Find an eigenvector corresponding to the eigenvalue λ = 3 for A = [[2, 1], [1, 2]].

Step 1: Use the eigenvector equation.

For eigenvalue λ = 3, an eigenvector v satisfies:
(A - 3I)v = 0

Step 2: Form A - 3I.

A = [[2, 1], [1, 2]]

Subtract 3 from the diagonal entries:

A - 3I = [[-1, 1], [1, -1]]

Step 3: Write the corresponding system.

Let v = (x, y).

Then:
-x + y = 0
x - y = 0

These equations are equivalent.

Step 4: Solve the equation.

From:
-x + y = 0

we get:
y = x

Therefore, x can be any nonzero scalar, and y must equal x.

Step 5: Choose a convenient value.

Let x = 1.

Then:
y = 1

So one eigenvector is:
v = (1, 1)

Step 6: Verify the result.

Multiply A by v:

[[2, 1], [1, 2]](1, 1) = (3, 3)

Also:
3(1, 1) = (3, 3)

Therefore, (1, 1) is an eigenvector corresponding to λ = 3. Any nonzero scalar multiple of (1, 1) is also an eigenvector.