An eigenvalue \lambda and eigenvector \mathbf{v} of a matrix A satisfy A\mathbf{v}=\lambda\mathbf{v}. In other words, A acts on \mathbf{v} as plain scalar multiplication. This course's own eigenvalue chapter starts from the exact-algebra procedure for finding them. Then it asks how to compute them once exact algebra becomes impractical.
An eigenvector is a direction a matrix doesn't rotate. The matrix only stretches or shrinks it. That stretch factor is called an eigenvalue. Finding them exactly means solving one polynomial equation. Then it means solving one linear system per root. Two more facts matter almost as much as the exact values. Multiply every stretch factor together, and you get one number attached to the whole matrix — its determinant. Add them together instead, and you get a different number — its trace. Both give a ready-made check on a computed answer. Now apply a matrix to almost any starting vector, over and over. The result eventually points almost entirely in one direction: the single strongest stretch factor. That happens fast when the strongest stretch factor clearly beats the second-strongest one. It happens slow when the two are close together.
For an n\times n matrix A, the eigenvalues are the roots of the characteristic equation \det(A-\lambda I)=0. This is a degree-n polynomial in \lambda. Each eigenvalue's eigenvectors are the nonzero solutions of (A-\lambda I)\mathbf{v}=\mathbf{0}. That equation follows from a short chain of steps. Start from A\mathbf{v}=\lambda\mathbf{v}. Rearrange it to (A-\lambda I)\mathbf{v}=\mathbf{0}. This has a nonzero solution \mathbf{v} exactly when A-\lambda I fails to be invertible. And a matrix fails to be invertible exactly when its determinant vanishes.
Three more facts about the full set of eigenvalues matter later in this course. The trace of A — the sum of its diagonal entries — always equals the sum of the eigenvalues. And \det(A) always equals their product. Both give a fast sanity check on any computed set. When A has n linearly independent eigenvectors, it is diagonalizable: A=V\Lambda V^{-1}. The eigenvalues fill the diagonal of \Lambda, and the eigenvectors fill the columns of V. Most important of all is the spectral radius \rho(A)=\max_i |\lambda_i|, the largest eigenvalue magnitude. It governs how fast repeated multiplication by A pulls a starting vector toward the eigenvector for that largest eigenvalue. The wider the gap between |\lambda_1| and the next-largest |\lambda_2|, the faster that convergence happens.
From matrix to characteristic polynomial to eigenvalues
Step through forming A-\lambda I, expanding its determinant, and solving the resulting polynomial for \lambda.
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Worked example
Eigenvalues of a diagonal matrix, by inspection
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1
Find the eigenvalues of D=\begin{bmatrix}2&0\\0&3\end{bmatrix}.
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2
\det(D-\lambda I) = (2-\lambda)(3-\lambda), already factored.
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3\boxed{\lambda=2 \text{ and } \lambda=3}
For a diagonal matrix, the eigenvalues are always just the diagonal entries. No elimination needed.
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Worked example
A matrix with a repeated eigenvalue
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Find the eigenvalues of A=\begin{bmatrix}4&1\\0&4\end{bmatrix}.
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2
\det(A-\lambda I) = (4-\lambda)^2 = 0.
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3\boxed{\lambda=4 \text{ (multiplicity 2)}}
Reporting "\lambda=4" once instead of twice silently drops information. This course's own methods depend on getting that right — deflation, especially.
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Worked example
Comparing convergence speed via the eigenvalue gap
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1
Matrix A has eigenvalues \lambda_1=5, \lambda_2=3. Matrix B has eigenvalues \lambda_1=5, \lambda_2=4.9. Repeated multiplication by which matrix separates the dominant eigen-direction faster?
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Both share the same largest eigenvalue. So the separation speed comes down to one ratio, |\lambda_2/\lambda_1|. That's 0.6 for A versus 0.98 for B.
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3\boxed{A \text{ converges far faster than } B}
A small ratio decays toward zero within a handful of multiplications. A ratio near 1 can take hundreds of them instead. That's exactly the situation this course's power-method chapter is built to speed up.
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Where this goes next
Why exact factoring stops working
Factoring a degree-n polynomial by hand works for n=2 or 3. Not for a 50\times50 matrix. Nobody expands a degree-50 characteristic polynomial by hand. Nobody factors it by hand either. This course's eigenvalue chapter builds other methods instead: the power method, the QR algorithm. They find eigenvalues without ever writing down that polynomial at all.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Ordinary Differential Equations
A 2\times2 matrix has trace 5 and determinant 6. What is its smaller eigenvalue?
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.
Why
\lambda^2-5\lambda+6=0 factors as (\lambda-2)(\lambda-3)=0, giving eigenvalues 2 and 3.
Correct answer
Also accepted:
Why
Use \lambda^2-(\operatorname{tr}A)\lambda+\det A=0 and factor the resulting quadratic.
From Ordinary Differential Equations
Write the characteristic polynomial of A=\begin{pmatrix}2&3\\1&4\end{pmatrix}, using L to stand for \lambda, as an expression in L.
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.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\operatorname{tr}A=6, \det A=8-3=5, giving \lambda^2-6\lambda+5=0.
Correct answer
Also accepted:
Why
Use L^2-(\operatorname{tr}A)L+\det A.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
For an n\times n matrix A, what must be true of A - \lambda I at every value of \lambda that is a genuine eigenvalue of 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.
Subtracting \lambda off the diagonal only changes the diagonal entries — it does nothing to any nonzero off-diagonal entries A already had, so A - \lambda I has no reason to become diagonal.
A-\lambda I inherits A's symmetry only if A itself was already symmetric; nothing in the eigenvalue condition requires that.
Invertible is the opposite of what's needed — an invertible A-\lambda I sends only \mathbf 0 to \mathbf 0, so no eigenvector could exist at that \lambda.
Why
\lambda is an eigenvalue exactly when \det(A-\lambda I)=0, i.e. when A - \lambda I is singular and therefore has a nonzero vector in its null space — that vector is the eigenvector belonging to \lambda.
Think about what it means for A - \lambda I to send some nonzero \mathbf v to \mathbf 0.