Av = λv rearranges to (A − λI)v = 0. A non-zero v solving that means the matrix A − λI crushes something.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
So it must be singular
Only a singular matrix crushes a non-zero vector. That forces the determinant of A − λI to be zero.
The characteristic equation
Setting that determinant to zero gives a polynomial in λ. Its roots are the eigenvalues.
The 2 by 2 shortcut
For a 2 by 2 the equation is λ² − (trace)λ + (determinant) = 0. The trace is the sum of the diagonal.
Two useful checks
The eigenvalues sum to the trace and multiply to the determinant. Both are worth verifying before moving on.
Then the vectors
With λ known, solve (A − λI)v = 0 for v. There will be a whole line of solutions, and any non-zero one will do.
The characteristic equation
Ax = λx rearranges to (A − λI)x = 0, which has a nonzero solution only when det(A − λI) = 0. That determinant condition is where the characteristic polynomial comes from.
Its degree is the size of the matrix
An n by n matrix gives a degree-n polynomial, so there are n eigenvalues counted with multiplicity, possibly complex. Counting them against n confirms none were missed.
Eigenvectors come afterwards
Once an eigenvalue is known, solve (A − λI)x = 0 for the eigenvectors. There is always a whole subspace of them, since any scalar multiple of an eigenvector is another.
Complex eigenvalues mean rotation
A real matrix can have complex eigenvalues, which correspond to rotational behaviour with no fixed real direction. They arise in conjugate pairs and are perfectly meaningful.
Step 2: Try It Yourself
Tap and try it out.
- i-hat lands on(2, 1)
- j-hat lands on(2, 3)
- Determinant4
- Eigenvalues4 and 1
The dashed lines are the eigen-directions. A vector on one of those lines still points the same way after the transformation, only longer or shorter, and the eigenvalue is that stretch factor.
The shaded parallelogram is the image of the unit square, and its area is 4. That is exactly what the determinant measures.
Step 3: Watch an Example
One step at a time.
Watch Hugo Solve for Eigenvalues
Hugo finds the eigenvalues of the matrix with rows [4, 1] and [2, 3].
- Step 1
He computes the trace, which is 4 + 3 = 7.
Step 4: Your Turn
Practice makes it stick.
The Trace
Problem 1 of 2
Rows [6, 2] and [1, 5]. What is the trace?
The Missing One
Problem 2 of 2
Trace 9 and one eigenvalue is 4. What is the other?
Solve for Lambda
1 of 8
Rows [3, 0] and [0, 8]. What is the trace?
2 of 8
Rows [3, 0] and [0, 8]. What is the larger eigenvalue?
3 of 8
Trace 10 and determinant 21. What is the larger eigenvalue?
4 of 8
Eigenvalues 6 and 2. What is the determinant?
5 of 8
Eigenvalues 6 and 2. What is the trace?
6 of 8
Trace 0 and determinant 1. Are the eigenvalues real? 1 yes, 0 no.
7 of 8
Order the steps for finding eigenvalues and eigenvectors.
- 1Solve it for the eigenvalues
- 2Substitute each eigenvalue back into A minus lambda I
- 3Solve that system for the eigenvector
- 4Write the characteristic equation from the trace and determinant
8 of 8
A triangular matrix with diagonal 2 and 9. What is the smaller eigenvalue?
Step 5: Quick Check
Show what you know.
Question 1 of 2
Trace 12 and determinant 35. What is the larger eigenvalue?
Question 2 of 2
Why must the determinant of A − λI be zero?
What You Learned
- Av = λv rearranges to (A − λI)v = 0, which forces det(A − λI) = 0.
- For a 2 by 2 the characteristic equation is λ² − (trace)λ + (determinant) = 0.
- Eigenvalues sum to the trace and multiply to the determinant, which makes checking easy.