P^{-1}AP\) necessarily share:
The same entries
The same eigenvalues, trace, and determinant
The same eigenvectors
Nothing
Q2 Medium
An \(n\times n\) matrix is diagonalizable iff it has:
A nonzero determinant
n linearly independent eigenvectors
All positive eigenvalues
Distinct rows
Q3 Medium
The matrix \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\) is:
Diagonalizable with distinct eigenvalues
Defective (not diagonalizable)
The identity
Singular

Practical Exercise

Let \(A=\begin{bmatrix}3&0\\1&2\end{bmatrix}\). (a) Find eigenvalues and eigenvectors. (b) Write \(A=PDP^{-1}\). (c) Use it to compute \(A^{3}\).

▶ Show full solution

(a) Triangular, so eigenvalues are the diagonal: \(\lambda=3,2\). For \(\lambda=3\): \((A-3I)=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\Rightarrow x=y\), eigenvector \((1,1)\). For \(\lambda=2\): \(\begin{bmatrix}1&0\\1&0\end{bmatrix}\Rightarrow x=0\), eigenvector \((0,1)\).

(b) \(P=\begin{bmatrix}1&0\\1&1\end{bmatrix},\ D=\begin{bmatrix}3&0\\0&2\end{bmatrix},\ P^{-1}=\begin{bmatrix}1&0\\-1&1\end{bmatrix}\).

(c) \(A^3=PD^3P^{-1}\) with \(D^3=\operatorname{diag}(27,8)\): \(A^3=\begin{bmatrix}27&0\\19&8\end{bmatrix}\) (since the (2,1) entry is \(27-8=19\)). Distinct eigenvalues guaranteed diagonalizability.

After the reveal, answer for yourself: Triangular matrices hand you the eigenvalues for free on the diagonal - always look there first.

Lesson Summary

Similar matrices are the same map in different coordinates and share all intrinsic invariants. A matrix with a full eigenbasis diagonalizes as \(A=PDP^{-1}\), turning powers and functions into elementwise operations on the diagonal. Repeated eigenvalues with deficient eigenspaces make a matrix defective and only Jordan-reducible.

Formula Sheet Additions

Function of a matrix
\[f(A)=P\,f(D)\,P^{-1},\quad A^k=PD^kP^{-1}\]
Diagonalize once, then act on the diagonal.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers
Q: What does diagonalization A=PDP^{-1} buy you?
A: Powers and analytic functions become elementwise on the diagonal: A^k=P D^k P^{-1}, f(A)=P f(D) P^{-1}.
Q: When is a matrix not diagonalizable?
A: When it is defective: some eigenvalue's geometric multiplicity is less than its algebraic multiplicity, so there is no full eigenbasis.

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Source References

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