This is the single most reused idea in the second half of this course. A system \mathbf{x}'=A\mathbf{x}'s eigenvalues decide whether every solution grows, decays, or oscillates, and its eigenvectors mark the special directions a trajectory bends toward or away from. Every phase portrait from Chapter 12 on is a picture of exactly these two numbers and two directions.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
In the derivation of the characteristic equation, why must the eigenvector \mathbf v be required nonzero?
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.
Matrix-vector multiplication is perfectly well-defined at \mathbf v=\mathbf 0 — it just always returns \mathbf 0, which is exactly the problem: it works for every \lambda at once.
Nothing in the definition restricts \lambda's sign — negative and zero eigenvalues are both legal, only \mathbf v=\mathbf 0 is excluded.
A-\lambda I is a fixed matrix regardless of what vector it's later applied to; it's perfectly defined at \mathbf v = \mathbf 0.
Why
\mathbf v=\mathbf 0 satisfies A\mathbf 0 = \lambda\mathbf 0 for literally every number \lambda, so it carries no information about which \lambda actually works. Excluding it is what makes the eigenvalue attached to a genuine direction.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Solve the characteristic equation of A = \begin{bmatrix} 4 & -3 \\ 6 & -5 \end{bmatrix} and enter the smaller (more negative) 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
\det(A-\lambda I) = (4-\lambda)(-5-\lambda) - (-3)(6) = \lambda^2 + \lambda - 20 + 18 = \lambda^2 + \lambda - 2 = (\lambda+2)(\lambda-1) = 0, so \lambda = -2 or \lambda = 1. The smaller one is -2.
Correct answer
Also accepted:
Why
Set up \det(A - \lambda I) = 0 as a quadratic in \lambda, solve for both roots, then pick the more negative one.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
For A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix}, expand \det(A - \lambda I) into standard polynomial form in \lambda.
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
\det(A-\lambda I) = (5-\lambda)(2-\lambda) - (4)(1) = \lambda^2 - 7\lambda + 10 - 4 = \lambda^2 - 7\lambda + 6.
Correct answer
Also accepted:
Why
Write out A - \lambda I first, then expand its 2\times 2 determinant using (a-\lambda)(d-\lambda) - bc.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
For the same C = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}, which vector lies in the eigenspace for C's other eigenvalue, \lambda = 2?
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.
(2,-1) isn't a scalar multiple of (1,-2) — check both rows of (C-2I)\mathbf v = \mathbf 0 against it, not just one.
This has the right numbers but the wrong sign relationship — plug it into (C-2I)\mathbf v and you won't get \mathbf 0.
This is -2 times (1,-2)'s first entry but not a consistent scalar multiple of the whole vector — it fails the second row of (C-2I)\mathbf v = \mathbf 0.
Why
C - 2I = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}, giving 2x_1 + x_2 = 0, i.e. x_2 = -2x_1. With x_1 = 1, the eigenspace is spanned by (1, -2).
Solve (C - 2I)\mathbf v = \mathbf 0 to find the direction of the eigenspace for \lambda = 2.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
For E = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, which vector spans the eigenspace for \lambda = 1?
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.
(1,-1) is the eigenvector for E's other eigenvalue, \lambda = -1 — check E(1,-1) = (-1, 1) = -1\cdot(1,-1), not 1\cdot(1,-1).
E(0,1) = (1,0), which isn't a scalar multiple of (0,1) at all — this vector isn't an eigenvector of E for any \lambda.
E(1,0) = (0,1), which isn't a scalar multiple of (1,0) — this vector isn't an eigenvector of E for any \lambda.
Why
E - I = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} gives -x_1 + x_2 = 0, so x_2 = x_1: the eigenspace for \lambda = 1 is spanned by (1,1).
Solve (E - I)\mathbf v = \mathbf 0 to find which direction is fixed by E.
An eigenvector is a direction a matrix respects. Pointed straight down that one line, multiplying by A only stretches or flips it, never turns it. Every other direction gets bent somewhere else entirely.
For a 2\times2 matrix A, the eigenvalues are the roots of the characteristic equation \det(A-\lambda I)=0, which expands to \lambda^2 - (\operatorname{tr}A)\lambda + \det A = 0 — a quadratic in \lambda built from just two numbers, the trace and the determinant. Each eigenvalue \lambda has a matching eigenvector \mathbf v satisfying A\mathbf v=\lambda\mathbf v: a direction A doesn't rotate, only stretches (or shrinks, or reverses) by the factor \lambda. This quadratic-in-\lambda shortcut, trace and determinant alone, is exactly what Chapter 13's trace–determinant plane turns into a single picture.
Find the two directions A respects
Drag v around the full circle. Av mostly points somewhere else -- find the two exact spots where it snaps back onto v's own line.
Loading interactive…
Recipe Find the eigenvalues and eigenvectors of a 2×2 matrix Use when: you have a 2\times2 matrix A — often the coefficient matrix of a system \mathbf x'=A\mathbf x — and need every \lambda and matching \mathbf v satisfying A\mathbf v=\lambda\mathbf v Goal: every eigenvalue of A, each paired with an eigenvector spanning its eigenspace
Quick recipe
-
1
Add the two diagonal entries to get \operatorname{tr}(A); compute \det(A)=ad-bc from all four entries.
Example
For B=\begin{pmatrix}5&2\\2&2\end{pmatrix}: \operatorname{tr}(B)=5+2=7, and \det(B)=(5)(2)-(2)(2)=10-4=6.
Why: for a 2\times2 matrix, these two numbers alone are enough to build the characteristic equation — there's no need to expand \det(A-\lambda I) by hand.
-
2
Substitute the two numbers from Step 1 into this quadratic and solve it — factor if it factors, otherwise fall back to the quadratic formula.
Example
\lambda^2-7\lambda+6=0 factors as (\lambda-6)(\lambda-1)=0, so \lambda=6 or \lambda=1.
- two distinct real roots — find a separate eigenvector for each one in Step 3.
- one repeated real root — Step 3 may turn up only one independent eigenvector direction instead of two; that's a genuine possibility for a repeated eigenvalue, not a sign of a mistake.
Why: the characteristic equation's roots are exactly the values of \lambda where A-\lambda I fails to be invertible, so they're the only candidates worth testing in Step 3.
-
3
Substitute \lambda, form A-\lambda I, then solve the resulting homogeneous system for a nonzero \mathbf v.
Example
For \lambda=6: B-6I=\begin{pmatrix}-1&2\\2&-4\end{pmatrix}. The first row gives -v_1+2v_2=0, i.e. v_1=2v_2; choosing v_2=1 gives \mathbf v=(2,1). For \lambda=1: B-1I=\begin{pmatrix}4&2\\2&1\end{pmatrix}. The first row gives 4v_1+2v_2=0, i.e. v_2=-2v_1; choosing v_1=1 gives \mathbf v=(1,-2).
Why: rearranging A\mathbf v=\lambda\mathbf v gives exactly (A-\lambda I)\mathbf v=\mathbf 0, so its nonzero solutions are the eigenvectors.
-
4
Multiply A by each eigenvector found and confirm it comes back as \lambda times that same vector.
Example
B(2,1)=(5\cdot2+2\cdot1,\ 2\cdot2+2\cdot1)=(12,6)=6(2,1). ✓ B(1,-2)=(5\cdot1+2\cdot(-2),\ 2\cdot1+2\cdot(-2))=(1,-2)=1(1,-2). ✓
Why: this catches an arithmetic slip in solving the homogeneous system before it carries into a system's general solution or phase portrait built on that eigenvector.
Common mistakes
- Treating the trace and determinant themselves as the eigenvalues, instead of solving the quadratic they build for \lambda.
- Substituting the wrong \lambda into A-\lambda I partway through the row reduction.
- Reporting only one eigenvector for a repeated eigenvalue that genuinely has two independent ones, or forcing a second independent one where only a single direction actually exists.
- Skipping the Step 4 check and carrying an eigenvector into later work that doesn't actually satisfy A\mathbf v=\lambda\mathbf v.
Worked example
Eigenvalues from trace and determinant
-
1
The same recipe as above, Steps 1 and 2, on a new matrix. Find the eigenvalues of A=\begin{pmatrix}3&1\\1&3\end{pmatrix}.
-
2
\operatorname{tr}A=6, \det A = 9-1=8, so \lambda^2-6\lambda+8=0.
-
3
Factoring: (\lambda-2)(\lambda-4)=0. \boxed{\lambda=2,\ \lambda=4}
Loading interactive…
Worked example
Finding the eigenvector for one eigenvalue
-
1
Steps 3 and 4 of the same recipe, continued. Find an eigenvector for \lambda=4 from the same matrix A above.
-
2
Solve (A-4I)\mathbf v=\mathbf 0: \begin{pmatrix}-1&1\\1&-1\end{pmatrix} \mathbf v=\mathbf 0 gives v_1=v_2.
-
3
Any nonzero scalar multiple of (1,1) works. \boxed{\mathbf v=(1,1)}
Loading interactive…
Where this goes next
Every system and phase portrait, ahead
Whether a system's trajectories spiral in, spiral out, or head straight for a saddle point is decided entirely by the sign and type of its eigenvalues — real or complex, positive or negative. Every classification in Chapters 12–14 is this section's computation, run on a different matrix each time.
Check your understanding (counts toward your progress)
\mathbf v is an eigenvector of A with eigenvalue \lambda=3. What does A\mathbf v look like?
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.
An eigenvector's whole defining property is that A does NOT rotate it off its own line.
Only \lambda=0 would send \mathbf v to the zero vector -- here \lambda=3\ne0.
Nothing in the eigenvector definition forces perpendicularity to the original vector.
Why
By definition, A\mathbf v=\lambda\mathbf v -- for \lambda=3, A\mathbf v=3\mathbf v, the same direction scaled by 3.
Recall the defining equation A\mathbf v=\lambda\mathbf v.
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.
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.
For A=\begin{pmatrix}4&0\\0&7\end{pmatrix}, what is the larger 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
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.
A student finds (A-\lambda I)\mathbf v=\mathbf 0 gives v_1=2v_2 and reports the single eigenvector \mathbf v=(2,1). Is a specific numeric vector really the ONLY correct answer?
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.
An eigenvector's direction is fixed, but its length and sign are not -- any nonzero scalar multiple works equally well.
(1,2) does not satisfy v_1=2v_2; (2,1) does.
A single eigenvalue with a one-dimensional eigenspace has only one independent direction, not two.
Why
Any nonzero scalar multiple of an eigenvector is itself an eigenvector for the same eigenvalue -- the defining equation A\mathbf v=\lambda\mathbf v is homogeneous in \mathbf v.
Any nonzero scalar multiple of an eigenvector satisfies the same defining equation.