Linear-Algebra-Done-Right · Unit 25 · Video 5 · Interactive Practice

Block Diagonal Form: One Block per Eigenvalue

IKey Formulas

Formula	Name	Hypotheses
$\mathcal{M}(T) = \begin{pmatrix} A_1 & & 0 \\ & \ddots & \\ 0 & & A_m \end{pmatrix}$ , $\ A_k = \begin{pmatrix} \lambda_k & & * \\ & \ddots & \\ 0 & & \lambda_k \end{pmatrix}$	Block diagonal form	$\mathbf{F} = \mathbf{C}$ , $T \in \mathcal{L}(V)$ , $V$ finite-dimensional; $\lambda_1, \dots, \lambda_m$ the distinct eigenvalues
$A_k$ is $d_k \times d_k$ , $\ d_k = \dim G(\lambda_k, T)$	Block sizes = multiplicities	same
$\displaystyle\sum_k \frac{d_k(d_k + 1)}{2} < \frac{n(n + 1)}{2}$	Fewer allowed nonzero positions	same, $n = \dim V$ , at least two distinct eigenvalues

Key Insight: List a basis of each generalized eigenspace that makes the nilpotent part strictly upper triangular; invariance of each $G(\lambda_k, T)$ forces every entry outside the blocks to be $0$ .

IIChange to a Basis of Generalized Eigenvectors

In the basis $(1,0,0)$ , $(0,1,0)$ , $(10,2,1)$ the entries $4$ and $2$ of the standard matrix become $0$ .

Basis Columns filled 3

💡 No basis of $G(6, T)$ makes the block $\begin{pmatrix} 6 & 3 \\ 0 & 6 \end{pmatrix}$ diagonal: that would make $T = 6I$ on the plane $G(6, T)$ , but $E(6, T)$ is only a line.

IIIHow Many Entries Can Be Nonzero?

Upper triangular allows $n(n+1)/2$ nonzero positions; block diagonal with upper-triangular blocks allows only $\sum_k d_k(d_k+1)/2$ .

d₁ 1 d₂ 2 d₃ 2

IVQuiz Questions

Problem 1 · Count the Allowed Positions

Given: $T \in \mathcal{L}(\mathbf{C}^8)$ has four distinct eigenvalues with multiplicities $4, 2, 1, 1$ — find the number of positions that may be nonzero in its block diagonal form with upper-triangular blocks.

✅ Correct!

\frac{4 \cdot 5}{2} + \frac{2 \cdot 3}{2} + 1 + 1 = 10 + 3 + 1 + 1 = 15

, against

\frac{8 \cdot 9}{2} = 36

for a general upper-triangular matrix.

❌ That is the count for an upper-triangular $8 \times 8$ matrix. In block form, every position outside the blocks is

0

❌ The blocks are upper triangular.

4^2 + 2^2 + 1 + 1 = 22

counts every entry of each block, including those below its diagonal.

❌ The blocks need not be diagonal. Entries above the diagonal inside a block may be nonzero.

Show solution

A $d \times d$ upper-triangular block has $d$ diagonal positions and $d(d-1)/2$ above the diagonal, $d(d+1)/2$ in all. The blocks have sizes equal to the multiplicities:

\frac{4 \cdot 5}{2} + \frac{2 \cdot 3}{2} + \frac{1 \cdot 2}{2} + \frac{1 \cdot 2}{2} = 10 + 3 + 1 + 1 = 15.

All other positions are $0$ , compared with $36$ allowed positions in a general upper-triangular $8 \times 8$ matrix.

Problem 2 · Put an Operator in Block Form

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with matrix $\begin{pmatrix} 2 & 1 & 3 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ in the standard basis, and the basis $e_1, e_2, w$ with $w$ spanning $G(-1, T)$ — find $w$ and the block of $\mathcal{M}(T)$ for the eigenvalue $2$ in this basis.

A vector $w$ spanning $G(-1, T)$

The block for the eigenvalue $2$

✅ Correct!

Tw = (2 - 3, 0, 1) = -w

Te_1 = 2e_1

Te_2 = e_1 + 2e_2

, so

\mathcal{M}(T) = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix}

❌ $e_3$ is not an eigenvector.

Te_3 = (3, 0, -1)

, which is not a multiple of

e_3

; that is exactly why the standard matrix has the entry

3

outside the blocks.

❌ Not quite. Solve

(T + I)w = 0

: the second row forces

w_2 = 0

, and the first row relates

w_1

and

w_3

❌ That block would make $T = 2I$ on $G(2, T)$ . But

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

is only a line, so no basis gives a diagonal block.

❌ Not quite. The block for

2

is the matrix of

T|_{G(2, T)}

in the basis

e_1, e_2

: compute

Te_1

and

Te_2

Show solution

The eigenvalues are $2$ (multiplicity $2$ ) and $-1$ (multiplicity $1$ ), read off the diagonal.

$(T + I)(z_1, z_2, z_3) = (3z_1 + z_2 + 3z_3,\ 3z_2,\ 0)$ , so $z_2 = 0$ and $z_1 = -z_3$ : $G(-1, T) = \operatorname{span}\big((1, 0, -1)\big)$ .

$(T - 2I)(z_1, z_2, z_3) = (z_2 + 3z_3,\ 0,\ -3z_3)$ and $(T - 2I)^2(z_1, z_2, z_3) = (-9z_3,\ 0,\ 9z_3)$ , so $G(2, T) = \operatorname{null}(T - 2I)^3 = \operatorname{span}(e_1, e_2)$ .

Te_1 = 2e_1, \quad Te_2 = (1, 2, 0) = e_1 + 2e_2, \quad T(1, 0, -1) = (-1, 0, 1) = -(1, 0, -1).

So in the basis $e_1, e_2, (1, 0, -1)$ , $\mathcal{M}(T) = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ , with the block $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ for the eigenvalue $2$ .

Problem 3 · Which Is Not the Theorem's Form?

Given: four $3 \times 3$ complex matrices — find the one that can NOT be the block diagonal form the theorem gives (one upper-triangular block per distinct eigenvalue, constant diagonal in each block).

✅ Correct! Its only possible blocks mix the eigenvalues

1

and

3

in one block, and the eigenvalue

1

also appears in a second block.

❌ That one has the right shape. Each eigenvalue occupies one block with that eigenvalue all down its diagonal, and every entry outside the blocks is

0

(a single block, or

1 \times 1

blocks, are allowed).

Show solution

(a) blocks $\begin{pmatrix} 4 & 1 \\ 0 & 4 \end{pmatrix}$ and $(-2)$ ; (c) one block $5I$ for the single eigenvalue $5$ ; (d) blocks $\begin{pmatrix} 0 & 3 \\ 0 & 0 \end{pmatrix}$ and $(2)$ . All have the theorem's shape.

In (b) the entry $2$ in position $(1, 2)$ forces rows and columns $1$ and $2$ into one block, whose diagonal is $1, 3$ , not constant. Moreover the eigenvalue $1$ appears in two different places, but the theorem puts all $d_1 = 2$ copies of $1$ into a single block.

Problem 4 · When a Block Is Diagonal

Given: $T \in \mathcal{L}(\mathbf{C}^4)$ has the eigenvalues $1$ and $5$ , with $\dim G(1, T) = 3$ and $\dim E(1, T) = 3$ — find the block for the eigenvalue $1$ in any block diagonal form of $T$ given by the theorem.

✅ Correct!

G(1, T) = E(1, T)

, so

T

is the identity on

G(1, T)

, and its matrix in every basis of that space is

I_3

❌ A nonzero entry is impossible here. Every vector of

G(1, T)

is an eigenvector, so

T v = v

for each basis vector

v

G(1, T)

❌ The basis cannot change it here.

T|_{G(1, T)} = I

, and the identity has the matrix

I_3

in every basis.

❌ The block size is the multiplicity.

\dim G(1, T) = 3

, so the block is

3 \times 3

Show solution

$E(1, T) \subseteq G(1, T)$ and both have dimension $3$ , so $G(1, T) = E(1, T)$ . Hence $Tv = v$ for every $v \in G(1, T)$ : the restriction $T|_{G(1, T)}$ is the identity, and the nilpotent part $(T - I)|_{G(1, T)}$ is $0$ .

The block for $1$ is the matrix of $T|_{G(1, T)}$ in the chosen basis of $G(1, T)$ , so it is $I_3$ whatever that basis is. The block for $5$ is the $1 \times 1$ block $(5)$ , since $\dim G(5, T) = 4 - 3 = 1$ .

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