How do you find the eigenvalues of the matrix [[2,1],[1,2]] by hand?

## From the definition to a solvable equation The eigen-condition is `Av = lambda v` with `v != 0`. Move everything to one side, being careful to insert the identity so the subtraction is matrix-minus-matrix rather than matrix-minus-scalar: ``` Av - lambda v = 0 -> (A - lambda I) v = 0 ``` This says the nonzero vector `v` lies in the null space of `A - lambda I`. A square matrix has a nonzero null space exactly when it is **singular**, and singular is exactly when the determinant vanishes. Hence the **characteristic equation**: ``` det(A - lambda I) = 0 ``` Every eigenvalue is a root of this equation, and every root is an eigenvalue. The left-hand side expanded is a degree-`n` polynomial in `lambda` called the characteristic polynomial. ## Working the example Take `A = [[2,1],[1,2]]`. Then ``` A - lambda I = [[2 - lambda, 1], [1, 2 - lambda]] ``` For a 2x2 matrix `[[a,b],[c,d]]` the determinant is `ad - bc`, so ``` det(A - lambda I) = (2 - lambda)(2 - lambda) - (1)(1) = lambda^2 - 4*lambda + 4 - 1 = lambda^2 - 4*lambda + 3 = (lambda - 3)(lambda - 1) ``` The roots are `lambda = 3` and `lambda = 1`. There is a useful shortcut for 2x2 matrices: the characteristic polynomial is always `lambda^2 - trace(A)*lambda + det(A)`. Here `trace = 2 + 2 = 4` and `det = 4 - 1 = 3`, which reproduces `lambda^2 - 4*lambda + 3` immediately. ## Getting the eigenvectors Each eigenvalue is substituted back and the resulting homogeneous system is solved. For `lambda = 3`: ``` A - 3I = [[-1, 1], [1, -1]] ``` Both rows say the same thing, `-v1 + v2 = 0`, i.e. `v1 = v2`. The eigenspace is the line spanned by `(1,1)`; normalized it is `(1,1)/sqrt(2)`. For `lambda = 1`: ``` A - I = [[1, 1], [1, 1]] ``` Both rows say `v1 + v2 = 0`, so `v1 = -v2` and the eigenspace is spanned by `(1,-1)`, or `(1,-1)/sqrt(2)` normalized. A reassuring structural detail: the rows of `A - lambda I` **must** turn out to be dependent. If solving the system leaves you with only `v = 0`, you made an arithmetic error in the eigenvalue, because a genuine eigenvalue makes the matrix singular by construction. Notice that `(1,1)` and `(1,-1)` have dot product `1*1 + 1*(-1) = 0` — they are orthogonal. That is not luck. `A` is symmetric, and eigenvectors of a real symmetric matrix belonging to different eigenvalues are always orthogonal. ## Two free sanity checks For any square matrix: - the sum of the eigenvalues (with multiplicity) equals the **trace**, the sum of the diagonal entries; - the product of the eigenvalues equals the **determinant**. Here `3 + 1 = 4 = trace` and `3 * 1 = 3 = det`. These take seconds and catch most sign and arithmetic slips. They also give quick intuition: a zero determinant forces at least one zero eigenvalue, since a product that is zero needs a zero factor. ## Multiplicities A root of the characteristic polynomial can repeat. Its multiplicity as a root is the **algebraic multiplicity**; the dimension of the corresponding eigenspace is the **geometric multiplicity**. Geometric multiplicity is always at least one and never exceeds algebraic multiplicity. In the example above both eigenvalues are simple, so each eigenspace is a single line. ## Why this is a hand method only The characteristic polynomial is the right way to *define* eigenvalues and a fine way to compute them for 2x2 and 3x3 matrices in an interview. It is a poor way to compute them numerically. The polynomial has degree `n`, so for `n >= 5` there is no general closed-form solution by radicals, and worse, polynomial roots can be violently sensitive to small changes in the coefficients, so forming the polynomial first loses accuracy that the original matrix still contained. Practical eigenvalue computation is iterative: it works on the matrix directly with orthogonal transformations, never on its characteristic polynomial. Saying this out loud is usually what separates a memorized answer from an informed one.