Linear Algebra

Finding an eigenvector

Once the stretch factors of a matrix are known, how do you find every direction that each one stretches, and how many independent directions does each factor bring?

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The characteristic equation and finding eigenvaluesLinear Algebra

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DiagonalizationLinear Algebra

Where you are 5 in Eigenvalues and eigenvectors

Linear algebra6

Eigenvalues and eigenvectors5

The characteristic equation and finding eigenvalues Complex eigenvalues and rotation-scaling matrices

Finding an eigenvector

Diagonalization Matrix powers, dynamical systems and steady states

When you reach for it

Javlon, Shoira and Kamila clamp a thin steel plate in a loading rig and mark arrows on it from a center point. Under the load, the point at \mathbf{x} moves to A\mathbf{x}, a Linear transformation. Most arrows come out turned. They want the ones that come out only stretched, shrunk or reversed: along those directions the plate is purely stretched.

The The characteristic equation and finding eigenvalues has already given the stretch factors, the eigenvalues \lambda, as the roots of \det(A - \lambda I) = 0. A number is not a direction, though. This lesson takes each eigenvalue, finds all the vectors it stretches, describes them with a basis, and counts the independent directions each eigenvalue brings. That count decides whether there are enough pure-stretch directions to describe every arrow, the question Diagonalization answers.

Day 1 finds eigenspaces of 2 \times 2 matrices and checks eigenvectors by one multiplication. Day 2 moves to 3 \times 3 matrices, compares the dimension of an eigenspace with the multiplicity of its eigenvalue, and uses the independence of eigenvectors for different eigenvalues. An inline (a, b) means the column vector \begin{bmatrix} a \\ b \end{bmatrix}.

The procedure

The eigenspace is a kernel

A nonzero \mathbf{v} is an eigenvector of the n \times n matrix A for \lambda when A\mathbf{v} = \lambda\mathbf{v}. Since \lambda\mathbf{v} = \lambda I\mathbf{v} and matrix products distribute over subtraction,

A\mathbf{v} = \lambda\mathbf{v} \iff (A - \lambda I)\mathbf{v} = \mathbf{0}.

Definition. For an eigenvalue \lambda of A, the eigenspace of A for \lambda is

E_\lambda = \ker(A - \lambda I) = \{\mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \lambda\mathbf{x}\}.

It holds every eigenvector for \lambda, and also \mathbf{0}, which is never called an eigenvector. Its dimension is the geometric multiplicity of \lambda. The algebraic multiplicity, from the The characteristic equation and finding eigenvalues, is the number of times \lambda is repeated as a root of p(\lambda) = \det(A - \lambda I).

Being the kernel of A - \lambda I, the eigenspace is a subspace of \mathbb{R}^n. So eigenvectors never come alone: with \mathbf{v} comes its whole line, and two eigenvectors for the same \lambda bring every combination of them.

Finding a basis

Finding E_\lambda is solving a homogeneous system, as in Homogeneous systems and solutions in parametric vector form. Subtract \lambda from each diagonal entry of A and row reduce A - \lambda I to Reduced row echelon form; row operations do not change the solution set. Writing each pivot variable in terms of the free variables puts the solution in parametric vector form,

\mathbf{x} = t_1\mathbf{b}_1 + t_2\mathbf{b}_2 + \dots + t_k\mathbf{b}_k,

one vector \mathbf{b}_i per free variable. These span E_\lambda, and they are independent because \mathbf{b}_i has a 1 in the slot of the ith free variable where every other \mathbf{b}_j has a 0 (Solving a system by row reduction: one, none or infinitely many solutions). So they are a basis, and the geometric multiplicity is the number of free variables, n - \operatorname{rank}(A - \lambda I) by Rank and nullity.

Two checks come free. There must be at least one free variable: \det(A - \lambda I) = 0, so A - \lambda I cannot reduce to I, and if it does, the eigenvalue or the arithmetic is wrong. And a proposed eigenvector is tested by one multiplication: compute A\mathbf{v} and see whether it equals \lambda\mathbf{v}.

For a 2 \times 2 matrix with A - \lambda I not zero, the rank is exactly 1, so either nonzero row carries the whole system. A nonzero row (a, b) says ax_1 + bx_2 = 0, so E_\lambda = \operatorname{span}\{(-b, a)\}.

Geometric against algebraic multiplicity

Result. If \lambda is an eigenvalue of A, then 1 \le \dim E_\lambda, and \dim E_\lambda is at most the algebraic multiplicity of \lambda.

The lower bound holds because an eigenvalue has an eigenvector. The upper bound says a simple root brings exactly a line of eigenvectors, while a repeated root may bring as many independent directions as it is repeated, or fewer; Examples 3, 4 and 5 show both.

Proof of the upper bound (optional). Let k = \dim E_\lambda, extend a basis \mathbf{v}_1, \dots, \mathbf{v}_k of E_\lambda to a basis \mathbf{v}_1, \dots, \mathbf{v}_n of \mathbb{R}^n, and let P have these columns. Since A\mathbf{v}_i = \lambda\mathbf{v}_i for i \le k,

P^{-1}AP = \begin{bmatrix} \lambda I_k & X \\ 0 & Y \end{bmatrix}

for some blocks X and Y. By the product rule of Properties of determinants: row operations, products, transposes,

\begin{aligned} \det(P^{-1}AP - tI) &= \det\big(P^{-1}(A - tI)P\big) \\ &= \det(A - tI), \end{aligned}

and expanding the left side down its first k columns gives (\lambda - t)^k\det(Y - tI). So (\lambda - t)^k divides p(t).

Eigenvectors for different eigenvalues

Result. Eigenvectors \mathbf{v}_1, \dots, \mathbf{v}_m for distinct eigenvalues \lambda_1, \dots, \lambda_m are linearly independent. More generally, a basis of each of E_{\lambda_1}, \dots, E_{\lambda_m}, joined together, is linearly independent.

Proof for two. Suppose c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}. Multiplying by A gives c_1\lambda_1\mathbf{v}_1 + c_2\lambda_2\mathbf{v}_2 = \mathbf{0}; multiplying by \lambda_2 instead gives c_1\lambda_2\mathbf{v}_1 + c_2\lambda_2\mathbf{v}_2 = \mathbf{0}. Subtracting leaves c_1(\lambda_1 - \lambda_2)\mathbf{v}_1 = \mathbf{0}, and since \lambda_1 \neq \lambda_2 and \mathbf{v}_1 \neq \mathbf{0}, c_1 = 0. Then c_2\mathbf{v}_2 = \mathbf{0} gives c_2 = 0.

For any number, by induction. Assume the result for m - 1 vectors and let c_1\mathbf{v}_1 + \dots + c_m\mathbf{v}_m = \mathbf{0}. Multiply by A, separately by \lambda_m, and subtract: the \mathbf{v}_m terms cancel, leaving \sum_{i < m} c_i(\lambda_i - \lambda_m)\mathbf{v}_i = \mathbf{0}. By the assumption each c_i(\lambda_i - \lambda_m) = 0, so c_i = 0 for i < m, and then c_m = 0.

The joined bases. Group a dependence among the joined vectors by eigenspace: \mathbf{w}_1 + \dots + \mathbf{w}_m = \mathbf{0} with \mathbf{w}_i in E_{\lambda_i}. Any nonzero \mathbf{w}_i would be eigenvectors for distinct eigenvalues adding to \mathbf{0}, which the result forbids. So each \mathbf{w}_i = \mathbf{0}, and since each is a combination of a basis of E_{\lambda_i}, all coefficients are 0.

So the geometric multiplicities add to at most n, and \mathbb{R}^n has a basis of eigenvectors of A exactly when they add to n. In particular, n distinct real eigenvalues always give one.

Worked examples

Example 1

Day 1: The plate’s two pure-stretch directions

Javlon’s rig loads the plate by A = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}, with p(\lambda) = \lambda^2 - 3\lambda - 10 = (\lambda - 5)(\lambda + 2), so the eigenvalues are 5 and -2. Find a basis of each eigenspace.

Solution

For \lambda = 5, subtract 5 down the diagonal, divide row 1 by -3, and subtract 4 times it from row 2:

A - 5I = \begin{bmatrix} -3 & 3 \\ 4 & -4 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}.

The zero row is there, as it must be for an eigenvalue. The system says x_1 = x_2 with x_2 free, so \mathbf{x} = x_2(1, 1) and E_5 = \operatorname{span}\{(1, 1)\}.

For \lambda = -2, subtracting -2 means adding 2:

A + 2I = \begin{bmatrix} 4 & 3 \\ 4 & 3 \end{bmatrix} \sim \begin{bmatrix} 1 & \frac{3}{4} \\ 0 & 0 \end{bmatrix}.

So \displaystyle x_1 = -\frac{3}{4}x_2 and \displaystyle \mathbf{x} = x_2\left(-\frac{3}{4}, 1\right). Any nonzero multiple spans the same line; x_2 = -4 clears the fraction, so E_{-2} = \operatorname{span}\{(3, -4)\}.

Check each by one multiplication: A(1, 1) = (5, 5) = 5(1, 1), and A(3, -4) = (6 - 12, 12 - 4) = (-6, 8) = -2(3, -4). Along E_5 the plate is stretched five times; along E_{-2} it is doubled and flipped.

Figure 1The plate map A = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} has two eigenspaces, both lines through the origin. Along E_5 every vector is stretched five times; along E_{-2} every vector is reversed and doubled.

Example 2

Day 1: Testing candidates by multiplication

Kamila is handed M = \begin{bmatrix} 1 & 2 \\ 3 & -4 \end{bmatrix} and the vectors \mathbf{u} = (2, 1), \mathbf{w} = (1, 1) and \mathbf{z} = (1, -3). Decide which are eigenvectors of M, for which eigenvalues, and whether 3 is an eigenvalue of M.

Solution

Compute M\mathbf{v} and ask whether it is a multiple of \mathbf{v}:

\begin{aligned} M\mathbf{u} &= (2 + 2,\ 6 - 4) = (4, 2) = 2\mathbf{u}, \\ M\mathbf{z} &= (1 - 6,\ 3 + 12) = (-5, 15) = -5\mathbf{z}. \end{aligned}

So \mathbf{u} is an eigenvector for 2 and \mathbf{z} one for -5. But M\mathbf{w} = (3, -1), and c(1, 1) always has equal entries, so \mathbf{w} is not an eigenvector.

A number is tested by the determinant, since 3 is an eigenvalue exactly when M - 3I is not invertible:

\det(M - 3I) = \det\begin{bmatrix} -2 & 2 \\ 3 & -7 \end{bmatrix} = 14 - 6 = 8 \neq 0.

So 3 is not an eigenvalue. A 2 \times 2 matrix has at most two eigenvalues, and 2 and -5 are both simple roots, so E_2 = \operatorname{span}\{\mathbf{u}\} and E_{-5} = \operatorname{span}\{\mathbf{z}\} with nothing left to find. As a check, 2 + (-5) = -3 = \operatorname{tr} M and 2 \cdot (-5) = -10 = \det M.

Example 3

Day 1: A repeated eigenvalue with one direction

Under a shearing load the plate moves by S = \begin{bmatrix} 5 & 1 \\ 0 & 5 \end{bmatrix}. Find the eigenvalues of S with their algebraic and geometric multiplicities, and compare with 5I.

Solution

S is triangular, so p(\lambda) = (5 - \lambda)^2: the only eigenvalue is 5, with algebraic multiplicity 2. Then

S - 5I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},

already reduced. Its one equation is x_2 = 0, with x_1 free, so E_5 = \operatorname{span}\{(1, 0)\}: geometric multiplicity 1, below the algebraic 2. Check: S(1, 0) = (5, 0), while S(0, 1) = (1, 5) is turned off its line.

The matrix 5I has the same polynomial (5 - \lambda)^2, but 5I - 5I is zero, both variables are free, and E_5 = \mathbb{R}^2. The polynomial cannot tell the two apart; the eigenspace can. Every eigenvector of S lies on the x_1 axis, so no basis of \mathbb{R}^2 is made of eigenvectors of S, the case Diagonalization cannot handle.

Example 4

Day 2: A plane of eigenvectors

The lab moves to a block loaded by B = \begin{bmatrix} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 1 & 0 & 4 \end{bmatrix}. Find its eigenvalues and a basis of each eigenspace.

Solution

Column 2 of B - \lambda I has one nonzero entry, 3 - \lambda, so a Determinants by cofactor expansion down it and a difference of squares give

\begin{aligned} p(\lambda) &= (3 - \lambda)\big((4 - \lambda)^2 - 1\big) \\ &= (3 - \lambda)^2(5 - \lambda). \end{aligned}

The eigenvalues are 3, algebraic multiplicity 2, and 5, multiplicity 1.

For \lambda = 3, every row of B - 3I is a multiple of the first:

\begin{aligned} B - 3I &= \begin{bmatrix} 1 & 0 & 1 \\ 2 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix} \\ &\sim \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{aligned}

The one equation x_1 = -x_3 leaves x_2 and x_3 free:

\begin{aligned} \mathbf{x} &= \begin{bmatrix} -x_3 \\ x_2 \\ x_3 \end{bmatrix} \\ &= x_2\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}. \end{aligned}

So (0, 1, 0), (-1, 0, 1) is a basis of E_3, a plane, and the geometric multiplicity 2 matches the algebraic.

For \lambda = 5, divide row 1 of B - 5I by -1, subtract 2 times it from row 2 and subtract it from row 3, then divide row 2 by -2:

\begin{aligned} B - 5I &= \begin{bmatrix} -1 & 0 & 1 \\ 2 & -2 & 2 \\ 1 & 0 & -1 \end{bmatrix} \\ &\sim \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}. \end{aligned}

So x_1 = x_3, x_2 = 2x_3, and E_5 = \operatorname{span}\{(1, 2, 1)\}.

Check: B(0, 1, 0) = (0, 3, 0), B(-1, 0, 1) = (-3, 0, 3) and B(1, 2, 1) = (5, 10, 5). By the joined-bases result the three vectors are independent, so they are a basis of \mathbb{R}^3 made of eigenvectors of B.

Example 5

Day 2: Too few eigenvectors

Find the eigenspaces of C = \begin{bmatrix} 1 & 1 & 0 \\ -1 & 3 & 0 \\ 2 & -1 & 4 \end{bmatrix} and decide whether \mathbb{R}^3 has a basis of eigenvectors of C.

Solution

Column 3 of C - \lambda I has only the entry 4 - \lambda, so

\begin{aligned} p(\lambda) &= (4 - \lambda)\big((1 - \lambda)(3 - \lambda) + 1\big) \\ &= (4 - \lambda)(\lambda - 2)^2. \end{aligned}

The eigenvalues are 2, algebraic multiplicity 2, and 4, multiplicity 1.

For \lambda = 2, multiply row 1 of C - 2I by -1, use it to clear column 1 (row 2 becomes zero, row 3 becomes (0, 1, 2)), swap the last two rows, and add the new row 2 to row 1:

\begin{aligned} C - 2I &= \begin{bmatrix} -1 & 1 & 0 \\ -1 & 1 & 0 \\ 2 & -1 & 2 \end{bmatrix} \\ &\sim \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}. \end{aligned}

Only x_3 is free, with x_1 = x_2 = -2x_3. Taking x_3 = -1 gives E_2 = \operatorname{span}\{(2, 2, -1)\}, a line: geometric multiplicity 1, below the algebraic 2.

For \lambda = 4, the rows of C - 4I are (-3, 1, 0), (-1, -1, 0) and (2, -1, 0). The second gives x_2 = -x_1, and then the first gives -4x_1 = 0, so x_1 = x_2 = 0 with x_3 free: E_4 = \operatorname{span}\{(0, 0, 1)\}.

Check: C(2, 2, -1) = (4, 4, -2) and C(0, 0, 1) = (0, 0, 4). The geometric multiplicities add to 1 + 1 = 2 < 3, so by the joined-bases result no independent set of eigenvectors has three vectors, and \mathbb{R}^3 has no basis of eigenvectors of C.

Example 6

Day 2: Tuning a load to get a full set

For each number k, let D_k = \begin{bmatrix} 3 & 0 & 0 \\ k & 3 & 0 \\ 1 & 2 & 5 \end{bmatrix}. For which k is \dim E_3 = 2? For that k, give a basis of \mathbb{R}^3 made of eigenvectors of D_k.

Solution

D_k is lower triangular, so its eigenvalues are 3, algebraic multiplicity 2, and 5, for every k. Then

D_k - 3I = \begin{bmatrix} 0 & 0 & 0 \\ k & 0 & 0 \\ 1 & 2 & 2 \end{bmatrix}.

If k \neq 0, the rows (k, 0, 0) and (1, 2, 2) are independent, the rank is 2, and \dim E_3 = 3 - 2 = 1 by Rank and nullity. If k = 0, the rank is 1 and \dim E_3 = 2, the most the multiplicity bound allows. So k = 0 is the one value.

For k = 0, the equation x_1 + 2x_2 + 2x_3 = 0 gives x_1 = -2x_2 - 2x_3, so E_3 has basis (-2, 1, 0), (-2, 0, 1). The rows of D_0 - 5I are (-2, 0, 0), (0, -2, 0) and (1, 2, 0), forcing x_1 = x_2 = 0, so E_5 = \operatorname{span}\{(0, 0, 1)\}.

Check: D_0(-2, 1, 0) = (-6, 3, 0), D_0(-2, 0, 1) = (-6, 0, 3) and D_0(0, 0, 1) = (0, 0, 5). The joined bases give three independent eigenvectors, (-2, 1, 0), (-2, 0, 1), (0, 0, 1), a basis of \mathbb{R}^3. For k \neq 0 the multiplicities add to 2 and no such basis exists.

Boundaries

Only for eigenvalues, only for square matrices. For a number that is not an eigenvalue, A - \lambda I is invertible and its kernel is \{\mathbf{0}\}. And A\mathbf{v} = \lambda\mathbf{v} compares a vector with its image, so both must lie in the same \mathbb{R}^n.
The zero vector is in every eigenspace but is not an eigenvector. Every nonzero vector of E_\lambda is an eigenvector for \lambda.
A basis is not unique. Any nonzero multiple of an eigenvector is another, and a plane of eigenvectors has many bases; one multiplication settles whether a vector works.
Real eigenvectors need real eigenvalues. For a complex root of p, the system (A - \lambda I)\mathbf{x} = \mathbf{0} has only complex solutions; see Complex eigenvalues and rotation-scaling matrices.
Independence needs distinct eigenvalues. Two eigenvectors for one eigenvalue may be dependent, such as \mathbf{v} and 2\mathbf{v}.

Where this goes next

When the geometric multiplicities add to n, the joined bases of the eigenspaces form a basis of \mathbb{R}^n, and in that basis the matrix acts by plain stretching along each direction: this is Diagonalization, and Examples 4 and 6 have already built the bases it needs. When they fall short, as for S and C, no such basis exists. Repeated application of a matrix, in Matrix powers, dynamical systems and steady states, becomes easy along eigenvectors, since A^k\mathbf{v} = \lambda^k\mathbf{v}.

Exercises

Day 1: eigenspaces of 2 by 2 matrices

The matrix \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} has eigenvalues 5 and 2. Find a basis of each eigenspace.
Find the eigenvalues of \begin{bmatrix} 1 & 4 \\ 1 & 1 \end{bmatrix} and a basis of each eigenspace.
Find the eigenvalues of \begin{bmatrix} 0 & 1 \\ -6 & 5 \end{bmatrix} and a basis of each eigenspace.
By one multiplication each, decide which of (1, 1), (1, 2) and (2, 1) are eigenvectors of \begin{bmatrix} 3 & -2 \\ 4 & -3 \end{bmatrix}, and give the eigenvalue for each that is.
Decide whether 4 and 3 are eigenvalues of \begin{bmatrix} 6 & 2 \\ 1 & 5 \end{bmatrix}, and give a basis of the eigenspace for each one that is.
Find the geometric multiplicity of the eigenvalue 3 for \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} and for \begin{bmatrix} 3 & 0 \\ 2 & 3 \end{bmatrix}, with a basis of each eigenspace.
Find the eigenvalues of \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} and an eigenvector for each.
Shoira loads a plate by \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}. Find the directions in which it is purely stretched, and the factor in each.
Find the eigenvalues of \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} and a basis of each eigenspace. Which eigenspace is the kernel of the matrix itself?
Find every number h for which (1, h) is an eigenvector of \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}, and the eigenvalue in each case.

Day 2: larger matrices and multiplicities

Find a basis of each eigenspace of \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 1 & -1 & 3 \end{bmatrix}.
For \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}, give each eigenvalue with its algebraic and geometric multiplicity and a basis of its eigenspace.
Find the characteristic polynomial of \begin{bmatrix} 3 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix}, its eigenvalues, and a basis of each eigenspace.
Let G = \begin{bmatrix} 2 & 1 & -1 \\ 1 & 2 & -1 \\ 1 & 1 & 0 \end{bmatrix}. By one multiplication each, decide which of (1, 1, 1), (1, -1, 0), (1, 0, 1) and (1, 0, 0) are eigenvectors of G. Then explain, without the characteristic polynomial, why the eigenvalue 1 has algebraic and geometric multiplicity both 2.
Find the algebraic and geometric multiplicity of the eigenvalue 2 of \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}, and a basis of its eigenspace.
Explain why the eigenvectors found in Exercise 13 form a basis of \mathbb{R}^3, write (1, 0, 0) as a combination of them, and use it to find the matrix times (1, 0, 0) without multiplying the matrix out.

Reasoning

Using only A\mathbf{x} = \lambda\mathbf{x}, without kernels, show that E_\lambda is closed under addition and under multiplication by scalars.
Prove that if \mathbf{v} is an eigenvector of A for \lambda, then it is an eigenvector of A^k for \lambda^k for every whole number k \ge 1.
Prove that if \lambda \neq \mu, then E_\lambda and E_\mu share only the zero vector.
Prove directly, from the two-vector case and without the induction, that eigenvectors \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 for three distinct eigenvalues are independent.
A 2 \times 2 matrix A has a single eigenvalue \lambda with geometric multiplicity 2. Prove that A = \lambda I.
Refute with a counterexample: “If a 3 \times 3 matrix has three independent eigenvectors, then it has three distinct eigenvalues.”
Let \lambda be an eigenvalue of a 2 \times 2 matrix A with A \neq \lambda I, and let (a, b) be a nonzero row of A - \lambda I. Explain why E_\lambda = \operatorname{span}\{(-b, a)\}.
A 4 \times 4 matrix has characteristic polynomial (\lambda - 1)^2(\lambda - 3)^2. What are the possible geometric multiplicities of each eigenvalue, and exactly when does \mathbb{R}^4 have a basis of eigenvectors of the matrix?

Answers

E_5 = \operatorname{span}\{(1, 1)\}, E_2 = \operatorname{span}\{(1, -2)\}. The rows (-1, 1) of A - 5I and (2, 1) of A - 2I give x_1 = x_2 and x_2 = -2x_1.
Eigenvalues 3 and -1, from \lambda^2 - 2\lambda - 3. E_3 = \operatorname{span}\{(2, 1)\} and E_{-1} = \operatorname{span}\{(-2, 1)\}, from the rows (-2, 4) and (2, 4). Check: A(-2, 1) = (2, -1).
Eigenvalues 2 and 3, from \lambda^2 - 5\lambda + 6. E_2 = \operatorname{span}\{(1, 2)\} and E_3 = \operatorname{span}\{(1, 3)\}, from the rows (-2, 1) and (-3, 1).
(1, 1) is an eigenvector for 1 and (1, 2) is an eigenvector for -1, since the products are (1, 1) and (-1, -2). (2, 1) is not: A(2, 1) = (4, 5).
4 is an eigenvalue and 3 is not: \det(A - 4I) = 2 - 2 = 0 but \det(A - 3I) = 6 - 2 = 4. E_4 = \operatorname{span}\{(1, -1)\}, from the row (2, 2).
Geometric multiplicity 2 for the first: A - 3I = 0, so E_3 = \mathbb{R}^2, basis (1, 0), (0, 1). Geometric multiplicity 1 for the second: its row (2, 0) gives x_1 = 0, so E_3 = \operatorname{span}\{(0, 1)\}.
\displaystyle \lambda = \pm\sqrt{5}, from \lambda^2 - 5. The row \displaystyle (2, -1 - \sqrt{5}) of \displaystyle A - \sqrt{5}I gives the eigenvector \displaystyle (1 + \sqrt{5}, 2); for \displaystyle -\sqrt{5}, \displaystyle (1 - \sqrt{5}, 2). Check: \displaystyle A(1 + \sqrt{5}, 2) = (5 + \sqrt{5}, 2\sqrt{5}).
Along (2, 1) the plate is stretched by 6; along (1, -2) it is unchanged. Here p(\lambda) = (\lambda - 6)(\lambda - 1), and the rows (-1, 2) of A - 6I and (4, 2) of A - I give the directions.
Eigenvalues 0 and 4 (trace 4, determinant 0). E_0 = \operatorname{span}\{(-2, 1)\} and E_4 = \operatorname{span}\{(2, 1)\}. E_0 = \ker(A - 0I) is the kernel of the matrix.
h = 2 with eigenvalue 5, and h = -1 with eigenvalue -1. A(1, h) = (1 + 2h, 4 + 3h) = \lambda(1, h) gives \lambda = 1 + 2h and 4 + 3h = h + 2h^2, so h^2 - h - 2 = 0.
E_2 has basis (1, 1, 0), (-1, 0, 1) and E_3 = \operatorname{span}\{(0, 0, 1)\}. A - 2I has the one nonzero row (1, -1, 1); A - 3I forces x_1 = x_2 = 0.
\lambda = 1: algebraic 2, geometric 1, E_1 = \operatorname{span}\{(0, 1, 0)\}, since A - I forces x_1 = x_3 = 0. \lambda = 2: algebraic and geometric 1, E_2 = \operatorname{span}\{(0, 0, 1)\}.
Expanding along row 3, p(\lambda) = (2 - \lambda)^2(4 - \lambda). E_2 has basis (-1, 1, 0), (0, 0, 1), from x_1 + x_2 = 0. E_4 = \operatorname{span}\{(1, 1, 0)\}, from x_1 = x_2, x_3 = 0.
(1, 1, 1) is an eigenvector for 2; (1, -1, 0) and (1, 0, 1) are eigenvectors for 1; (1, 0, 0) is not, since G(1, 0, 0) = (2, 1, 1). Two independent vectors in E_1 make the algebraic multiplicity of 1 at least 2, and the eigenvalue 2 uses at least one of the three roots, so it is exactly 2; then 2 \le \dim E_1 \le 2.
Algebraic multiplicity 3, geometric multiplicity 2, basis (1, 0, 0), (0, 0, 1): A - 2I has the one nonzero row (0, 1, 0).
They are joined bases of E_2 and E_4, so independent, and three independent vectors in \mathbb{R}^3 are a basis. Solving gives (1, 0, 0) = -\tfrac{1}{2}(-1, 1, 0) + \tfrac{1}{2}(1, 1, 0), so the product is (3, 1, 0), the first column of the matrix, worked out below.

\begin{aligned} A(1, 0, 0) &= -\tfrac{1}{2}A(-1, 1, 0) + \tfrac{1}{2}A(1, 1, 0) \\ &= -(-1, 1, 0) + 2(1, 1, 0) \\ &= (3, 1, 0). \end{aligned}

If A\mathbf{u} = \lambda\mathbf{u} and A\mathbf{w} = \lambda\mathbf{w}, then A(\mathbf{u} + \mathbf{w}) = \lambda\mathbf{u} + \lambda\mathbf{w} = \lambda(\mathbf{u} + \mathbf{w}) and A(c\mathbf{u}) = cA\mathbf{u} = \lambda(c\mathbf{u}), using only linearity of matrix times vector.
If A^k\mathbf{v} = \lambda^k\mathbf{v}, then A^{k+1}\mathbf{v} = A(\lambda^k\mathbf{v}) = \lambda^k A\mathbf{v} = \lambda^{k+1}\mathbf{v}. The case k = 1 is given, so induction covers every k; and \mathbf{v} \neq \mathbf{0}.
If \mathbf{x} is in both, \lambda\mathbf{x} = A\mathbf{x} = \mu\mathbf{x}, so (\lambda - \mu)\mathbf{x} = \mathbf{0}, and \lambda - \mu \neq 0 forces \mathbf{x} = \mathbf{0}.
From c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0}, multiply by A, subtract \lambda_3 times the original: c_1(\lambda_1 - \lambda_3)\mathbf{v}_1 + c_2(\lambda_2 - \lambda_3)\mathbf{v}_2 = \mathbf{0}. The two-vector case makes both coefficients 0, and the differences are nonzero, so c_1 = c_2 = 0; then c_3 = 0.
\dim E_\lambda = 2 means E_\lambda = \mathbb{R}^2, so A\mathbf{e}_1 = \lambda\mathbf{e}_1 and A\mathbf{e}_2 = \lambda\mathbf{e}_2. These are the columns of A, so A = \lambda I.
The 3 \times 3 identity: \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 are independent eigenvectors, all for the one eigenvalue 1.
\det(A - \lambda I) = 0 makes the rank below 2, and A - \lambda I \neq 0 makes it 1, so \dim E_\lambda = 1 and every row is a multiple of (a, b). The system is ax_1 + bx_2 = 0, which the nonzero vector (-b, a) satisfies, so it spans the line E_\lambda.
Each of \dim E_1, \dim E_3 is 1 or 2, by the multiplicity bound. A basis exists exactly when both are 2. If both are 2, the joined bases give 4 independent eigenvectors. If not, an independent set of eigenvectors has at most \dim E_1 + \dim E_3 vectors by the joined-bases result, and a basis of \mathbb{R}^4 needs 4.