The Central Idea
A square matrix \(A\) is diagonalizable when there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that
\[ \boxed{A=PDP^{-1}}. \]Equivalently,
\[ \boxed{P^{-1}AP=D}. \]The columns of \(P\) are eigenvectors of \(A\), and the diagonal entries of \(D\) are the corresponding eigenvalues.
Why Diagonal Matrices Are Special
A diagonal matrix has the form
\[ D= \begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}. \]It scales each coordinate direction independently:
\[ D \begin{bmatrix} x_1\\ \vdots\\ x_n \end{bmatrix} = \begin{bmatrix} \lambda_1x_1\\ \vdots\\ \lambda_nx_n \end{bmatrix}. \]There is no mixing between coordinates.
Diagonalization Is a Change of Basis
The matrix \(A\) may look complicated in the standard basis.
If the space has a basis of eigenvectors, then using that eigenbasis as the coordinate system changes the matrix representation into a diagonal matrix.
Thus diagonalization is fundamentally a basis-selection problem.
Eigenvectors Become Coordinate Axes
Suppose
\[ Av_i=\lambda_i v_i. \]If the eigenvectors
\[ v_1,\ldots,v_n \]form a basis of \(\mathbb{R}^n\), then every vector can be written uniquely as
\[ x=c_1v_1+\cdots+c_nv_n. \]Applying \(A\) gives
\[ Ax=c_1\lambda_1v_1+\cdots+c_n\lambda_nv_n. \]In the eigenbasis, the transformation simply scales each coordinate.
Constructing the Matrix \(P\)
Let
\[ v_1,\ldots,v_n \]be linearly independent eigenvectors of \(A\).
Form
\[ \boxed{ P= \begin{bmatrix} \vert&&\vert\\ v_1&\cdots&v_n\\ \vert&&\vert \end{bmatrix}}. \]Because the eigenvectors are independent, \(P\) is invertible.
Constructing the Matrix \(D\)
Place the eigenvalues in the same order as their corresponding eigenvectors:
\[ \boxed{ D= \begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}}. \]The ordering matters.
If \(v_j\) is column \(j\) of \(P\), then \(\lambda_j\) must occupy position \(j\) on the diagonal of \(D\).
Why \(AP=PD\)
Multiplying \(A\) by \(P\) applies \(A\) to every column:
\[ AP= \begin{bmatrix} Av_1&\cdots&Av_n \end{bmatrix}. \]Since each \(v_i\) is an eigenvector,
\[ AP= \begin{bmatrix} \lambda_1v_1&\cdots&\lambda_nv_n \end{bmatrix}. \]But right-multiplying \(P\) by \(D\) scales column \(i\) by \(\lambda_i\), so
\[ \boxed{AP=PD}. \]Deriving \(A=PDP^{-1}\)
Starting from
\[ AP=PD, \]multiply on the right by \(P^{-1}\):
\[ APP^{-1}=PDP^{-1}. \]Therefore,
\[ \boxed{A=PDP^{-1}}. \]Equivalent Form \(P^{-1}AP=D\)
Starting from
\[ A=PDP^{-1}, \]multiply on the left by \(P^{-1}\) and on the right by \(P\):
\[ \boxed{P^{-1}AP=D}. \]This shows directly that \(D\) is the matrix representation of the same transformation in the eigenvector basis.
Example: A Simple Diagonalizable Matrix
Consider
\[ A= \begin{bmatrix} 2&1\\ 0&3 \end{bmatrix}. \]Its eigenvalues are
\[ \lambda_1=2, \qquad \lambda_2=3. \]Corresponding eigenvectors may be chosen as
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]Build \(P\) and \(D\)
Place the eigenvectors into \(P\):
\[ P= \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}. \]Place the corresponding eigenvalues into \(D\):
\[ D= \begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}. \]Then
\[ \boxed{A=PDP^{-1}}. \]When Is a Matrix Diagonalizable?
An \(n\times n\) matrix is diagonalizable exactly when it has
\[ \boxed{n\text{ linearly independent eigenvectors}}. \]Equivalently, the vector space must possess a basis consisting entirely of eigenvectors of \(A\).
Distinct Eigenvalues Give Independence
Eigenvectors corresponding to distinct eigenvalues are linearly independent.
Therefore, if an \(n\times n\) matrix has \(n\) distinct eigenvalues, then it automatically has \(n\) independent eigenvectors.
Hence
\[ \boxed{ n\text{ distinct eigenvalues} \Longrightarrow A\text{ is diagonalizable}}. \]Distinct Eigenvalues Are Sufficient, Not Necessary
A matrix can still be diagonalizable when some eigenvalues repeat.
The real issue is not whether eigenvalues repeat, but whether enough independent eigenvectors exist.
Repeated eigenvalues require closer inspection of their eigenspaces.
Algebraic Multiplicity
The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial
\[ \det(A-\lambda I)=0. \]For example, if
\[ (\lambda-2)^3(\lambda+1)=0, \]then \(\lambda=2\) has algebraic multiplicity \(3\).
Geometric Multiplicity
The geometric multiplicity of an eigenvalue \(\lambda\) is the dimension of its eigenspace:
\[ \boxed{ \dim\ker(A-\lambda I)}. \]It counts how many linearly independent eigenvectors are available for that eigenvalue.
Multiplicity Bounds
For every eigenvalue,
\[ \boxed{ 1 \le \text{geometric multiplicity} \le \text{algebraic multiplicity}}. \]A repeated eigenvalue may therefore contribute one independent eigenvector or several.
Diagonalizability with Repeated Eigenvalues
A matrix is diagonalizable when the total number of independent eigenvectors supplied by all eigenspaces equals \(n\).
Equivalently, for each eigenvalue, its eigenspace must be large enough so that the geometric multiplicities collectively sum to \(n\).
Example: Repeated but Diagonalizable
Consider
\[ A= \begin{bmatrix} 2&0\\ 0&2 \end{bmatrix}. \]The only eigenvalue is
\[ \lambda=2, \]with algebraic multiplicity \(2\).
But every nonzero vector in \(\mathbb{R}^2\) is an eigenvector, so the eigenspace has dimension \(2\).
Therefore \(A\) is diagonalizable.
Example: Repeated and Not Diagonalizable
Consider
\[ A= \begin{bmatrix} 2&1\\ 0&2 \end{bmatrix}. \]The only eigenvalue is
\[ \lambda=2 \]with algebraic multiplicity \(2\).
However,
\[ A-2I= \begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}, \]whose null space is one-dimensional.
Only one independent eigenvector exists, so the matrix is not diagonalizable.
Defective Matrices
A square matrix that does not have enough independent eigenvectors to form a basis is sometimes called defective.
A defective matrix cannot be diagonalized.
More advanced methods replace diagonal form with structures such as Jordan form.
Diagonal Matrices Are Already Diagonalized
If \(A\) is already diagonal, we may simply choose
\[ P=I \]and
\[ D=A. \]Then
\[ A=IDI^{-1}=D. \]Scalar Matrices
A scalar matrix
\[ A=\lambda I \]acts by scaling every vector by the same factor.
Every nonzero vector is an eigenvector, so every basis is an eigenbasis.
Such a matrix is diagonalizable in infinitely many ways.
Diagonalization Is Not Unique
The factorization
\[ A=PDP^{-1} \]is generally not unique.
Eigenvectors may be rescaled, eigenvectors within an eigenspace may be replaced by another basis, and the ordering of eigenvectors may be changed if the diagonal entries of \(D\) are reordered consistently.
Rescaling Eigenvectors
If \(v\) is an eigenvector, then any nonzero scalar multiple
\[ cv \]is also an eigenvector for the same eigenvalue.
Therefore rescaling a column of \(P\) does not destroy a valid diagonalization as long as the matrix remains invertible.
Reordering Eigenvectors
Suppose
\[ P= \begin{bmatrix} v_1&v_2 \end{bmatrix}, \qquad D= \begin{bmatrix} \lambda_1&0\\ 0&\lambda_2 \end{bmatrix}. \]If the columns of \(P\) are swapped, the diagonal entries of \(D\) must be swapped as well.
The pairing between each eigenvector and its eigenvalue must remain intact.
Diagonalization and Coordinates
Let the eigenbasis be
\[ B=(v_1,\ldots,v_n). \]Then \(P\) converts \(B\)-coordinates into standard coordinates:
\[ x=P[x]_B. \]Therefore
\[ [x]_B=P^{-1}x. \]The Transformation in Eigenbasis Coordinates
Start with standard coordinates \(x\).
Convert to eigenbasis coordinates:
\[ P^{-1}x. \]Apply the diagonal scaling:
\[ D(P^{-1}x). \]Convert back:
\[ P(DP^{-1}x). \]Hence
\[ Ax=PDP^{-1}x. \]Why Matrix Powers Become Easy
Suppose
\[ A=PDP^{-1}. \]Then
\[ A^2 = (PDP^{-1})(PDP^{-1}). \]Since
\[ P^{-1}P=I, \]we obtain
\[ A^2=PD^2P^{-1}. \]General Power Formula
Repeating the cancellation gives
\[ \boxed{A^k=PD^kP^{-1}}. \]Because \(D\) is diagonal,
\[ D^k= \begin{bmatrix} \lambda_1^k&0&\cdots&0\\ 0&\lambda_2^k&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n^k \end{bmatrix}. \]This is far easier than multiplying \(A\) by itself repeatedly.
Example: Powers of a Diagonalizable Matrix
Suppose
\[ A=P \begin{bmatrix} 2&0\\ 0&3 \end{bmatrix} P^{-1}. \]Then
\[ \boxed{ A^{10} = P \begin{bmatrix} 2^{10}&0\\ 0&3^{10} \end{bmatrix} P^{-1}}. \]Only the eigenvalues need to be raised to the tenth power.
Negative Powers
If \(A\) is invertible and diagonalizable, then no diagonal entry of \(D\) is zero.
Therefore,
\[ A^{-1}=PD^{-1}P^{-1}, \]where
\[ D^{-1} = \operatorname{diag} \left( \frac1{\lambda_1}, \ldots, \frac1{\lambda_n} \right). \]Functions of a Diagonalizable Matrix
Diagonalization also simplifies many matrix functions.
If a suitable scalar function \(f\) is applied through a matrix functional calculus, then
\[ \boxed{ f(A)=Pf(D)P^{-1}}, \]where
\[ f(D)= \operatorname{diag} \bigl(f(\lambda_1),\ldots,f(\lambda_n)\bigr). \]Matrix Exponentials
One especially important example is the matrix exponential.
If
\[ A=PDP^{-1}, \]then
\[ \boxed{ e^{At} = Pe^{Dt}P^{-1}}, \]where
\[ e^{Dt} = \operatorname{diag} \left( e^{\lambda_1t}, \ldots, e^{\lambda_nt} \right). \]This is central to systems of differential equations.
Discrete Dynamical Systems
Consider
\[ x_{k+1}=Ax_k. \]After \(k\) steps,
\[ x_k=A^kx_0. \]If \(A\) is diagonalizable,
\[ x_k=PD^kP^{-1}x_0. \]The long-term behavior is therefore governed by powers of the eigenvalues.
Dominant Eigenvalues
In repeated applications of a diagonalizable matrix, eigenvalues with larger absolute value tend to dominate the long-term behavior.
If
\[ |\lambda_1|>|\lambda_2|, \]then the component along the first eigendirection may eventually dominate, provided the initial vector has a nonzero component in that direction.
Decay, Stability, and Growth
In a discrete system:
- if \(|\lambda|<1\), that eigendirection decays;
- if \(|\lambda|>1\), that eigendirection grows in magnitude;
- if \(|\lambda|=1\), its magnitude may persist;
- negative eigenvalues introduce sign reversal between iterations.
Continuous Systems
For a differential system
\[ x'(t)=Ax(t), \]diagonalization reduces the system into independent scalar equations in eigenbasis coordinates.
Each coordinate evolves according to
\[ y_i'(t)=\lambda_i y_i(t). \]Hence
\[ y_i(t)=c_ie^{\lambda_it}. \]Diagonalization Decouples a System
A matrix with off-diagonal entries mixes coordinates.
A diagonal matrix does not.
Thus changing into an eigenbasis converts a coupled linear system into independent one-dimensional scaling laws.
This process is often called decoupling.
Symmetric Matrices
Real symmetric matrices have an especially strong diagonalization property.
Every real symmetric matrix can be diagonalized using an orthonormal basis of eigenvectors.
This leads to the spectral theorem, developed more fully alongside inner products and orthogonality.
Orthogonal Diagonalization
If the eigenvectors can be chosen orthonormally, form
\[ Q= \begin{bmatrix} \vert&&\vert\\ q_1&\cdots&q_n\\ \vert&&\vert \end{bmatrix}. \]Then
\[ Q^{-1}=Q^T. \]The diagonalization becomes
\[ \boxed{A=QDQ^T}. \]Similarity
Two matrices \(A\) and \(B\) are similar if
\[ \boxed{B=P^{-1}AP} \]for some invertible matrix \(P\).
Similar matrices represent the same linear transformation relative to different bases.
Diagonalization as Similarity
A matrix is diagonalizable exactly when it is similar to a diagonal matrix.
That is,
\[ \boxed{ A\text{ is diagonalizable} \iff A\sim D \text{ for some diagonal }D}. \]Quantities Preserved by Similarity
Similar matrices share important structural quantities, including:
- the same characteristic polynomial;
- the same eigenvalues with algebraic multiplicities;
- the same determinant;
- the same trace;
- the same rank.
They may look different numerically because the coordinate system has changed.
Trace and Diagonalization
If
\[ A=PDP^{-1}, \]then
\[ \operatorname{tr}(A)=\operatorname{tr}(D). \]Since \(D\) contains the eigenvalues on its diagonal,
\[ \boxed{ \operatorname{tr}(A) = \lambda_1+\cdots+\lambda_n} \]when eigenvalues are counted with algebraic multiplicity.
Determinant and Diagonalization
Similar matrices have equal determinants.
Therefore,
\[ \det(A)=\det(D). \]Since the determinant of a diagonal matrix is the product of its diagonal entries,
\[ \boxed{ \det(A)=\lambda_1\lambda_2\cdots\lambda_n}. \]Zero Eigenvalues and Invertibility
If one diagonal entry of \(D\) is zero, then
\[ \det(D)=0. \]Therefore \(A\) is singular.
Thus an invertible diagonalizable matrix has no zero eigenvalues.
A Reliable Procedure for Diagonalizing a Matrix
- Find the eigenvalues of \(A\).
- For each eigenvalue \(\lambda\), solve \((A-\lambda I)v=0\).
- Find a basis for each eigenspace.
- Count the total number of independent eigenvectors.
- If fewer than \(n\) exist, the matrix is not diagonalizable.
- If \(n\) independent eigenvectors exist, place them into the columns of \(P\).
- Place the corresponding eigenvalues into the diagonal of \(D\) in matching order.
- Verify \(AP=PD\) or \(P^{-1}AP=D\).
A Reliable Procedure with Distinct Eigenvalues
- Compute the characteristic polynomial.
- Find all \(n\) distinct eigenvalues.
- Find one eigenvector for each eigenvalue.
- Place those eigenvectors into \(P\).
- Place the matching eigenvalues into \(D\).
Because eigenvectors for distinct eigenvalues are automatically independent, no additional independence test is necessary.
A Reliable Procedure with Repeated Eigenvalues
- Record each eigenvalue's algebraic multiplicity.
- Find the eigenspace \(\ker(A-\lambda I)\).
- Determine its dimension, the geometric multiplicity.
- Collect a basis of eigenvectors from each eigenspace.
- Add the geometric multiplicities.
- If the total is \(n\), the matrix is diagonalizable.
- If the total is less than \(n\), it is not diagonalizable.
Common Error: Putting Eigenvectors in \(D\)
The diagonal matrix \(D\) contains eigenvalues, not eigenvectors.
Eigenvectors form the columns of \(P\).
The correct structure is
\[ \boxed{ P=[v_1\ \cdots\ v_n], \qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)}. \]Common Error: Mismatching Eigenvectors and Eigenvalues
If column \(j\) of \(P\) is an eigenvector for \(\lambda_j\), then position \(j\) on the diagonal of \(D\) must contain \(\lambda_j\).
Reordering is allowed only when both matrices are reordered consistently.
Common Error: Assuming Every Square Matrix Is Diagonalizable
Squareness is necessary for ordinary diagonalization, but it is not sufficient.
The matrix must possess enough independent eigenvectors to form a basis.
Common Error: Assuming Repeated Eigenvalues Prevent Diagonalization
Repeated eigenvalues do not automatically cause failure.
A repeated eigenvalue is acceptable if its eigenspace supplies enough independent eigenvectors.
Common Error: Counting Eigenvalues Instead of Eigenvectors
A matrix may have an eigenvalue of algebraic multiplicity \(3\) but only one independent eigenvector associated with it.
Diagonalizability depends on the total number of independent eigenvectors, not just on root multiplicities.
Common Error: Writing \(A=P^{-1}DP\)
The standard diagonalization identity is
\[ \boxed{A=PDP^{-1}}. \]Equivalently,
\[ \boxed{D=P^{-1}AP}. \]Reversing the order changes the transformation.
Common Error: Assuming \(P\) Is Orthogonal
In a general diagonalization,
\[ P^{-1} \]is not necessarily equal to
\[ P^T. \]That simplification occurs when \(P\) is orthogonal, as in orthogonal diagonalization of real symmetric matrices.
Common Error: Computing \(A^k\) Directly After Diagonalizing
Once
\[ A=PDP^{-1} \]is known, use
\[ \boxed{A^k=PD^kP^{-1}}. \]Repeated multiplication of \(A\) defeats one of the main computational advantages of diagonalization.
Example: Full Diagonalization Workflow
Consider
\[ A= \begin{bmatrix} 4&1\\ 2&3 \end{bmatrix}. \]Its characteristic polynomial is
\[ \det(A-\lambda I) = \begin{vmatrix} 4-\lambda&1\\ 2&3-\lambda \end{vmatrix}. \]Therefore,
\[ (4-\lambda)(3-\lambda)-2 = \lambda^2-7\lambda+10. \]Factor:
\[ (\lambda-5)(\lambda-2)=0. \]So the eigenvalues are
\[ \lambda_1=5, \qquad \lambda_2=2. \]Find the Eigenvector for \(\lambda=5\)
Solve
\[ (A-5I)v=0. \]We have
\[ A-5I= \begin{bmatrix} -1&1\\ 2&-2 \end{bmatrix}. \]Thus
\[ -x+y=0, \]so one eigenvector is
\[ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]Find the Eigenvector for \(\lambda=2\)
Solve
\[ (A-2I)v=0. \]Then
\[ A-2I= \begin{bmatrix} 2&1\\ 2&1 \end{bmatrix}. \]Thus
\[ 2x+y=0. \]One eigenvector is
\[ v_2= \begin{bmatrix} 1\\ -2 \end{bmatrix}. \]Assemble the Diagonalization
Place the eigenvectors into
\[ P= \begin{bmatrix} 1&1\\ 1&-2 \end{bmatrix}. \]Place the matching eigenvalues into
\[ D= \begin{bmatrix} 5&0\\ 0&2 \end{bmatrix}. \]Therefore,
\[ \boxed{A=PDP^{-1}}. \]Use the Diagonalization to Compute Powers
For the matrix above,
\[ A^k = P \begin{bmatrix} 5^k&0\\ 0&2^k \end{bmatrix} P^{-1}. \]The complicated repeated action of \(A\) has been reduced to raising two scalars to powers.
Why Diagonalization Matters in Multidimensional Systems
A coupled multidimensional system may mix several variables at every step.
Diagonalization identifies special directions in which the transformation acts independently.
In that coordinate system, repeated transformations, differential systems, long-term growth, stability, and modal behavior become much easier to analyze.
The method underlies applications in dynamics, vibration analysis, population models, economics, control systems, quantum mechanics, data analysis, and numerical simulation.
Practice 1: Identify \(P\) and \(D\)
Suppose \(A\) has eigenvectors
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \]with eigenvalues \(2\) and \(4\), respectively.
Write valid matrices \(P\) and \(D\).
Practice 2: Distinct Eigenvalues
A \(3\times3\) matrix has three distinct eigenvalues. Must it be diagonalizable?
Practice 3: Repeated Eigenvalue
A \(2\times2\) matrix has only the eigenvalue \(\lambda=5\), with algebraic multiplicity \(2\), and its eigenspace has dimension \(1\).
Is the matrix diagonalizable?
Practice 4: Matrix Power
Suppose
\[ A=P \begin{bmatrix} 3&0\\ 0&-2 \end{bmatrix} P^{-1}. \]Write \(A^6\) in diagonalized form.
Practice 5: Similarity
If
\[ D=P^{-1}AP, \]what relationship holds between \(A\) and \(D\)?
Practice 6: Determinant from Eigenvalues
A diagonalizable \(3\times3\) matrix has eigenvalues
\[ 2,\quad -1,\quad 4. \]Find its determinant.
Practice 7: Trace from Eigenvalues
Using the same eigenvalues
\[ 2,\quad -1,\quad 4, \]find the trace.
Practice 8: Invertibility
A diagonalizable matrix has eigenvalues
\[ 5,\quad 2,\quad 0. \]Is it invertible?
Practice 9: Eigenbasis Coordinates
If
\[ x=P[x]_B, \]how do you recover \([x]_B\) from standard coordinates \(x\)?
Practice 10: Orthogonal Diagonalization
If \(Q\) is orthogonal, what is \(Q^{-1}\)?
Practice 11: Identify the Error
A student finds two eigenvalues for a \(3\times3\) matrix and concludes that the matrix cannot be diagonalizable because there are not three distinct eigenvalues.
Explain the error.
Practice 12: Long-Term Behavior
A diagonalizable discrete system has eigenvalues
\[ 0.5,\qquad 3. \]Which eigendirection tends to dominate repeated applications for a generic initial vector?
Practice Answers
Practice 1
\[ \boxed{ P= \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}}, \qquad \boxed{ D= \begin{bmatrix} 2&0\\ 0&4 \end{bmatrix}}. \]Practice 2
Yes. Eigenvectors corresponding to distinct eigenvalues are linearly independent, so three distinct eigenvalues provide three independent eigenvectors in \(\mathbb{R}^3\).
Practice 3
No. The eigenspace contributes only one independent eigenvector, but a \(2\times2\) matrix needs two independent eigenvectors to be diagonalizable.
Practice 4
\[ \boxed{ A^6 = P \begin{bmatrix} 3^6&0\\ 0&(-2)^6 \end{bmatrix} P^{-1}}. \]Practice 5
The matrices are similar:
\[ \boxed{A\sim D}. \]They represent the same linear transformation in different bases.
Practice 6
\[ \det(A) = 2(-1)(4) = \boxed{-8}. \]Practice 7
\[ \operatorname{tr}(A) = 2+(-1)+4 = \boxed{5}. \]Practice 8
No. A zero eigenvalue makes the determinant zero, so the matrix is singular.
Practice 9
\[ \boxed{[x]_B=P^{-1}x}. \]Practice 10
\[ \boxed{Q^{-1}=Q^T}. \]Practice 11
Distinct eigenvalues are sufficient for diagonalization but not necessary. A repeated eigenvalue may still provide several independent eigenvectors. The student must count independent eigenvectors, not merely distinct eigenvalues.
Practice 12
The eigendirection corresponding to
\[ \boxed{\lambda=3} \]tends to dominate because \(3^k\) grows while \(0.5^k\) decays.
What to Remember
Diagonalization rewrites a square matrix as
\[ \boxed{A=PDP^{-1}}. \]- The columns of \(P\) are linearly independent eigenvectors.
- The diagonal entries of \(D\) are the corresponding eigenvalues.
- A matrix is diagonalizable exactly when it has a basis of eigenvectors.
- \(n\) distinct eigenvalues guarantee diagonalizability for an \(n\times n\) matrix.
- Repeated eigenvalues require enough independent eigenvectors in their eigenspaces.
- \(A^k=PD^kP^{-1}\).
- Diagonalization decouples a multidimensional linear transformation into independent scalar actions.
- Similar matrices represent the same transformation in different bases.
- Real symmetric matrices admit orthogonal diagonalization.
The structural idea is:
\[ \boxed{ \text{find a basis in which the transformation acts by independent scaling}}. \]Where We Go Next
Diagonalization is especially powerful when eigenvectors can be chosen orthogonally. To develop that structure carefully, we next need the geometry supplied by inner products, lengths, angles, projections, and orthogonality.
The next lesson, Inner Products and Orthogonality, develops dot products, norms, orthogonal complements, orthonormal bases, projections, and the geometric machinery that supports least squares and spectral methods.