Linear Algebra

Diagonalization

Most matrices turn almost every arrow they act on. When can a matrix be redescribed, in a well-chosen basis, as nothing but a separate stretch along each axis, and what does that make easy?

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Finding an eigenvectorLinear Algebra Coordinates in a basis and change of basisLinear Algebra

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Matrix powers, dynamical systems and steady statesLinear 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

Malika, Bakyt and Mihi make mosaics. A border strip one unit wide and n units long is laid from white square tiles (1 by 1) and dominoes (1 by 2) that come in two colors, blue and gold. Malika wants to know how many different strips of each length there are. Sorting the strips by their last piece gives a rule: a strip ending in a square is a strip of length n - 1 plus that square, and a strip ending in a domino is a strip of length n - 2 plus one of two dominoes (Setting up a recurrence from a counting problem). So

a_n = a_{n-1} + 2a_{n-2}, \qquad a_1 = 1, \quad a_2 = 3.

The rule is linear, so one step is a matrix: (a_{n+1}, a_n) = M(a_n, a_{n-1}) with M = \begin{bmatrix} 1 & 2 \\ 1 & 0 \end{bmatrix}, and a strip of length 30 needs M^{28} applied to (a_2, a_1). Powers of a general matrix show no pattern; powers of a diagonal matrix are the powers of its diagonal entries.

This lesson finds when a matrix is a diagonal matrix in disguise, seen through a basis of eigenvectors. Day 1 proves the diagonalization theorem and reads it as a change of basis; day 2 handles repeated eigenvalues, powers, and the strip count.

Definition

A square matrix D is diagonal when every entry off its main diagonal is 0; write D = \operatorname{diag}(\lambda_1, \dots, \lambda_n).

An n \times n matrix A is diagonalizable when there are an invertible n \times n matrix P and a diagonal matrix D with

A = PDP^{-1}.

Writing A this way is a diagonalization of A.

Two n \times n matrices A and B are similar when B = P^{-1}AP for some invertible P. Then also A = PBP^{-1}, so the relation runs both ways. A matrix is diagonalizable exactly when it is similar to a diagonal matrix. Throughout, P and D are real unless complex numbers are named.

Intuition

Take A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}. Its eigenvalues are 5 and -1, with eigenvectors \mathbf{v}_1 = (1, 1) and \mathbf{v}_2 = (-2, 1) (Example 1). These are independent, so they form a basis \mathcal{B} of \mathbb{R}^2, and every \mathbf{x} is c_1\mathbf{v}_1 + c_2\mathbf{v}_2 in exactly one way. Linearity does the rest:

A\mathbf{x} = c_1A\mathbf{v}_1 + c_2A\mathbf{v}_2 = 5c_1\mathbf{v}_1 - c_2\mathbf{v}_2.

In the coordinates of \mathcal{B}, the map sends (c_1, c_2) to (5c_1, -c_2): stretch the first coordinate by 5, flip the second. That is D = \operatorname{diag}(5, -1) at work. The figure shows it for \mathbf{x} = (-1, 2), whose coordinates are (1, 1).

Figure 1For A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}, the vector \mathbf{x} = \mathbf{v}_1 + \mathbf{v}_2 has eigenvector coordinates (1, 1). The map multiplies the first coordinate by 5 and the second by -1, so A\mathbf{x} = 5\mathbf{v}_1 - \mathbf{v}_2.

In the slanted grid of eigenvectors, A only stretches along each grid line. The formula A = PDP^{-1} says this in three moves: P^{-1} converts to \mathcal{B}-coordinates, D scales each one, and P converts back (Coordinates in a basis and change of basis).

Results

1. The diagonalization theorem. An n \times n matrix A is diagonalizable exactly when A has n linearly independent eigenvectors. In that case A = PDP^{-1} with the eigenvectors as the columns of P, and the diagonal entries of D are their eigenvalues, in the same order.

2. Distinct eigenvalues suffice. An n \times n matrix with n distinct real eigenvalues is diagonalizable.

3. The multiplicity test. Let \lambda_1, \dots, \lambda_r be the distinct real eigenvalues of A. Then A is diagonalizable exactly when \dim E_{\lambda_1} + \dots + \dim E_{\lambda_r} = n. Equivalently: every root of the characteristic polynomial is real, and each eigenvalue’s geometric multiplicity equals its algebraic multiplicity.

4. Similarity. Similar matrices have the same characteristic polynomial, so the same eigenvalues with the same algebraic multiplicities, the same trace and the same determinant. If \mathcal{B} is a basis of eigenvectors and P = P_{\mathcal{B}}, then

[A\mathbf{x}]_{\mathcal{B}} = D[\mathbf{x}]_{\mathcal{B}} \quad \text{for every } \mathbf{x}:

D is A seen in the eigenvector basis.

5. Powers. If A = PDP^{-1} with D = \operatorname{diag}(\lambda_1, \dots, \lambda_n), then for every whole number k,

A^k = PD^kP^{-1}, \qquad D^k = \operatorname{diag}(\lambda_1^k, \dots, \lambda_n^k).

Why it is true

Result 1. Let P = \begin{bmatrix} \mathbf{p}_1 & \cdots & \mathbf{p}_n \end{bmatrix} and D = \operatorname{diag}(d_1, \dots, d_n). By the column view of matrix multiplication, column i of AP is A\mathbf{p}_i, and column i of PD is P(d_i\mathbf{e}_i) = d_i\mathbf{p}_i. So

AP = PD \iff A\mathbf{p}_i = d_i\mathbf{p}_i \text{ for each } i.

Suppose A has independent eigenvectors \mathbf{p}_1, \dots, \mathbf{p}_n with eigenvalues d_1, \dots, d_n. Then AP = PD. The n columns of the square matrix P are independent, so P is invertible by the The invertible matrix theorem, and multiplying AP = PD on the right by P^{-1} gives A = PDP^{-1}.

Conversely, suppose A = PDP^{-1}. Multiplying on the right by P gives AP = PD, so A\mathbf{p}_i = d_i\mathbf{p}_i for each i. Since P is invertible, its columns are independent, and in particular none is \mathbf{0}. So they are n independent eigenvectors of A.

Result 2. Eigenvectors for distinct eigenvalues are linearly independent (Finding an eigenvector). With n distinct eigenvalues, one eigenvector each gives n independent eigenvectors, and Result 1 applies.

Result 3. Write g_i = \dim E_{\lambda_i} and let m_i be the algebraic multiplicity. From Finding an eigenvector: g_i \le m_i, and bases of the eigenspaces joined together stay independent. The characteristic polynomial has degree n, so m_1 + \dots + m_r \le n, with equality exactly when all its roots are real. If g_1 + \dots + g_r = n, the joined bases are n independent eigenvectors. If instead n independent eigenvectors exist, sort them by eigenvalue: those for \lambda_i are independent vectors in E_{\lambda_i}, so there are at most g_i of them, and n \le g_1 + \dots + g_r. So A is diagonalizable exactly when the g_i add to n. Since g_1 + \dots + g_r \le m_1 + \dots + m_r \le n, the g_i add to n exactly when both inequalities are equalities: every g_i = m_i and every root is real.

Result 4. Using P^{-1}(\lambda I)P = \lambda I and the product rule of Properties of determinants: row operations, products, transposes,

\begin{aligned} \det(P^{-1}AP - \lambda I) &= \det\big(P^{-1}(A - \lambda I)P\big) \\ &= \det(P^{-1})\det(A - \lambda I)\det P \\ &= \det(A - \lambda I). \end{aligned}

Trace and determinant are read off the characteristic polynomial (The characteristic equation and finding eigenvalues), so they agree too. For the coordinate statement, [\mathbf{x}]_{\mathcal{B}} = P^{-1}\mathbf{x}, so

[A\mathbf{x}]_{\mathcal{B}} = P^{-1}A\mathbf{x} = (P^{-1}AP)(P^{-1}\mathbf{x}) = D[\mathbf{x}]_{\mathcal{B}}.

The same computation with any basis shows that P_{\mathcal{B}}^{-1}AP_{\mathcal{B}} is the matrix of \mathbf{x} \mapsto A\mathbf{x} in \mathcal{B}-coordinates, so similar matrices are one transformation described in two bases.

Result 5. For k = 0 both sides are I. If A^k = PD^kP^{-1}, then A^{k+1} = PD^kP^{-1}PDP^{-1} = PD^{k+1}P^{-1}, because P^{-1}P = I in the middle. A product of diagonal matrices multiplies matching diagonal entries, so D^k = \operatorname{diag}(\lambda_1^k, \dots, \lambda_n^k).

Worked examples

Example 1

Day 1: Diagonalizing a 2 by 2 matrix

Diagonalize A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}.

Solution

The trace is 4 and the determinant is 3 - 8 = -5, so the The characteristic equation and finding eigenvalues is \lambda^2 - 4\lambda - 5 = (\lambda - 5)(\lambda + 1) = 0. The eigenvalues 5 and -1 are distinct, so by Result 2 A is diagonalizable.

For \lambda = 5, A - 5I = \begin{bmatrix} -4 & 4 \\ 2 & -2 \end{bmatrix}; its first row says -4x_1 + 4x_2 = 0, so E_5 = \operatorname{span}\{(1, 1)\}. For \lambda = -1, A + I = \begin{bmatrix} 2 & 4 \\ 2 & 4 \end{bmatrix}; the row says x_1 + 2x_2 = 0, so E_{-1} = \operatorname{span}\{(-2, 1)\}. Put the eigenvectors in P and their eigenvalues in D in the same order:

\begin{aligned} P &= \begin{bmatrix} 1 & -2 \\ 1 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 5 & 0 \\ 0 & -1 \end{bmatrix}, \\ P^{-1} &= \frac{1}{3}\begin{bmatrix} 1 & 2 \\ -1 & 1 \end{bmatrix}, \end{aligned}

where \det P = 1 + 2 = 3 and P^{-1} comes from Inverse of a 2 by 2 matrix, and solving with it. Check AP = PD rather than the triple product: both equal \begin{bmatrix} 5 & 2 \\ 5 & -1 \end{bmatrix}. So A = PDP^{-1}.

Listing the eigenvectors in the other order swaps the diagonal of D to match, and either column may be replaced by a nonzero multiple of itself, so P is not unique. The diagonal entries of D are fixed up to order: by Result 4 they are the eigenvalues.

Example 2

Day 1: The same map in eigenvector coordinates

With A, P and D as in Example 1, find the coordinates of \mathbf{x} = (-1, 2) in the eigenvector basis \mathcal{B} = \{(1, 1), (-2, 1)\}, use D to find A\mathbf{x}, and check by multiplying directly.

Solution

By Coordinates in a basis and change of basis, [\mathbf{x}]_{\mathcal{B}} = P^{-1}\mathbf{x} = \tfrac{1}{3}(-1 + 4,\ 1 + 2) = (1, 1), so \mathbf{x} = \mathbf{v}_1 + \mathbf{v}_2. By Result 4, [A\mathbf{x}]_{\mathcal{B}} = D(1, 1) = (5, -1), so

A\mathbf{x} = 5\begin{bmatrix} 1 \\ 1 \end{bmatrix} - \begin{bmatrix} -2 \\ 1 \end{bmatrix} = \begin{bmatrix} 7 \\ 4 \end{bmatrix}.

Directly, A(-1, 2) = (-1 + 8,\ -2 + 6) = (7, 4), as the figure shows. In \mathcal{B}-coordinates the matrix of the map is P^{-1}AP = D: one transformation, two descriptions, so A and D are similar.

Example 3

Day 2: A repeated eigenvalue that still works

Decide whether C = \begin{bmatrix} 4 & -1 & 1 \\ 0 & 3 & 0 \\ 1 & -1 & 4 \end{bmatrix} is diagonalizable, and diagonalize it if so.

Solution

Row 2 of C - \lambda I is (0,\ 3 - \lambda,\ 0), so Determinants by cofactor expansion along it has one term:

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

The eigenvalue 3 has algebraic multiplicity 2, so Result 2 does not apply, and Result 3 asks for \dim E_3. Here

C - 3I = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 1 & -1 & 1 \end{bmatrix}

has rank 1: one equation x_1 - x_2 + x_3 = 0 with two free variables, so \dim E_3 = 2, with basis (1, 1, 0) and (-1, 0, 1). For \lambda = 5, C - 5I = \begin{bmatrix} -1 & -1 & 1 \\ 0 & -2 & 0 \\ 1 & -1 & -1 \end{bmatrix} gives x_2 = 0 and x_1 = x_3: E_5 = \operatorname{span}\{(1, 0, 1)\}. The dimensions add to 2 + 1 = 3, so C is diagonalizable:

P = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}, \qquad D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix}.

Check: C(1, 1, 0) = (3, 3, 0), C(-1, 0, 1) = (-3, 0, 3) and C(1, 0, 1) = (5, 0, 5). Expanding along row 2 gives \det P = -1 \cdot \det\begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} = 2 \neq 0, as the independence result promised. Geometrically, C stretches a whole plane by 3 and the line through (1, 0, 1) by 5.

Example 4

Day 2: A repeated eigenvalue that does not

Show that F = \begin{bmatrix} 3 & 0 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} is not diagonalizable.

Solution

Expand \det(F - \lambda I) along row 1, whose only nonzero entry is 3 - \lambda:

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

As for C, the eigenvalue 3 has algebraic multiplicity 2. But

F - 3I = \begin{bmatrix} 0 & 0 & 0 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}.

Subtracting row 2 from row 3 leaves (0, 2, -2), so the rank is 2: x_2 = x_3, then x_1 = x_2 - x_3 = 0, and E_3 = \operatorname{span}\{(0, 1, 1)\} has dimension 1. The eigenvalue 1 is simple, so \dim E_1 = 1. The dimensions add to 1 + 1 = 2 < 3, and by Result 3 no basis of \mathbb{R}^3 is made of eigenvectors of F. The characteristic polynomial of F has the same shape as that of C; only the eigenspace tells them apart.

Example 5

Day 2: A power in closed form

For A = \begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix} of Example 1, find a formula for A^k and use it to find A^4.

Solution

By Result 5, A^k = PD^kP^{-1} with D^k = \operatorname{diag}(5^k, (-1)^k). First

PD^k = \begin{bmatrix} 5^k & -2(-1)^k \\ 5^k & (-1)^k \end{bmatrix},

since multiplying by a diagonal matrix on the right scales the columns. Then multiply by P^{-1} = \tfrac13\begin{bmatrix} 1 & 2 \\ -1 & 1 \end{bmatrix}:

\begin{aligned} A^k &= \frac{5^k}{3}\begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} + \frac{(-1)^k}{3}\begin{bmatrix} 2 & -2 \\ -1 & 1 \end{bmatrix}. \end{aligned}

Each eigenvalue’s power multiplies a fixed matrix, an eigenvector column times a row of P^{-1}.

At k = 0 this is I, and at k = 1 it is \tfrac13\begin{bmatrix} 3 & 12 \\ 6 & 9 \end{bmatrix} = A. With 5^4 = 625 and (-1)^4 = 1,

A^4 = \frac{1}{3}\begin{bmatrix} 627 & 1248 \\ 624 & 1251 \end{bmatrix} = \begin{bmatrix} 209 & 416 \\ 208 & 417 \end{bmatrix}.

Check: A^2 = \begin{bmatrix} 9 & 16 \\ 8 & 17 \end{bmatrix}, and A^2A^2 has top row (81 + 128,\ 144 + 272) = (209, 416). The entries grow like 5^k; the long run is the subject of Matrix powers, dynamical systems and steady states.

Example 6

Day 2: Counting mosaic strips

Find a closed formula for Malika’s strip counts a_n = a_{n-1} + 2a_{n-2}, a_1 = 1, a_2 = 3, check it against a list of the short strips, and find the number of strips of length 10.

Solution

Write S for a square, B and G for a blue or gold domino. Length 1: S. Length 2: SS, B, G. Length 3: SSS, SB, SG, BS, GS. So a_1 = 1, a_2 = 3, a_3 = 5, and indeed a_3 = a_2 + 2a_1. Setting a_0 = 1 (one empty strip) makes the rule hold at n = 2 as well: 3 = 1 + 2 \cdot 1.

Let \mathbf{u}_n = (a_{n+1}, a_n). The rule says \mathbf{u}_n = M\mathbf{u}_{n-1} with M = \begin{bmatrix} 1 & 2 \\ 1 & 0 \end{bmatrix}, so \mathbf{u}_n = M^n\mathbf{u}_0 with \mathbf{u}_0 = (1, 1). The trace of M is 1 and its determinant -2, so \lambda^2 - \lambda - 2 = (\lambda - 2)(\lambda + 1), with eigenvectors (2, 1) for 2 and (1, -1) for -1. Rather than form M^n, write \mathbf{u}_0 in the eigenvector basis: solving c_1(2, 1) + c_2(1, -1) = (1, 1) gives c_1 = \tfrac23, c_2 = -\tfrac13. Each eigenvector is only scaled by M, so

\mathbf{u}_n = \tfrac23 \cdot 2^n\begin{bmatrix} 2 \\ 1 \end{bmatrix} - \tfrac13(-1)^n\begin{bmatrix} 1 \\ -1 \end{bmatrix}.

The second entry is a_n:

a_n = \frac{2^{n+1} + (-1)^n}{3}.

Check: n = 1, 2, 3 give \tfrac{4 - 1}{3} = 1, \tfrac{8 + 1}{3} = 3 and \tfrac{16 - 1}{3} = 5, matching the list. For length 10, a_{10} = \tfrac{2048 + 1}{3} = 683, and the rule confirms it step by step: 5, 11, 21, 43, 85, 171, 341, 683. The same roots come out of the power-guess method of Solving and checking a two-term recurrence, whose quadratic is the characteristic polynomial of M.

Boundaries

Not every matrix is diagonalizable. The shear \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} has the single eigenvalue 1 with a one-dimensional eigenspace, and F in Example 4 fails the same way.
Real or complex. The quarter-turn \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} has no real eigenvector, so no real P and D exist (Exercise 22). With complex entries it can be diagonalized; Complex eigenvalues and rotation-scaling matrices takes that route.
Distinct eigenvalues are enough, not necessary. The identity has one eigenvalue and is already diagonal; C in Example 3 has a repeated eigenvalue and is diagonalizable.
Diagonalizable and invertible are separate questions. \operatorname{diag}(0, 1) is diagonalizable and singular; the shear above is invertible and not diagonalizable.
The order matters. The columns of P are eigenvectors and A = PDP^{-1}; the matrix P^{-1}DP is in general a different matrix.
Negative powers. A^k = PD^kP^{-1} holds for negative k exactly when A is invertible (Exercise 17).
Symmetric matrices. Every real symmetric matrix is diagonalizable, even with a P whose columns are perpendicular unit vectors; this is the spectral theorem, stated here without proof (The transpose and symmetric matrices).

Where this goes next

A diagonalization turns repeated application of a matrix into powers of numbers, which is how Matrix powers, dynamical systems and steady states reads the long run of a process. When the eigenvalues are complex, Complex eigenvalues and rotation-scaling matrices finds the next best thing, a similar matrix that rotates and scales.

Exercises

Day 1: diagonalizing and similarity

Diagonalize \begin{bmatrix} 5 & -2 \\ 1 & 2 \end{bmatrix}.
Diagonalize \begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}.
Diagonalize \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix}.
Decide whether \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} is diagonalizable.
Without finding any eigenvector, explain why L = \begin{bmatrix} 2 & 0 & 0 \\ 1 & -1 & 0 \\ 3 & 2 & 4 \end{bmatrix} is diagonalizable, and give a diagonal matrix similar to it.
Find the 2 \times 2 matrix that has (1, 2) as an eigenvector for 3 and (1, 1) as an eigenvector for -1.
For the matrix A of Exercise 1 and the basis \mathcal{B} = \{(1, 1), (2, 1)\}, find [\mathbf{x}]_{\mathcal{B}} for \mathbf{x} = (3, 2), use D to find [A\mathbf{x}]_{\mathcal{B}}, then find A\mathbf{x} and check it directly.
(a) Explain why \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} and \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix} are not similar. (b) Show that \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} and \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} are similar, and find an invertible Q with \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} = Q\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}Q^{-1}.
For the matrix of Exercise 1, give (a) a diagonalization with the eigenvalues in the other order, and (b) a diagonalization with the same D as your first answer but a different P.

Day 2: repeated eigenvalues, powers and recurrences

Diagonalize \begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}.
Show that \begin{bmatrix} 2 & 1 & 1 \\ 0 & 3 & 0 \\ 0 & -1 & 2 \end{bmatrix} is not diagonalizable.
For which numbers a is \begin{bmatrix} a & 1 \\ 0 & 2 \end{bmatrix} diagonalizable?
For the matrix A of Exercise 1, find a formula for A^k, and use it to find A^3.
For A = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} of Exercise 3 and \mathbf{x} = (2, 3), find A^k\mathbf{x} and A^{10}\mathbf{x}.
Bakyt’s strips use one kind of square and dominoes in six colors, so b_n = b_{n-1} + 6b_{n-2} with b_1 = 1, b_2 = 7. Write the step as a matrix, diagonalize it, and find a closed formula for b_n and the value b_8.
Mihi’s strips use squares in two colors and dominoes in three, so c_n = 2c_{n-1} + 3c_{n-2} with c_1 = 2, c_2 = 7. Find a closed formula for c_n and the value c_6.

Reasoning

Suppose A = PDP^{-1} and A is invertible. Show that no diagonal entry of D is 0, and that A^{-1} = PD^{-1}P^{-1}.
Show that similar matrices have the same determinant and the same trace. Use this to show that \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} and \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} are not similar.
Show that a diagonalizable n \times n matrix whose only eigenvalue is \lambda equals \lambda I. Conclude that \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} is not diagonalizable.
Show that if A is diagonalizable, then so are A^{T} and A + cI for every number c.
Show that if A is diagonalizable and every eigenvalue of A is 0 or 1, then A^2 = A.
Explain why \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} cannot be written as PDP^{-1} with P and D real.

Answers

P = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}, D = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}. Trace 7, determinant 12: (\lambda - 3)(\lambda - 4). A - 3I has row (2, -2), giving (1, 1); A - 4I has row (1, -2), giving (2, 1).
P = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, D = \begin{bmatrix} 6 & 0 \\ 0 & -2 \end{bmatrix}. Trace 4, determinant 4 - 16 = -12: (\lambda - 6)(\lambda + 2). The rows (-4, 4) of A - 6I and (4, 4) of A + 2I give (1, 1) and (1, -1).
P = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}, D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}. Trace 3, determinant 2: (\lambda - 1)(\lambda - 2). The rows (-1, 1) of A - I and (-2, 1) of A - 2I give (1, 1) and (1, 2).
Not diagonalizable. Trace 6, determinant 9: (\lambda - 3)^2. A - 3I = \begin{bmatrix} 1 & 1 \\ -1 & -1 \end{bmatrix} has rank 1, so \dim E_3 = 1 < 2.
L is lower triangular, so its eigenvalues are 2, -1 and 4, distinct; by Result 2 it is diagonalizable, and it is similar to \operatorname{diag}(2, -1, 4).
\begin{bmatrix} -5 & 4 \\ -8 & 7 \end{bmatrix}, as PDP^{-1} with P = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}, D = \operatorname{diag}(3, -1), P^{-1} = \begin{bmatrix} -1 & 1 \\ 2 & -1 \end{bmatrix}. Check: (-5 + 8,\ -8 + 14) = (3, 6).
[\mathbf{x}]_{\mathcal{B}} = (1, 1), since (1, 1) + (2, 1) = (3, 2). Then [A\mathbf{x}]_{\mathcal{B}} = (3, 4) and A\mathbf{x} = 3(1, 1) + 4(2, 1) = (11, 7). Directly, (15 - 4,\ 3 + 4) = (11, 7).
(a) Their traces are 4 and 5, and similar matrices have equal traces. (b) Both have the distinct eigenvalues 1 and 3, so both equal P_iDP_i^{-1} with D = \operatorname{diag}(1, 3), P_1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} and P_2 = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}. Then D = P_2^{-1}(\text{second})P_2, so Q = P_1P_2^{-1} = \begin{bmatrix} 1 & 0 \\ \tfrac12 & \tfrac12 \end{bmatrix}.
(a) P = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, D = \operatorname{diag}(4, 3). (b) Scale the columns, for example P = \begin{bmatrix} -1 & 6 \\ -1 & 3 \end{bmatrix} with D = \operatorname{diag}(3, 4).
P = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix}, D = \operatorname{diag}(3, 3, 1). Eigenvalues 1, 3, 3 from the diagonal. A - 3I has rank 1 and says x_1 = 0, so E_3 has basis (0, 1, 0), (0, 0, 1). A - I gives x_3 = 0, x_1 = -x_2: (1, -1, 0).
Expanding along column 1, p(\lambda) = (2 - \lambda)^2(3 - \lambda). A - 2I = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & -1 & 0 \end{bmatrix} has rank 2, so \dim E_2 = 1 < 2, and the dimensions add to 2 < 3.
Every a except a = 2. For a \neq 2 the eigenvalues a and 2 are distinct. For a = 2, A - 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} has rank 1, so \dim E_2 = 1 < 2.
A^k = 3^k\begin{bmatrix} -1 & 2 \\ -1 & 2 \end{bmatrix} + 4^k\begin{bmatrix} 2 & -2 \\ 1 & -1 \end{bmatrix}, from PD^k times P^{-1} = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}. With 3^3 = 27 and 4^3 = 64, A^3 = \begin{bmatrix} 101 & -74 \\ 37 & -10 \end{bmatrix}.
\mathbf{x} = (1, 1) + (1, 2), eigenvectors for 1 and 2, so A^k\mathbf{x} = (1, 1) + 2^k(1, 2) = (1 + 2^k,\ 1 + 2^{k+1}), and A^{10}\mathbf{x} = (1{,}025,\ 2{,}049).
(b_{n+1}, b_n) = N(b_n, b_{n-1}) with N = \begin{bmatrix} 1 & 6 \\ 1 & 0 \end{bmatrix} and b_0 = 1. N = PDP^{-1} with P = \begin{bmatrix} 3 & -2 \\ 1 & 1 \end{bmatrix}, D = \operatorname{diag}(3, -2), from \lambda^2 - \lambda - 6. Since (1, 1) = \tfrac35(3, 1) + \tfrac25(-2, 1), \displaystyle b_n = \frac{3^{n+1} + 2(-2)^n}{5}, and \displaystyle b_8 = \frac{19{,}683 + 512}{5} = 4{,}039.
\displaystyle c_n = \frac{3^{n+1} + (-1)^n}{4} and c_6 = 547. The matrix \begin{bmatrix} 2 & 3 \\ 1 & 0 \end{bmatrix} has eigenvalues 3 and -1, eigenvectors (3, 1) and (1, -1), and with c_0 = 1, (c_1, c_0) = (2, 1) = \tfrac34(3, 1) - \tfrac14(1, -1). Then c_6 = \tfrac{2187 + 1}{4}.
The diagonal entries of D are eigenvalues of A, and 0 is not an eigenvalue of an invertible matrix. So D^{-1} = \operatorname{diag}(1/\lambda_1, \dots, 1/\lambda_n) exists, and (PDP^{-1})(PD^{-1}P^{-1}) = PDD^{-1}P^{-1} = I.
If B = P^{-1}AP, then \det B = \det(P^{-1})\det A\det P = \det A, and Result 4 gives equal characteristic polynomials, from which the trace is read. The determinants -2 and 6 differ, so the two matrices are not similar.
If A = PDP^{-1} with every diagonal entry of D equal to \lambda, then D = \lambda I and A = \lambda PP^{-1} = \lambda I. The matrix \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} has only the eigenvalue 3 but is not 3I.
If A = PDP^{-1}, then A^{T} = (P^{-1})^{T}D^{T}P^{T} = QDQ^{-1} with Q = (P^{T})^{-1}, since D^{T} = D. And A + cI = P(D + cI)P^{-1}, with D + cI diagonal.
D has only 0s and 1s on its diagonal, so D^2 = D, and A^2 = PD^2P^{-1} = PDP^{-1} = A.
Its characteristic polynomial \lambda^2 + 1 has no real root, so it has no real eigenvector, while by Result 1 a real diagonalization needs two (the columns of P).