Linear-Algebra-Done-Right · Unit 13 · Video 1 · Interactive Practice

Diagonalizable Operators and Direct Sums of Eigenspaces

IKey Formulas

Formula	Name	Hypotheses
$\mathcal{M}\big(T, (v_1, \ldots, v_n)\big) = \begin{pmatrix} \lambda_1 & & 0 \\ & \ddots & \\ 0 & & \lambda_n \end{pmatrix} \iff Tv_k = \lambda_k v_k$ for each $k$	Diagonal matrix; $T$ is diagonalizable if this happens for some basis, and then the eigenvalues of $T$ are exactly $\lambda_1, \ldots, \lambda_n$	$T \in \mathcal{L}(V)$ , $v_1, \ldots, v_n$ a basis of $V$
$E(\lambda, T) = \text{null}(T - \lambda I) = \{v \in V : Tv = \lambda v\}$	Eigenspace: the eigenvectors for $\lambda$ together with $0$ ; $\lambda$ is an eigenvalue $\iff E(\lambda, T) \neq \{0\}$	$T \in \mathcal{L}(V)$ , $\lambda \in \mathbf{F}$
$E(\lambda_1, T) + \cdots + E(\lambda_m, T)$ is a direct sum	Eigenspaces for distinct eigenvalues form a direct sum	$\lambda_1, \ldots, \lambda_m$ distinct eigenvalues of $T$
$\dim E(\lambda_1, T) + \cdots + \dim E(\lambda_m, T) \le \dim V$	The eigenspace dimensions add to at most $\dim V$	Same, with $V$ finite-dimensional

Key Insight: Diagonalizable describes the operator, not one of its matrices: a diagonal matrix appears exactly when every basis vector $v_k$ satisfies $Tv_k = \lambda_k v_k$ , so the search for one is a search inside the eigenspaces.

IIDiagonal in Some Basis

The standard matrix of $T(x, y) = (41x + 7y,\ -20x + 74y)$ is not diagonal, yet $T$ is diagonalizable.

eigenspace lines

IIIEigenspaces as Null Spaces

With $\mathcal{M}\big(T, (v_1, v_2, v_3)\big) = \begin{pmatrix} 8 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$ , which vectors does $T - \lambda I$ send to $0$ ?

λ 5

IVEigenspace Dimensions Add to at Most dim V

For an upper-triangular $\mathcal{M}(T)$ , a repeated diagonal entry need not give an eigenspace of that dimension.

💡 Challenge: with diagonal $8, 5, 5$ , find the entries above the diagonal that can be made nonzero while $\dim E(5, T)$ stays $2$ .

VQuiz Questions

Problem 1 · Reading an Eigenspace off a Diagonal Matrix

Given: $T \in \mathcal{L}(\mathbf{R}^4)$ and a basis $v_1, v_2, v_3, v_4$ of $\mathbf{R}^4$ with $\mathcal{M}\big(T, (v_1, v_2, v_3, v_4)\big) = \begin{pmatrix} 3 & 0 & 0 & 0 \\ 0 & -2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}$ — find $E(3, T)$ .

✅ Correct!

(T - 3I)v = -5a_2v_2

vanishes exactly when

a_2 = 0

, so

E(3, T)

is spanned by the three basis vectors with diagonal entry

3

and has dimension

3

❌ Only the first $3$ was used.

Tv_3 = 3v_3

and

Tv_4 = 3v_4

as well, so

v_3, v_4 \in E(3, T)

: the eigenspace is larger than a line.

❌ That is the coordinate that must vanish.

Tv_2 = -2v_2

, so

(T - 3I)v_2 = -5v_2 \neq 0

; in fact

\text{span}(v_2) = E(-2, T)

❌ That would make $T = 3I$ .

Tv_2 = -2v_2 \neq 3v_2

, so

v_2 \notin E(3, T)

❌ Not quite. Write

v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4

and compute

(T - 3I)v

column by column.

Show solution

Column $k$ gives $Tv_k$ : $Tv_1 = 3v_1$ , $Tv_2 = -2v_2$ , $Tv_3 = 3v_3$ , $Tv_4 = 3v_4$ . For $v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4$ ,

(T - 3I)v = 0a_1v_1 - 5a_2v_2 + 0a_3v_3 + 0a_4v_4 = -5a_2v_2.

Since $v_2 \neq 0$ , this is $0$ exactly when $a_2 = 0$ , while $a_1, a_3, a_4$ are arbitrary. Hence

E(3, T) = \text{span}(v_1, v_3, v_4), \qquad \dim E(3, T) = 3.

Likewise $E(-2, T) = \text{span}(v_2)$ , and $\dim E(3, T) + \dim E(-2, T) = 3 + 1 = 4 = \dim \mathbf{R}^4$ .

Problem 2 · A Repeated Entry Above a Nonzero Entry

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ , $T(x, y, z) = (4x + y,\ 4y,\ 7z)$ , whose standard matrix $\begin{pmatrix} 4 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 7 \end{pmatrix}$ has $4$ twice on the diagonal — find $E(4, T)$ .

✅ Correct!

(T - 4I)(x, y, z) = (y, 0, 3z)

0

exactly when

y = z = 0

: the entry

4

appears twice, but

\dim E(4, T) = 1

❌ The repeated diagonal entry misled you.

(T - 4I)(0, 1, 0) = (1, 0, 0) \neq 0

, so

(0, 1, 0) \notin E(4, T)

; the entry

1

above the diagonal is what stops it.

❌ The roles of the coordinates are swapped. The equations

y = 0

and

3z = 0

constrain

y

and

z

;

x

is the free coordinate, and

(T - 4I)(0, 1, 0) = (1, 0, 0) \neq 0

❌ $(0, 0, 1)$ belongs to the other eigenvalue.

T(0, 0, 1) = (0, 0, 7)

, so

(0, 0, 1) \in E(7, T)

; eigenspaces for the distinct eigenvalues

4

and

7

meet only in

0

❌ Not quite. Compute

(T - 4I)(x, y, z)

and solve for the vectors it sends to

0

Show solution

Subtract $4$ times the identity:

(T - 4I)(x, y, z) = (4x + y - 4x,\ 4y - 4y,\ 7z - 4z) = (y,\ 0,\ 3z).

This is $0$ exactly when $y = 0$ and $z = 0$ , with $x$ free, so

E(4, T) = \text{span}\big((1, 0, 0)\big), \qquad \dim E(4, T) = 1.

For the other eigenvalue, $(T - 7I)(x, y, z) = (-3x + y,\ -3y,\ 0)$ vanishes exactly when $y = 0$ and $x = 0$ , so $E(7, T) = \text{span}\big((0, 0, 1)\big)$ . The dimensions add to $1 + 1 = 2$ , strictly less than $\dim \mathbf{R}^3 = 3$ : the inequality in the theorem can be strict.

Problem 3 · Counting Eigenspace Dimensions in $\mathbf{R}^4$

Given: $T \in \mathcal{L}(\mathbf{R}^4)$ , $T(x_1, x_2, x_3, x_4) = (2x_1 + x_3,\ 2x_2,\ 2x_3,\ 5x_4)$ , whose standard matrix is upper triangular with diagonal entries $2, 2, 2, 5$ , so its eigenvalues are $2$ and $5$ — find $\dim E(2, T)$ and compare the eigenspace dimensions with $\dim \mathbf{R}^4$ .

What is $\dim E(2, T)$ ?

Which statement is true?

✅ Correct!

E(2, T) = \text{span}(e_1, e_2)

and

E(5, T) = \text{span}(e_4)

; their direct sum has dimension

3

, one short of

\mathbf{R}^4

❌ Too small.

(T - 2I)x = (x_3, 0, 0, 3x_4)

leaves both

x_1

and

x_2

free:

e_1

and

e_2

are both in

E(2, T)

❌ Three diagonal $2$ s do not give three dimensions.

(T - 2I)e_3 = e_1 \neq 0

, so

e_3 \notin E(2, T)

❌ That would make $T = 2I$ .

Te_4 = 5e_4 \neq 2e_4

❌ Not quite. Compute

(T - 2I)(x_1, x_2, x_3, x_4)

and count the free coordinates.

❌ The number of diagonal entries is $\dim V$ , not the sum. Here

\dim E(2, T) = 2

and

\dim E(5, T) = 1

, so the sum is

3

❌ The theorem has no exception for repeats.

2

and

5

are distinct eigenvalues, so

E(2, T) + E(5, T)

is always a direct sum.

❌ The theorem says $\le \dim V$ . A sum of

3

in a space of dimension

4

is allowed; equality is not promised.

❌ Not quite. Compare

\dim E(2, T) + \dim E(5, T)

with

4

, keeping in mind that the sum of the eigenspaces is direct.

Show solution

Eigenvalue $2$ :

(T - 2I)(x_1, x_2, x_3, x_4) = (x_3,\ 0,\ 0,\ 3x_4),

which is $0$ exactly when $x_3 = x_4 = 0$ . So $E(2, T) = \text{span}(e_1, e_2)$ and $\dim E(2, T) = 2$ .

Eigenvalue $5$ :

(T - 5I)(x_1, x_2, x_3, x_4) = (-3x_1 + x_3,\ -3x_2,\ -3x_3,\ 0),

which is $0$ exactly when $x_3 = 0$ , $x_2 = 0$ and then $x_1 = 0$ . So $E(5, T) = \text{span}(e_4)$ and $\dim E(5, T) = 1$ .

The sum $E(2, T) \oplus E(5, T)$ is direct, so joining the bases gives the linearly independent list $e_1, e_2, e_4$ , and

\dim E(2, T) + \dim E(5, T) = 2 + 1 = 3 \lt 4 = \dim \mathbf{R}^4.

Three vectors cannot span $\mathbf{R}^4$ , so this list is not a basis.

Problem 4 · What the Inequality Allows

Given: $T \in \mathcal{L}(\mathbf{R}^4)$ has an upper-triangular matrix with respect to some basis, with diagonal entries $7, 7, 7, -3$ — find every value that $\dim E(7, T)$ can take, over all such $T$ .

✅ Correct!

\dim E(7, T) \ge 1

because

7

is an eigenvalue, and

\dim E(7, T) \le 4 - \dim E(-3, T) \le 3

; each of

1

2

3

occurs.

❌ Repetition on the diagonal does not fix the dimension. With entries

1

above the diagonal, as in Problem 2,

\dim E(7, T)

can drop to

2

1

❌ $0$ is impossible. The diagonal entries of an upper-triangular matrix are eigenvalues (Unit 12), and

\lambda

is an eigenvalue exactly when

E(\lambda, T) \neq \{0\}

❌ $4$ is impossible.

-3

is also an eigenvalue, so

\dim E(-3, T) \ge 1

and

\dim E(7, T) \le 4 - 1 = 3

❌ Not quite. Bound

\dim E(7, T)

below using that

7

is an eigenvalue, and above using the theorem together with

\dim E(-3, T) \ge 1

Show solution

Eigenvalues. The eigenvalues of an operator with an upper-triangular matrix are its diagonal entries (Unit 12), so $7$ and $-3$ are the eigenvalues of $T$ , and both eigenspaces are nonzero: $\dim E(7, T) \ge 1$ and $\dim E(-3, T) \ge 1$ .

Upper bound. $7 \neq -3$ , so the theorem gives

\dim E(7, T) + \dim E(-3, T) \le \dim \mathbf{R}^4 = 4, \qquad\text{hence}\qquad \dim E(7, T) \le 4 - 1 = 3.

Every value occurs. Take the standard basis $e_1, e_2, e_3, e_4$ :

$\mathcal{M}(T) = \begin{pmatrix} 7 & 0 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}$ : $E(7, T) = \text{span}(e_1, e_2, e_3)$ , dimension $3$ .
$\mathcal{M}(T) = \begin{pmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}$ : $(T - 7I)x = (x_2, 0, 0, -10x_4)$ , so $E(7, T) = \text{span}(e_1, e_3)$ , dimension $2$ .
$\mathcal{M}(T) = \begin{pmatrix} 7 & 1 & 0 & 0 \\ 0 & 7 & 1 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & -3 \end{pmatrix}$ : $(T - 7I)x = (x_2, x_3, 0, -10x_4)$ , so $E(7, T) = \text{span}(e_1)$ , dimension $1$ .

So $\dim E(7, T)$ can be exactly $1$ , $2$ or $3$ .

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