Linear Algebra

The characteristic equation and finding eigenvalues

A linear map stretches a rubber sheet and turns almost every arrow drawn on it. Which directions come out only stretched or shrunk, never turned, and how can the stretch factors be found without guessing one direction at a time?

You need

When a matrix or transformation can be undoneLinear Algebra The quadratic formulaGrades 9 to 10

Unlocks

Complex eigenvalues and rotation-scaling matricesLinear Algebra Finding an eigenvectorLinear 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

Why this exists

Sarnai, Tuvshin and Odval clamp a square sheet of rubber membrane in an adjustable frame and draw arrows on it from a center point. Pulling the frame moves the point at (x, y) to A\begin{bmatrix} x \\ y \end{bmatrix}, a Linear transformation with matrix

A = \begin{bmatrix} 7 & -4 \\ 2 & 1 \end{bmatrix}.

After the pull, almost every arrow points a new way. Odval asks whether any arrow comes out pointing along its old line, only longer, shorter or reversed. Along such a direction the map does the simplest thing a matrix can do: it multiplies by one number. But there are infinitely many directions to try, and testing them one by one never ends.

This lesson turns the search into one polynomial equation in one unknown number. Its roots are the stretch factors, called eigenvalues, and they can be found before any direction is known. The lesson runs over two days: day 1 builds the equation and solves it for 2 \times 2 matrices; day 2 takes it to larger matrices and reads the trace and determinant off the eigenvalues.

Definition

Let A be an n \times n matrix. A number \lambda is an eigenvalue of A when some nonzero vector \mathbf{v} has

A\mathbf{v} = \lambda\mathbf{v}.

Such a \mathbf{v} is an eigenvector of A for \lambda. In plain words, \mathbf{v} is a direction the map keeps on its own line, and \lambda is the factor applied along it: a stretch when \lambda > 1, a shrink when 0 < \lambda < 1, a reversal when \lambda < 0, and a collapse to \mathbf{0} when \lambda = 0.

The vector must be nonzero, because A\mathbf{0} = \lambda\mathbf{0} holds for every number \lambda; allowing \mathbf{0} would make every number an eigenvalue. The number \lambda = 0 itself is allowed.

The characteristic polynomial of A is

p(\lambda) = \det(A - \lambda I),

where A - \lambda I is A with \lambda subtracted from each diagonal entry. The characteristic equation is \det(A - \lambda I) = 0. If \lambda_1 is a root and the factor (\lambda_1 - \lambda) appears exactly m times in p(\lambda), then m is the algebraic multiplicity of \lambda_1. The trace of A is the sum of its diagonal entries, \operatorname{tr} A = a_{11} + a_{22} + \dots + a_{nn}.

Some books use \det(\lambda I - A) instead, which is (-1)^n p(\lambda): the same roots with the same multiplicities. In this lesson an eigenvalue means a real one unless complex numbers are named; complex roots belong to Complex eigenvalues and rotation-scaling matrices.

Intuition

The figure shows the membrane map at work. The arrow \mathbf{v} = (1, 1) lands at (3, 3), on its own line, three times as long. The arrow \mathbf{w} = (1, 0) lands at (7, 2), turned. The question is which numbers \lambda have an arrow like \mathbf{v}.

Figure 1The membrane map A = \begin{bmatrix} 7 & -4 \\ 2 & 1 \end{bmatrix} sends \mathbf{v} = (1, 1) to (3, 3), three times as long on the same line, so \mathbf{v} is an eigenvector for 3. It sends \mathbf{w} = (1, 0) to (7, 2), which is turned off its line.

Move everything in A\mathbf{v} = \lambda\mathbf{v} to one side: (A - \lambda I)\mathbf{v} = \mathbf{0}. So \mathbf{v} is a nonzero vector in the kernel of the shifted matrix A - \lambda I. A matrix with a nonzero kernel vector flattens that direction to \mathbf{0}, and by What the determinant means a map flattens the plane exactly when its determinant is 0.

So slide \lambda along the number line and watch \det(A - \lambda I). For the membrane matrix it is (7 - \lambda)(1 - \lambda) + 8. At \lambda = 0 it is 15, the area factor of A itself. At \lambda = 3 it is 0: the shifted map flattens, and a kept direction appears. At \lambda = 4 it is -1, and at \lambda = 5 it is 0 again. The eigenvalues are the moments when the shifted map flattens.

Results

Eigenvalues are roots. A number \lambda is an eigenvalue of the square matrix A exactly when \det(A - \lambda I) = 0.

The 2 \times 2 formula. For A = \begin{bmatrix} a & b \\ c & d \end{bmatrix},

\begin{aligned} p(\lambda) &= \lambda^2 - (a + d)\lambda + (ad - bc) \\ &= \lambda^2 - (\operatorname{tr} A)\lambda + \det A. \end{aligned}

Degree n. For an n \times n matrix, p(\lambda) is a polynomial of degree n with leading term (-1)^n\lambda^n. So A has at most n eigenvalues, and their algebraic multiplicities add to at most n; counting complex roots, they add to exactly n.

Triangular matrices. The eigenvalues of a triangular matrix are its diagonal entries, and each one’s algebraic multiplicity is the number of times it appears on the diagonal.

Trace and determinant. For a 2 \times 2 matrix with eigenvalues \lambda_1, \lambda_2 (repeated or complex roots included),

\lambda_1 + \lambda_2 = \operatorname{tr} A, \qquad \lambda_1\lambda_2 = \det A.

For an n \times n matrix the same holds: the n roots of p, listed with multiplicity, add to \operatorname{tr} A and multiply to \det A.

Why it is true

Eigenvalues are roots. Since I\mathbf{v} = \mathbf{v} and matrix products distribute over subtraction,

\begin{aligned} A\mathbf{v} = \lambda\mathbf{v} &\iff A\mathbf{v} - \lambda I\mathbf{v} = \mathbf{0} \\ &\iff (A - \lambda I)\mathbf{v} = \mathbf{0}. \end{aligned}

So \lambda is an eigenvalue exactly when \ker(A - \lambda I) holds a nonzero vector. By the square test of When a matrix or transformation can be undone, that happens exactly when A - \lambda I is not invertible, which is exactly when \det(A - \lambda I) = 0.

The 2 \times 2 formula. By Computing a determinant,

\begin{aligned} \det\begin{bmatrix} a - \lambda & b \\ c & d - \lambda \end{bmatrix} &= (a - \lambda)(d - \lambda) - bc \\ &= \lambda^2 - (a + d)\lambda + (ad - bc). \end{aligned}

Degree n, in outline. Unwinding Determinants by cofactor expansion all the way down writes \det(A - \lambda I) as a sum of signed products of n entries, one from each row and each column. Each entry is a number or a number minus \lambda, so each product has degree at most n, and degree n only when all n entries are diagonal. A product that uses n - 1 diagonal entries leaves one row and one column with the same index, so its last entry is diagonal too. Hence only the diagonal product (a_{11} - \lambda)\cdots(a_{nn} - \lambda) reaches degree n, giving the leading term (-1)^n\lambda^n; every other product has degree at most n - 2.

Triangular matrices. If A is triangular, so is A - \lambda I, with diagonal entries a_{ii} - \lambda. The determinant of a triangular matrix is the product of its diagonal (Determinants by cofactor expansion, day 2), so p(\lambda) = (a_{11} - \lambda)(a_{22} - \lambda)\cdots(a_{nn} - \lambda), whose roots are the diagonal entries.

Trace and determinant, for 2 \times 2. By the The quadratic formula, the quadratic p(\lambda) has roots \lambda_1, \lambda_2 (possibly equal, possibly complex) and factors as

\begin{aligned} p(\lambda) &= (\lambda - \lambda_1)(\lambda - \lambda_2) \\ &= \lambda^2 - (\lambda_1 + \lambda_2)\lambda + \lambda_1\lambda_2. \end{aligned}

Matching coefficients with the 2 \times 2 formula gives \lambda_1 + \lambda_2 = \operatorname{tr} A and \lambda_1\lambda_2 = \det A.

For n \times n, stated with a sketch. Over the complex numbers p(\lambda) = (\lambda_1 - \lambda)\cdots(\lambda_n - \lambda) (the fundamental theorem of algebra, taken without proof). Setting \lambda = 0 gives \det A = p(0) = \lambda_1\cdots\lambda_n. For the trace, the \lambda^{n-1} term comes only from the diagonal product, by the outline above, so comparing \lambda^{n-1} coefficients gives a_{11} + \dots + a_{nn} = \lambda_1 + \dots + \lambda_n.

Worked examples

Example 1

Day 1: The membrane’s two kept directions

Find the eigenvalues of the membrane matrix A = \begin{bmatrix} 7 & -4 \\ 2 & 1 \end{bmatrix}, and check one eigenvector for each.

Solution

Subtract \lambda down the diagonal and take the determinant:

\begin{aligned} \det(A - \lambda I) &= (7 - \lambda)(1 - \lambda) - (-4)(2) \\ &= \lambda^2 - 8\lambda + 15 \\ &= (\lambda - 3)(\lambda - 5). \end{aligned}

The eigenvalues are 3 and 5. The 2 \times 2 formula agrees: \operatorname{tr} A = 8 and \det A = 7 + 8 = 15.

As a check, A(1, 1) = (3, 3) = 3(1, 1) and A(2, 1) = (10, 5) = 5(2, 1). The membrane triples every arrow along (1, 1) and makes every arrow along (2, 1) five times as long. How such vectors are found is the next lesson, Finding an eigenvector.

A number that is not a root gets no direction: \det(A - 4I) = (3)(-3) + 8 = -1 \neq 0, so A - 4I is invertible, only \mathbf{0} solves (A - 4I)\mathbf{v} = \mathbf{0}, and no arrow is exactly quadrupled.

Example 2

Day 1: Roots that need the quadratic formula

Tuvshin resets the frame to M = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}. Find its eigenvalues.

Solution

By the 2 \times 2 formula, \operatorname{tr} M = 4 and \det M = 3 - 2 = 1, so p(\lambda) = \lambda^2 - 4\lambda + 1. This has no integer roots. The quadratic formula, \displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} for a\lambda^2 + b\lambda + c, gives

\lambda = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}.

These are about 3.73 and 0.27: one direction is stretched, another shrunk. Check: the sum is 4 = \operatorname{tr} M and the product is \displaystyle (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 = \det M. One eigenvector as a check: with \displaystyle \mathbf{v} = (1 + \sqrt{3},\ 1),

\begin{aligned} M\mathbf{v} &= (5 + 3\sqrt{3},\ 2 + \sqrt{3}), \\ (2 + \sqrt{3})\mathbf{v} &= (2 + 3\sqrt{3} + 3,\ 2 + \sqrt{3}), \end{aligned}

and the two agree.

Example 3

Day 1: One eigenvalue counted twice

Odval tries N = \begin{bmatrix} 6 & -1 \\ 4 & 2 \end{bmatrix}. Find its eigenvalues and their algebraic multiplicities.

Solution

\operatorname{tr} N = 8 and \det N = 12 + 4 = 16, so

p(\lambda) = \lambda^2 - 8\lambda + 16 = (\lambda - 4)^2.

The only eigenvalue is 4, with algebraic multiplicity 2. The discriminant 8^2 - 4 \cdot 16 is 0, which is why the two roots coincide; in general the discriminant of p is (\operatorname{tr} A)^2 - 4\det A. As a check, N(1, 2) = (4, 8) = 4(1, 2).

Note that N has the same trace as the membrane matrix A of Example 1 but a larger determinant, and that alone pushed the two roots together. The multiplicity counts a repeated root; whether the root brings one kept direction or a whole plane of them is decided in Finding an eigenvector.

Example 4

Day 1: A setting that keeps no direction

The frame can also twist. Show that C = \begin{bmatrix} 1 & -5 \\ 1 & -3 \end{bmatrix} has no real eigenvalue.

Solution

\operatorname{tr} C = -2 and \det C = -3 + 5 = 2, so

p(\lambda) = \lambda^2 + 2\lambda + 2.

Its discriminant is 2^2 - 4 \cdot 2 = -4 < 0, so p has no real root, and no real \lambda makes C - \lambda I singular. Every arrow on the sheet is turned. The quadratic formula still returns the complex roots \lambda = -1 \pm i, and Complex eigenvalues and rotation-scaling matrices reads them as a rotation combined with a scaling. A quarter-turn rotation of the frame, from Finding the matrix of a transformation, behaves the same way: its characteristic polynomial is \lambda^2 + 1.

Example 5

Day 2: A rubber block, expanded along a row

Sarnai stretches a block of the same rubber in space by B = \begin{bmatrix} 3 & 1 & 1 \\ 0 & 2 & 0 \\ 1 & 1 & 3 \end{bmatrix}. Find the characteristic polynomial, the eigenvalues and their algebraic multiplicities.

Solution

Row 2 of B - \lambda I is (0,\ 2 - \lambda,\ 0), so Determinants by cofactor expansion along row 2 has a single term, with sign (-1)^{2+2} = +1:

\begin{aligned} \det(B - \lambda I) &= (2 - \lambda)\det\begin{bmatrix} 3 - \lambda & 1 \\ 1 & 3 - \lambda \end{bmatrix} \\ &= (2 - \lambda)\big((3 - \lambda)^2 - 1\big). \end{aligned}

The bracket is a difference of squares, (3 - \lambda - 1)(3 - \lambda + 1) = (2 - \lambda)(4 - \lambda), so

\begin{aligned} p(\lambda) &= (2 - \lambda)^2(4 - \lambda) \\ &= -\lambda^3 + 8\lambda^2 - 20\lambda + 16. \end{aligned}

The eigenvalues are 2, with algebraic multiplicity 2, and 4, with multiplicity 1. The multiplicities add to 3, the size of B, and the leading term is (-1)^3\lambda^3, as the degree result says.

The other results check out. The eigenvalues with multiplicity add to 2 + 2 + 4 = 8 = 3 + 2 + 3 = \operatorname{tr} B, and they multiply to 16 = p(0) = \det B. One eigenvector each: B(1, 0, 1) = (4, 0, 4) and B(1, -1, 0) = (2, -2, 0).

Example 6

Day 2: A triangular matrix needs no expansion

Find the eigenvalues of

L = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 3 & -2 & 0 & 0 \\ 5 & 0 & 1 & 0 \\ 2 & 7 & 4 & 6 \end{bmatrix}.

Solution

L is lower triangular, so L - \lambda I is too, and its determinant is the product of its diagonal:

p(\lambda) = (1 - \lambda)^2(-2 - \lambda)(6 - \lambda).

The eigenvalues are 1 (algebraic multiplicity 2), -2 and 6. The entries below the diagonal play no part. Check: 1 + 1 - 2 + 6 = 6 = \operatorname{tr} L, and the product 1 \cdot 1 \cdot (-2) \cdot 6 = -12 is \det L. One eigenvector is easy to see: the last column of L is 6\mathbf{e}_4, so L\mathbf{e}_4 = 6\mathbf{e}_4.

Example 7

Day 2: Tuning the frame to a chosen stretch

Odval wants the setting K = \begin{bmatrix} k & 2 \\ 3 & 1 \end{bmatrix} to make some direction exactly four times as long. Find k, then the other eigenvalue without solving a second quadratic.

Solution

4 must be a root of the characteristic equation:

\begin{aligned} \det(K - 4I) &= (k - 4)(1 - 4) - 2 \cdot 3 \\ &= -3k + 6 = 0, \end{aligned}

so k = 2. Now \operatorname{tr} K = 2 + 1 = 3, and the eigenvalues add to the trace, so the other one is 3 - 4 = -1. The determinant confirms it: \det K = 2 - 6 = -4 = 4 \cdot (-1). In full, p(\lambda) = \lambda^2 - 3\lambda - 4 = (\lambda - 4)(\lambda + 1).

The eigenvalue -1 reverses a direction without changing lengths. Checks: K(1, 1) = (4, 4), and K(2, -3) = (4 - 6,\ 6 - 3), which is (-2, 3) = -1 \cdot (2, -3).

Boundaries

Square matrices only. A\mathbf{v} and \lambda\mathbf{v} must lie in the same space, so a matrix that is not square has no eigenvalues, and \det(A - \lambda I) is not even defined.
The vector is nonzero; the number may be zero. \lambda = 0 is an eigenvalue exactly when A is singular (Exercise 20).
Real eigenvalues may not exist. A real 2 \times 2 matrix can have none (Example 4). Its complex roots are eigenvalues for complex vectors, the subject of Complex eigenvalues and rotation-scaling matrices. A real matrix of odd size always has at least one real eigenvalue (Exercise 22).
Row operations change eigenvalues. Row reduction keeps the kernel and keeps whether the determinant is zero, but not the eigenvalues. Subtracting \tfrac{2}{7} of row 1 from row 2 of the membrane matrix gives a triangular matrix with diagonal 7 and \tfrac{15}{7}, while the eigenvalues of A are 3 and 5.
The triangular shortcut needs a triangular matrix. The diagonal of B in Example 5 is 3, 2, 3, yet its eigenvalues are 2, 2, 4; only the sum agrees, because both add to the trace.
Trace and determinant pin down only two eigenvalues. For n \geq 3 two numbers cannot fix n roots, and at no size do they say which directions are kept.
Multiplicity counts roots, not directions. The number of independent eigenvectors for a repeated eigenvalue can be smaller than its algebraic multiplicity; that count is the business of Finding an eigenvector.

Where this goes next

Each eigenvalue is a number \lambda that makes A - \lambda I singular, so its kept directions are the nonzero vectors in \ker(A - \lambda I); Finding an eigenvector solves that system. When an n \times n matrix has enough independent eigenvectors, Diagonalization rewrites it as plain scaling along them. A negative discriminant leads to Complex eigenvalues and rotation-scaling matrices, where rotation enters. And powers A^k\mathbf{v} = \lambda^k\mathbf{v} along an eigenvector are what Matrix powers, dynamical systems and steady states uses to see where a repeated process settles.

Exercises

Day 1: eigenvalues of 2 by 2 matrices

Find the eigenvalues of \begin{bmatrix} 6 & 2 \\ 2 & 3 \end{bmatrix}.
Find the eigenvalues of \begin{bmatrix} 1 & 6 \\ 1 & 0 \end{bmatrix}.
Find the eigenvalues of \begin{bmatrix} 5 & 2 \\ -2 & 1 \end{bmatrix} and their algebraic multiplicities.
Find the eigenvalues of \begin{bmatrix} 2 & 5 \\ 1 & -2 \end{bmatrix}.
Find the eigenvalues of \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix} exactly.
Show that \begin{bmatrix} 3 & -5 \\ 2 & -1 \end{bmatrix} has no real eigenvalue, and give its complex roots.
Let P = \begin{bmatrix} 5 & -2 \\ 3 & 0 \end{bmatrix}. Decide whether (2, 3) and (1, 2) are eigenvectors of P, then find both eigenvalues of P.
Without finding all the eigenvalues first, decide whether 5 and 2 are eigenvalues of A = \begin{bmatrix} 3 & 4 \\ 1 & 3 \end{bmatrix}. Then find both eigenvalues.
For which numbers k does \begin{bmatrix} 1 & k \\ 1 & 1 \end{bmatrix} have real eigenvalues? For which k is there one repeated eigenvalue?
(a) A 2 \times 2 matrix has trace 7 and determinant 12. Find its eigenvalues. (b) Another has eigenvalue 2 and determinant 6. Find its other eigenvalue and its trace.
A 2 \times 2 matrix A sends (1, 1) to (-2, -2) and (1, -1) to (3, -3). Find its eigenvalues, trace and determinant without finding A. Then find A and confirm.

Day 2: larger matrices, triangular matrices, trace and determinant

Find the eigenvalues of \begin{bmatrix} 5 & 0 & 0 \\ 1 & 2 & 1 \\ 3 & 1 & 2 \end{bmatrix} by expanding along row 1.
Find the eigenvalues of \begin{bmatrix} 1 & 2 & 0 \\ 3 & 2 & 0 \\ 1 & 1 & 4 \end{bmatrix} and their algebraic multiplicities.
Find the eigenvalues of \begin{bmatrix} 3 & 7 & -2 \\ 0 & -1 & 5 \\ 0 & 0 & 3 \end{bmatrix}, and use them to give its trace and determinant.
A matrix has characteristic polynomial p(\lambda) = (\lambda - 2)^3(\lambda + 1)^2(\lambda - 6). What size is it? Give its eigenvalues with their algebraic multiplicities, its trace and its determinant.
Find the eigenvalues of \begin{bmatrix} 4 & 1 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 2 \end{bmatrix} and confirm the trace and determinant results.
For which k is 0 an eigenvalue of \begin{bmatrix} 2 & k \\ 3 & 6 \end{bmatrix}? For that k, find the other eigenvalue.
Show that G = \begin{bmatrix} 0 & 0 & 6 \\ 1 & 0 & -11 \\ 0 & 1 & 6 \end{bmatrix} has p(\lambda) = -\lambda^3 + 6\lambda^2 - 11\lambda + 6, and find its eigenvalues.

Reasoning

Show that A and its transpose A^{T} have the same characteristic polynomial, so the same eigenvalues.
Show that 0 is an eigenvalue of a square matrix A exactly when A is singular.
Suppose A\mathbf{v} = \lambda\mathbf{v} with \mathbf{v} \neq \mathbf{0}. Show that \mathbf{v} is an eigenvector of A + cI for \lambda + c, and, if A is invertible, an eigenvector of A^{-1} for \tfrac{1}{\lambda}.
Explain why every real 3 \times 3 matrix has at least one real eigenvalue.
Show that if A^2 = A, then every eigenvalue of A is 0 or 1.
Show that a real 2 \times 2 matrix \begin{bmatrix} a & b \\ c & d \end{bmatrix} with bc \geq 0 has real eigenvalues. Conclude that every real symmetric 2 \times 2 matrix does.

Answers

2 and 7. Trace 9, determinant 18 - 4 = 14, so p(\lambda) = \lambda^2 - 9\lambda + 14 = (\lambda - 2)(\lambda - 7).
3 and -2. Trace 1, determinant -6: p(\lambda) = \lambda^2 - \lambda - 6 = (\lambda - 3)(\lambda + 2).
3, with algebraic multiplicity 2. Trace 6, determinant 5 + 4 = 9: p(\lambda) = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2.
3 and -3. Trace 0, determinant -4 - 5 = -9: p(\lambda) = \lambda^2 - 9.
\displaystyle \lambda = \frac{5 \pm \sqrt{21}}{2}. Trace 5, determinant 4 - 3 = 1: p(\lambda) = \lambda^2 - 5\lambda + 1, and the quadratic formula gives \displaystyle \frac{5 \pm \sqrt{25 - 4}}{2}.
Trace 2, determinant -3 + 10 = 7: p(\lambda) = \lambda^2 - 2\lambda + 7, with discriminant 4 - 28 = -24 < 0, so no real root. The complex roots are \displaystyle 1 \pm i\sqrt{6}.
P(2, 3) = (10 - 6,\ 6) = (4, 6) = 2(2, 3): an eigenvector for 2. P(1, 2) = (1, 3), not a multiple of (1, 2): not an eigenvector. The eigenvalues are 2 and 3, from p(\lambda) = \lambda^2 - 5\lambda + 6.
\det(A - 5I) = (-2)(-2) - 4 = 0, so 5 is an eigenvalue. \det(A - 2I) = 1 \cdot 1 - 4 = -3 \neq 0, so 2 is not. Trace 6 and determinant 5 give p(\lambda) = (\lambda - 1)(\lambda - 5): the eigenvalues are 1 and 5.
p(\lambda) = (1 - \lambda)^2 - k, so \displaystyle \lambda = 1 \pm \sqrt{k}. Real eigenvalues exactly when k \geq 0; one repeated eigenvalue, 1, when k = 0. For k < 0 the roots are the complex numbers \displaystyle 1 \pm i\sqrt{-k}.
(a) p(\lambda) = \lambda^2 - 7\lambda + 12 = (\lambda - 3)(\lambda - 4): eigenvalues 3 and 4. (b) The product is 6, so the other eigenvalue is 3, and the trace is 2 + 3 = 5.
A(1, 1) = -2(1, 1) and A(1, -1) = 3(1, -1), so -2 and 3 are eigenvalues, and a 2 \times 2 matrix has no others. Trace -2 + 3 = 1, determinant -6. With A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, the conditions read a + b = -2, c + d = -2, a - b = 3, c - d = -3, so A = \begin{bmatrix} \tfrac12 & -\tfrac52 \\ -\tfrac52 & \tfrac12 \end{bmatrix}, with trace 1 and determinant \tfrac14 - \tfrac{25}{4} = -6.
1, 3 and 5. Row 1 of the shifted matrix is (5 - \lambda,\ 0,\ 0), so p(\lambda) = (5 - \lambda)\big((2 - \lambda)^2 - 1\big), which is (5 - \lambda)(1 - \lambda)(3 - \lambda).
4 with algebraic multiplicity 2, and -1. Expand along column 3, whose only nonzero entry is 4 - \lambda: p(\lambda) = (4 - \lambda)\big((1 - \lambda)(2 - \lambda) - 6\big), which is (4 - \lambda)(\lambda^2 - 3\lambda - 4), or (4 - \lambda)^2(-1 - \lambda).
Upper triangular, so the eigenvalues are 3 (algebraic multiplicity 2) and -1. Trace 3 - 1 + 3 = 5; determinant 3 \cdot (-1) \cdot 3 = -9.
Degree 6, so 6 \times 6. Eigenvalues 2 (multiplicity 3), -1 (multiplicity 2), 6 (multiplicity 1). Trace 3 \cdot 2 + 2 \cdot (-1) + 6 = 10; determinant p(0) = (-2)^3 \cdot 1^2 \cdot (-6) = 48, the same as 2^3 \cdot (-1)^2 \cdot 6.
Expand along row 3: p(\lambda) = (2 - \lambda)\big((4 - \lambda)^2 - 1\big), which is (2 - \lambda)(3 - \lambda)(5 - \lambda). Eigenvalues 2, 3, 5. They add to 10 = 4 + 4 + 2 and multiply to 30 = 2(16 - 1), the determinant.
k = 4. 0 is an eigenvalue exactly when \det = 12 - 3k = 0. Then the trace is 8, so the other eigenvalue is 8 - 0 = 8.
Expand \det(G - \lambda I) along row 1, (-\lambda, 0, 6): -\lambda\big(-\lambda(6 - \lambda) + 11\big) + 6 \cdot 1, which expands to -\lambda^3 + 6\lambda^2 - 11\lambda + 6. This is -(\lambda - 1)(\lambda - 2)(\lambda - 3), so the eigenvalues are 1, 2 and 3; their sum 6 is \operatorname{tr} G and their product 6 is \det G = p(0).
(A - \lambda I)^{T} = A^{T} - \lambda I, since I^{T} = I, and a matrix and its transpose have the same determinant (Properties of determinants: row operations, products, transposes). So \det(A^{T} - \lambda I) = \det(A - \lambda I) for every \lambda.
0 is an eigenvalue exactly when \det(A - 0I) = \det A = 0, which is exactly when A is singular. Equivalently, A\mathbf{v} = 0\mathbf{v} = \mathbf{0} for a nonzero \mathbf{v} means \ker A \neq \{\mathbf{0}\}.
(A + cI)\mathbf{v} = A\mathbf{v} + c\mathbf{v} = (\lambda + c)\mathbf{v}. If A is invertible, \lambda \neq 0 by Exercise 20; applying A^{-1} to A\mathbf{v} = \lambda\mathbf{v} gives \mathbf{v} = \lambda A^{-1}\mathbf{v}, so A^{-1}\mathbf{v} = \tfrac{1}{\lambda}\mathbf{v}.
p(\lambda) is a cubic with real coefficients and leading term -\lambda^3. It is positive for large negative \lambda and negative for large positive \lambda, and a polynomial is continuous, so it crosses 0 somewhere: that crossing is a real eigenvalue.
If A\mathbf{v} = \lambda\mathbf{v} with \mathbf{v} \neq \mathbf{0}, then A^2\mathbf{v} = A(\lambda\mathbf{v}) = \lambda^2\mathbf{v}. Since A^2 = A, \lambda^2\mathbf{v} = \lambda\mathbf{v}, so (\lambda^2 - \lambda)\mathbf{v} = \mathbf{0}, and as \mathbf{v} \neq \mathbf{0}, \lambda(\lambda - 1) = 0.
The discriminant of p(\lambda) = \lambda^2 - (a + d)\lambda + (ad - bc) is (a + d)^2 - 4(ad - bc) = (a - d)^2 + 4bc, a square plus a number that is not negative, so it is at least 0 and both roots are real. A symmetric matrix has c = b, so bc = b^2 \geq 0.