"Eigenvalue" is about to change meaning. Right now it means "a number attached to a matrix." Soon it will also mean "a number attached to a boundary value problem." The two uses share more than just a name. A matrix eigenvalue problem, Av=\lambda v, asks one question: which vectors does a linear map only stretch, never turn? A differential eigenvalue problem asks that same question one dimension up. A differential operator stands in for the matrix. A function stands in for the vector. Diagonalization is the finite-dimensional picture worth having in hand before that jump.
An eigenvector is a direction. The matrix can hit it and it survives unchanged in kind, only in size. Point along that direction, apply A, and you're still pointing the same way. You're just stretched or shrunk, by the eigenvalue. Diagonalizing means re-describing A's entire action. We switch to the coordinate system built from those special directions. In that system, A stops looking like a matrix that mixes coordinates together. Instead it looks like one that just rescales each coordinate on its own, by its own eigenvalue. That's also why the eigenvector count matters. Here P is built by lining up eigenvectors as its columns. A matrix whose columns aren't independent can't be undone at all. It can't be inverted. So "enough independent eigenvectors" and "this matrix diagonalizes" turn out to be the same fact, just seen from two angles.
An eigenvalue \lambda of a square matrix A is a scalar. It makes Av=\lambda v true for some nonzero vector v. We find \lambda from the characteristic equation \det(A-\lambda I)=0. The matching eigenvector v is a direction A only scales, never rotates. Say A has size n\times n. When A has n linearly independent eigenvectors, it diagonalizes: A=PDP^{-1}. Here D is diagonal, with the eigenvalues on its diagonal. P's columns are the matching eigenvectors, in the same order. This independence requirement is not a technicality. P's columns are the eigenvectors themselves, and P must be invertible for P^{-1} to exist at all. That happens exactly when its columns are linearly independent. So "enough independent eigenvectors" and "diagonalizable" are the same condition, stated two ways.
Assembling P and D from A's own eigenvectors
Reveal each eigenpair to fill in a column of P and an entry of D, then confirm P, D, and P inverse multiply back to A.
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Recipe Diagonalize a matrix Use when: you have a square matrix A and need P, D with A=PDP^{-1} Goal: the eigenvalues and a matching set of independent eigenvectors, assembled into P and D
Quick recipe
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1
Form A-\lambda I by subtracting \lambda from each diagonal entry. Then solve \det(A-\lambda I)=0 for \lambda.
Example
For A=\begin{pmatrix}5&4\\1&2\end{pmatrix}, \det(A-\lambda I)=(5-\lambda)(2-\lambda)-4=\lambda^2-7\lambda+6=(\lambda-6)(\lambda-1), so \lambda=6 or \lambda=1.
Why: an eigenvalue is exactly a \lambda that makes A-\lambda I singular. A square matrix is singular in exactly one case. That case is when its determinant is zero.
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2
For each \lambda, solve (A-\lambda I)v=0 for a nonzero v.
Example
For \lambda=6: (A-6I)v=0 reduces to v=(4,1). For \lambda=1: (A-I)v=0 reduces to v=(-1,1).
Why: rearranging Av=\lambda v gives (A-\lambda I)v=0. Its nonzero solutions are exactly A's eigenvectors.
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3
A 2\times2 matrix needs 2 independent eigenvectors to diagonalize. An n\times n matrix needs n.
Example
A has two distinct eigenvalues, 6 and 1. So its two eigenvectors (4,1) and (-1,1) are automatically independent.
- distinct eigenvalues — their eigenvectors are automatically independent. So continue.
- a repeated eigenvalue can have fewer independent eigenvectors than its multiplicity — then A cannot be diagonalized.
Why: P needs a full set of independent columns. Only then is P invertible. Otherwise PDP^{-1} is undefined.
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4
Put each eigenvector into a column of P. Put its eigenvalue in the matching spot on D's diagonal.
Example
Column 1 holds \lambda=6 and its eigenvector (4,1). Column 2 holds \lambda=1 and its eigenvector (-1,1): P=\begin{pmatrix}4&-1\\1&1\end{pmatrix},\qquad D=\begin{pmatrix}6&0\\0&1\end{pmatrix}
Why: Av_i=\lambda_i v_i has to hold, column by column. Only then is AP=PD true.
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5
Compute P^{-1}. Then check that PDP^{-1} multiplies back out to A.
⚡ Shortcut
For a 2\times2 matrix P=\begin{pmatrix}a&b\\c&d\end{pmatrix}, P^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}. That's faster than general row reduction.
Why: this step actually confirms A is diagonalizable. It doesn't just assume that.
Common mistakes
- Subtracting \lambda from every entry of A instead of only the diagonal.
- Writing P^{-1}DP instead of PDP^{-1}.
- Putting an eigenvector in a column that doesn't match its eigenvalue's column in D.
- Assuming a repeated eigenvalue always blocks diagonalization. Check its eigenvector count instead.
Worked example
Eigenvalues from the characteristic equation
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1
Find the eigenvalues of A=\begin{pmatrix}5&4\\1&2\end{pmatrix}, the same matrix as recipe Step 1 above.
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2
\det(A-\lambda I) = (5-\lambda)(2-\lambda) - 4 = \lambda^2-7\lambda+6 = (\lambda-6)(\lambda-1).
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3\boxed{\lambda = 6,\ \lambda = 1}
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Worked example
Diagonalizing A
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1
Diagonalize A=\begin{pmatrix}5&4\\1&2\end{pmatrix} using its eigenvalues 6 and 1, continuing the recipe's Steps 2 and 4.
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2
For \lambda=6: (A-6I)v=0 gives v=(4,1). For \lambda=1: (A-I)v=0 gives v=(-1,1).
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3\boxed{A = PDP^{-1},\quad P=\begin{pmatrix}4&-1\\1&1\end{pmatrix},\ D=\begin{pmatrix}6&0\\0&1\end{pmatrix}}
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Where this goes next
Sturm-Liouville eigenvalue problems
A Sturm-Liouville problem is different. It hands back infinitely many eigenvalues, not a finite list. Each eigenvalue comes with its own eigenfunction. But the defining property is identical. Apply the operator to the eigenfunction. You get back a scalar multiple of the same function. Ch.03, "Fourier series and eigenfunction expansions," builds on this directly. There, an arbitrary function gets expanded in that eigenfunction basis. That's the same idea as before, just in infinite dimensions. It's the analogue of writing a vector in an eigenbasis. That's the same eigenbasis P diagonalizes A into, here. Ch.04, "Sturm-Liouville theory," is where that infinite eigenvalue list gets built in general.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Which single equation, solved for \lambda, gives every eigenvalue of a square matrix A?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That equation asks the whole matrix A-\lambda I to equal the zero matrix — far stronger than needed. We only need A-\lambda I to be singular, i.e. its determinant to be zero, not every one of its entries to vanish.
\det(A) is a single fixed number that doesn't depend on \lambda at all, so \det(A)\lambda=0 would only ever give \lambda=0 (or say nothing, if \det A = 0). The characteristic equation needs \lambda inside the determinant, subtracted from A's diagonal first.
This says A and \lambda A commute, which is always true for any scalar \lambda and gives no information about eigenvalues at all.
Why
Starting from A\mathbf v=\lambda\mathbf v, rearranging gives (A-\lambda I)\mathbf v=\mathbf 0. A nonzero \mathbf v exists exactly when A-\lambda I is singular, i.e. \det(A-\lambda I)=0 — the characteristic equation.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Which equation is the correct matrix form of A\mathbf v = \lambda \mathbf v used to derive the characteristic equation?
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A - \lambda subtracts a plain number from a matrix, which isn't defined — \lambda has to be turned into the matrix \lambda I first so both terms are 'a matrix times \mathbf v' before they can combine.
A\mathbf v is a vector, but \lambda alone is a scalar — the two can't be subtracted; \lambda needs to multiply a vector (namely \mathbf v) first.
This isn't a rearrangement of A\mathbf v = \lambda\mathbf v at all — it introduces A on the wrong side and doesn't reduce to the original relation.
Why
Writing \lambda\mathbf v as \lambda I \mathbf v lets both terms combine: A\mathbf v - \lambda I \mathbf v = (A-\lambda I)\mathbf v = \mathbf 0.
Start from A\mathbf v = \lambda\mathbf v and move every term to one side before factoring out \mathbf v.
From Ordinary Differential Equations
For A=\begin{pmatrix}4&0\\0&7\end{pmatrix}, what is the larger eigenvalue?
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Why
For a diagonal matrix, the eigenvalues are exactly the diagonal entries: 4 and 7.
Correct answer
Also accepted:
Why
For a diagonal matrix, the eigenvalues are simply the diagonal entries themselves.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A diagonalizable 3\times3 matrix has eigenvalues 1, -2, and 5. Write its characteristic polynomial in standard form (leading coefficient -1, matching the sign convention for a 3\times3 matrix).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
p_A(λ) = -(λ-1)(λ+2)(λ-5). Expanding (λ-1)(λ+2) = λ²+λ-2, then multiplying by (λ-5) gives λ³-4λ²-7λ+10, and negating gives -λ³+4λ²+7λ-10.
Correct answer
Also accepted:
Why
Build the polynomial from (\lambda-1)(\lambda+2)(\lambda-5), expand it, and apply the requested leading-coefficient sign.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
For A = \begin{bmatrix}5 & 4\\ 1 & 2\end{bmatrix}, which vector is a genuine eigenvector for the eigenvalue \lambda = 6, suitable as a column of P?
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(1,4) swaps the components of the real eigenvector — check directly: A(1,4)^T = (5+16,\ 1+8) = (21,\ 9), which is not a scalar multiple of (1,4).
(1,-1) is an eigenvector of this matrix, but for the other eigenvalue, \lambda = 1, not \lambda = 6.
(4,-1) isn't an eigenvector at all here: A(4,-1)^T = (20-4,\ 4-2) = (16,\ 2), which is not a scalar multiple of (4,-1).
Why
Solving (A - 6I)v = 0 gives [[-1,4],[1,-4]]v = 0, so v₁ = 4v₂ — v = (4,1) is the eigenvector. Check: A(4,1)ᵀ = (20+4, 4+2) = (24,6) = 6(4,1)ᵀ.
Solve (A - 6I)\mathbf v = \mathbf 0 and check which of the choices actually satisfies it.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A diagonalizable 3\times3 matrix has eigenvalues 2, -3, and 5. What is \det(A)?
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Why
det(A) = det(PDP⁻¹) = det(P)det(D)det(P)⁻¹ = det(D), the product of the diagonal entries of D — which are exactly A's eigenvalues. So det(A) = (2)(-3)(5) = -30.
Correct answer
Also accepted:
Why
Recall that the determinant of a diagonalizable matrix equals the product of its eigenvalues.