Finding eigenvalues: solve det(A − λI) = 0
Slide 1 tested arrows we had guessed. Now we want a recipe that finds them. Think of tuning an old radio: at most settings of the dial you hear only noise, and at a few settings a station comes in clearly. Here the dial is the number λ. For each λ, look at the matrix A − λI (I is the identity matrix). At most settings it moves the plane without squashing anything. At a few special settings it squashes a whole line of arrows flat onto zero, and that line is an eigenvector line. Lesson 6 gave us the test for "squashed flat": the determinant is zero. So we solve det(A − λI) = 0, the characteristic equation, to get the eigenvalues first. Then we find each eigenvector by elimination.
Lesson 7 · slide 2det(A − λI) = 0- Ax = λx is the same as (A − λI)x = 0.
- A nonzero solution x exists only when A − λI is singular: det(A − λI) = 0. This is the characteristic equation.
- For 2 by 2 it is λ² − (trace)λ + det = 0, where the trace is the sum of the diagonal entries.
- For each λ, the eigenvectors are the nonzero vectors in the null space of A − λI.
From Ax = λx to (A − λI)x = 0
Move everything to one side. To subtract λx as a matrix times x, write λx = λIx, where I is the identity matrix (lesson 2):
A − λI is just A with λ taken off each diagonal entry. A small example with A = 4123 and λ = 5:
So 11 is sent to zero by A − 5I. That says A11 = 5·11: an eigenvector with λ = 5.
A − λI must squash a line flat
If (A − λI)x = 0 for a nonzero x, then A − λI sends a whole line to the origin. It squashes the plane flat, so its determinant is 0 (lesson 6). Watch A − 5I do exactly that:
Turning the dial
Try several values of λ and compute det(A − λI) each time. It is zero only at λ = 2 and λ = 5. Those are the eigenvalues.
The characteristic equation
Instead of trying values, keep λ as a letter and compute the determinant once:
Check one value from the dial: at λ = 3, 9 − 21 + 10 = −2, the same as in the picture.
Then the eigenvectors
For each λ, solve (A − λI)x = 0 by elimination, like a null space in lesson 4. The matrix is singular, so one row turns into zeros:
Two quick checks: trace and determinant
For A = abcd the equation is always λ² − (a + d)λ + (ad − bc) = 0. The sum a + d of the diagonal is called the trace. So the two eigenvalues add up to the trace and multiply to the determinant:
Use these two checks to catch mistakes. They also give the equation fast. For 5412: trace 7, det 10 − 4 = 6, so λ² − 7λ + 6 = (λ − 6)(λ − 1) and λ = 6, 1.
Triangular matrices: read the diagonal
If A is triangular, so is A − λI, and its determinant is the product of the diagonal (lesson 6). So the eigenvalues are simply the diagonal entries:
Keep reading with the full course
Finish lesson 7 and you will be able to: tell whether a system settles down or blows up.
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Examples
Example 1 · A triangular matrix
A = 1203.
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For triangular matrices the eigenvalues sit on the diagonal.
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