Linear-Algebra-Done-Right · Unit 13 · Video 2 · Interactive Practice

Four Equivalent Tests for Diagonalizability

IKey Formulas

Formula	Statement	Hypotheses
$V = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$	Condition (3); equivalent to (1) $T$ is diagonalizable and (2) $V$ has a basis of eigenvectors of $T$	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $\lambda_1, \ldots, \lambda_m$ the distinct eigenvalues of $T$
$\dim V = \dim E(\lambda_1, T) + \cdots + \dim E(\lambda_m, T)$	Condition (4): the inequality $\le$ , which always holds, is an equality	Same
$T$ has $\dim V$ distinct eigenvalues $\Rightarrow$ $T$ diagonalizable	Sufficient, not necessary: $T(x, y, z) = (6x, 6y, 7z)$ has two eigenvalues and is diagonalizable	$V$ finite-dimensional, $T \in \mathcal{L}(V)$
$T^k(a_1 v_1 + \cdots + a_n v_n) = a_1 \lambda_1^k v_1 + \cdots + a_n \lambda_n^k v_n$	Powers act coordinate by coordinate in an eigenvector basis, since $T^k v_j = \lambda_j^k v_j$	$T v_j = \lambda_j v_j$ for each $j$

Key Insight: Having an eigenvalue is not having enough eigenvectors. For $T(a, b, c) = (b, c, 0)$ on $\mathbf{F}^3$ the only eigenspace is $E(0, T) = \{(a, 0, 0) : a \in \mathbf{F}\}$ , and $1 \ne 3$ , so all four conditions fail together.

IICondition (4) Is a Count

For an upper-triangular operator on $\mathbf{F}^3$ , the eigenspace dimensions either add up to $3$ or fall short, and all four conditions follow.

operator a₁₁ 2 a₂₂ 5 a₃₃ 8 a₁₂ 1 a₂₃ 3 a₁₃ 0

💡 Challenge: with diagonal entries $5$ , $5$ , $8$ , find exactly which off-diagonal entries keep $T$ diagonalizable.

IIIPowers Through the Eigenvector Basis

In the basis $(1, 0, 0)$ , $(1, 3, 0)$ , $(1, 6, 6)$ , applying $T^k$ only multiplies the coordinates by $2^k$ , $5^k$ and $8^k$ .

x 0 y 0 z 1 k 1

IVQuiz Questions

Problem 1 · An Eigenvector Basis of R²

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x, y) = (4x + y,\ 2x + 3y)$ . Which list is a basis of $\mathbf{R}^2$ consisting of eigenvectors of $T$ ?

✅ Correct!

T(1, 1) = (5, 5) = 5(1, 1)

and

T(1, -2) = (2, -4) = 2(1, -2)

; two linearly independent eigenvectors in

\mathbf{R}^2

form a basis.

❌ Not quite.

T(1, 0) = (4, 2)

is not a scalar multiple of

(1, 0)

, so the standard basis vectors are not eigenvectors; the matrix of

T

in that basis is not diagonal.

❌ Check the second vector.

T(1, 2) = (6, 8)

, which is not a scalar multiple of

(1, 2)

. Solve

4x + y = 2x

for the eigenvalue

2

❌ Both are eigenvectors, but not a basis.

(2, 2) = 2(1, 1)

, so the list is linearly dependent; both vectors lie in the same eigenspace

E(5, T)

Show solution

Test each candidate with $T v = \lambda v$ :

T(1, 1) = (4 + 1,\ 2 + 3) = (5, 5) = 5(1, 1), \qquad T(1, -2) = (4 - 2,\ 2 - 6) = (2, -4) = 2(1, -2).

So $(1, 1)$ is an eigenvector for $5$ and $(1, -2)$ is an eigenvector for $2$ . Eigenvectors for distinct eigenvalues are linearly independent, and a linearly independent list of length $2$ in $\mathbf{R}^2$ is a basis. With respect to it, $\mathcal{M}(T) = \operatorname{diag}(5, 2)$ .

The others fail: $T(1, 0) = (4, 2)$ and $T(1, 2) = (6, 8)$ are not multiples of the input, and $(1, 1)$ , $(2, 2)$ is linearly dependent.

Problem 2 · Differentiation on P₂(R)

Given: $D \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}))$ defined by $Dp = p'$ . Is $D$ diagonalizable, and why?

✅ Correct! The eigenvectors are the nonzero constants, a

1

-dimensional eigenspace in a

3

-dimensional space, so condition (4) fails and with it all four.

❌ Upper triangular is not diagonal. The matrix with respect to

1, x, x^2

has

1

and

2

above the diagonal; that gives the eigenvalues (only

0

), not a basis of eigenvectors.

❌ One eigenvalue is not enough. An eigenvalue guarantees one eigenvector, not a basis of them. Count:

\dim E(0, D)

must reach

\dim \mathcal{P}_2(\mathbf{R}) = 3

❌ Not quite.

D1 = 0 = 0 \cdot 1

, so the constant

1

is an eigenvector with eigenvalue

0

Show solution

Write $p = a_0 + a_1 x + a_2 x^2$ , so $Dp = a_1 + 2a_2 x$ . The equation $Dp = \lambda p$ compares coefficients:

a_1 = \lambda a_0, \qquad 2a_2 = \lambda a_1, \qquad 0 = \lambda a_2.

If $\lambda \ne 0$ : the third equation gives $a_2 = 0$ , then $a_1 = 0$ , then $a_0 = 0$ , so $p = 0$ and $\lambda$ is not an eigenvalue. If $\lambda = 0$ : $a_1 = 0$ and $a_2 = 0$ with $a_0$ free, so $E(0, D)$ is the set of constant polynomials, of dimension $1$ .

The only eigenvalue is $0$ , and $\dim E(0, D) = 1 \ne 3 = \dim \mathcal{P}_2(\mathbf{R})$ . Condition (4) fails, so $D$ is not diagonalizable. In the basis $1, x, x^2$ the matrix of $D$ is a shift with entries $1, 2$ just above the diagonal, the same shape as $(b, c, 0)$ .

Problem 3 · Tᵏ Through Eigenvectors

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x, y) = (x + 2y,\ 3x + 2y)$ , with eigenvectors $T(2, 3) = 4(2, 3)$ and $T(1, -1) = -(1, -1)$ . Find $(5, 0)$ in this eigenvector basis and a formula for $T^k(5, 0)$ .

Write (5, 0) = a(2, 3) + b(1, −1)

Formula for Tᵏ(5, 0), k ≥ 1

✅ Correct!

T^k(5, 0) = 4^k(2, 3) + 3(-1)^k(1, -1)

; for

k = 3

this is

(125, 195)

, matching three direct applications of

T

❌ Check the coordinates. Solve

2a + b = 5

and

3a - b = 0

; adding them gives

5a = 5

❌ Swapped.

3(2, 3) + 1(1, -1) = (7, 8)

. The coefficient of

(2, 3)

a = 1

❌ Check the formula at $k = 1$ . It must give

T(5, 0) = (5, 15)

❌ The eigenvalue is $-1$ , not $1$ . The second term is

3(-1)^k(1, -1)

; your formula is right only for even

k

(at

k = 1

it gives

(11, 9)

, not

(5, 15)

❌ $(5, 0)$ is not an eigenvector.

T(5, 0) = (5, 15)

, not a multiple of

(5, 0)

, so

T^k v = \lambda^k v

does not apply to it directly; expand in the eigenvector basis first.

Show solution

Step 1: coordinates. $(5, 0) = a(2, 3) + b(1, -1)$ means $2a + b = 5$ and $3a - b = 0$ . Adding: $5a = 5$ , so $a = 1$ and $b = 3$ . Check: $(2, 3) + 3(1, -1) = (5, 0)$ .

Step 2: apply $T^k$ term by term. Since $T^k(2, 3) = 4^k(2, 3)$ and $T^k(1, -1) = (-1)^k(1, -1)$ ,

T^k(5, 0) = 4^k(2, 3) + 3(-1)^k(1, -1) = \bigl(2 \cdot 4^k + 3(-1)^k,\ 3 \cdot 4^k - 3(-1)^k\bigr).

Check. $k = 1$ : $(8 - 3,\ 12 + 3) = (5, 15) = T(5, 0)$ . $k = 3$ : $(128 - 3,\ 192 + 3) = (125, 195)$ , and directly $T(5, 0) = (5, 15)$ , $T(5, 15) = (35, 45)$ , $T(35, 45) = (125, 195)$ .

Problem 4 · A Family of Operators

Given: for each $t \in \mathbf{R}$ , $T_t \in \mathcal{L}(\mathbf{R}^2)$ defined by $T_t(x, y) = (2x + y,\ t y)$ . For which $t$ is $T_t$ diagonalizable?

✅ Correct! For

t \ne 2

there are

2 = \dim \mathbf{R}^2

distinct eigenvalues; for

t = 2

the only eigenspace is

E(2, T_2) = \{(x, 0)\}

, of dimension

1

❌ Check $t = 2$ . Then

T_2(x, y) = (2x + y, 2y)

has only the eigenvalue

2

, and

E(2, T_2) = \{(x, 0) : x \in \mathbf{R}\}

has dimension

1 \ne 2

❌ Reversed. A repeated eigenvalue is exactly where the guarantee is lost: at

t = 2

the eigenspace dimensions add to

1

, not

2

❌ Not quite. For

t \ne 2

, the vector

(1, t - 2)

satisfies

T_t(1, t - 2) = t(1, t - 2)

, so

(1, 0)

(1, t - 2)

is an eigenvector basis; the off-diagonal entry only changes which vectors are eigenvectors.

Show solution

Solve $T_t(x, y) = \lambda(x, y)$ : $(2 - \lambda)x + y = 0$ and $(t - \lambda)y = 0$ . If $y \ne 0$ the second equation forces $\lambda = t$ ; if $y = 0$ then $x \ne 0$ and the first forces $\lambda = 2$ . So the eigenvalues are $2$ and $t$ , with $T_t(1, 0) = 2(1, 0)$ .

$t \ne 2$ : two distinct eigenvalues on the $2$ -dimensional space $\mathbf{R}^2$ , so $T_t$ is diagonalizable. An eigenvector for $t$ : $2x + y = tx$ gives $(1, t - 2)$ , and indeed $T_t(1, t - 2) = (t,\ t(t - 2))$ .

$t = 2$ : the only eigenvalue is $2$ , and $(2 - 2)x + y = 0$ gives $y = 0$ , so $E(2, T_2) = \{(x, 0) : x \in \mathbf{R}\}$ has dimension $1 \ne 2$ . Condition (4) fails: $T_2$ is not diagonalizable.

So $T_t$ is diagonalizable exactly when $t \ne 2$ .

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