Linear-Algebra-Done-Right · Unit 26 · Video 2 · Interactive Practice

Jordan Bases: Strings and Ones Above the Diagonal

IKey Formulas

Formula	Name	Hypotheses
$T^{m-1}v,\ \dots,\ Tv,\ v$ with $T^{m-1}v \neq 0$ , $T^m v = 0$	A string, listed from its last nonzero vector back to $v$	$T \in \mathcal{L}(V)$ nilpotent, $v \in V$
$\begin{pmatrix} 0 & 1 & & 0 \\ & 0 & \ddots & \\ & & \ddots & 1 \\ 0 & & & 0 \end{pmatrix}$ ( $m \times m$ )	The block of one string	$T$ sends each listed vector to the one before it, the first to $0$
$\mathcal{M}(T) = \begin{pmatrix} A_1 & & 0 \\ & \ddots & \\ 0 & & A_p \end{pmatrix}$ , $\ A_k = \begin{pmatrix} \lambda_k & 1 & & 0 \\ & \lambda_k & \ddots & \\ & & \ddots & 1 \\ 0 & & & \lambda_k \end{pmatrix}$	Jordan basis for $T$	$T \in \mathcal{L}(V)$ ; each $\lambda_k$ an eigenvalue of $T$ , not necessarily distinct; $1 \times 1$ blocks allowed

Key Insight: Each string gives one block, its length is the block's size, and the order of the list puts the $1$ 's above the diagonal rather than below it.

IIStrings Line Up into Blocks

A nilpotent operator on $\mathbf{C}^6$ built from strings of lengths adding to $6$ has one block per string.

String lengths Order in each string Basis vector 2

IIIAn Eigenvalue on Each Block

In a Jordan basis, the first vector of each block is an eigenvector and every other one picks up the vector before it.

size of A₁ 2 λ₁ 3 size of A₂ 2 λ₂ 3

💡 Two blocks may carry the same eigenvalue; the entry above the diagonal between them is still $0$ .

IVQuiz Questions

Problem 1 · Block Sizes from Strings

Given: $N \in \mathcal{L}(\mathbf{C}^4)$ defined by $N(z_1, z_2, z_3, z_4) = (z_3,\ z_4,\ 0,\ 0)$ — find the sizes of the blocks in the matrix of $N$ with respect to a Jordan basis.

✅ Correct!

e_3 \mapsto e_1 \mapsto 0

and

e_4 \mapsto e_2 \mapsto 0

: two strings of length

2

, so the basis

e_1, e_3, e_2, e_4

gives two

2 \times 2

blocks.

❌ No string has length $4$ .

N^2(z_1, z_2, z_3, z_4) = N(z_3, z_4, 0, 0) = 0

, so every string stops after two vectors.

❌ A block of size $3$ needs a string $N^2 v \neq 0$ . Here

N^2 = 0

❌ Sizes $2, 1, 1$ allow only one string of length $2$ . Both

e_3

and

e_4

start strings of length

2

, and together they already fill

\mathbf{C}^4

Show solution

$Ne_3 = e_1$ , $Ne_1 = 0$ , $Ne_4 = e_2$ , $Ne_2 = 0$ . So $e_3$ gives the string $Ne_3, e_3 = e_1, e_3$ and $e_4$ gives $Ne_4, e_4 = e_2, e_4$ . Together $e_1, e_3, e_2, e_4$ is a basis of $\mathbf{C}^4$ , and

\mathcal{M}\bigl(N, (e_1, e_3, e_2, e_4)\bigr) = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}:

two $2 \times 2$ blocks.

Problem 2 · Choose the Starting Vector

Given: $N \in \mathcal{L}(\mathbf{C}^3)$ defined by $N(z_1, z_2, z_3) = (z_3,\ 0,\ z_2)$ — find the vector $u$ for which $N^2 u,\ Nu,\ u$ is a Jordan basis for $N$ .

✅ Correct!

Ne_2 = e_3

and

N^2 e_2 = Ne_3 = e_1

, so the list is

e_1, e_3, e_2

, a basis, and

N

sends each vector to the one before it.

❌ $Ne_1 = 0$ . The list would be

0, 0, e_1

, which contains the zero vector.

❌ $N^2 e_3 = Ne_1 = 0$ . The string from

e_3

has length

2

, not

3

❌ $N(e_1 + e_3) = e_1$ and $N^2(e_1 + e_3) = 0$ . That string has length

2

Show solution

$Ne_1 = 0$ , $Ne_2 = e_3$ , $Ne_3 = e_1$ . So $N^2 \neq 0$ and $N^3 = 0$ , and a string of length $3$ needs $N^2 u \neq 0$ . Writing $u = (a, b, c)$ : $N^2 u = N(c, 0, b) = (b, 0, 0)$ , so $b \neq 0$ . Among the options only $e_2$ has $b \neq 0$ .

With $u = e_2$ the list $N^2 u, Nu, u = e_1, e_3, e_2$ is a basis, and

\mathcal{M}\bigl(N, (e_1, e_3, e_2)\bigr) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.

Problem 3 · Read the Operator off the Matrix

Given: $v_1, \dots, v_5$ is a Jordan basis for $T$ in which $\mathcal{M}(T) = \begin{pmatrix} 4 & 1 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 & -1 \end{pmatrix}$ — find $Tv_5$ and the number of $v_k$ 's that are eigenvectors of $T$ .

$Tv_5 =$

Number of eigenvectors among $v_1, \dots, v_5$

✅ Correct! Column

5

reads

(0, 0, 0, 1, -1)

, so

Tv_5 = v_4 - v_5

; the eigenvectors are

v_1

v_3

v_4

, the first vector of each block.

❌ The diagonal entry is only part of column $5$ . The

1

above it adds

v_4

❌ That reads row $4$ instead of column $5$ . Column

k

\mathcal{M}(T)

holds the coefficients of

Tv_k

❌ Not quite. Read column

5

: the entries are the coefficients of

Tv_5

v_1, \dots, v_5

❌ $v_2$ and $v_5$ are not eigenvectors.

Tv_2 = v_1 + 4v_2

is not a multiple of

v_2

❌ Not quite. A column with a

1

above its diagonal entry is not the column of an eigenvector; count the others.

Show solution

Column $k$ gives $Tv_k$ :

Tv_1 = 4v_1, \quad Tv_2 = v_1 + 4v_2, \quad Tv_3 = 4v_3, \quad Tv_4 = -v_4, \quad Tv_5 = v_4 - v_5.

The blocks are $\begin{pmatrix} 4 & 1 \\ 0 & 4 \end{pmatrix}$ , $(4)$ and $\begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix}$ . The first vector of each block, $v_1$ , $v_3$ , $v_4$ , is an eigenvector; $v_2$ and $v_5$ are not, since $v_1, \dots, v_5$ is linearly independent. That makes $3$ eigenvectors.

Problem 4 · Not a Jordan Matrix

Given: four $3 \times 3$ matrices — find the one that is NOT the matrix of an operator with respect to a Jordan basis.

✅ Correct! The

1

in position

(1, 2)

puts rows

1

and

2

in the same block, whose diagonal would hold both

2

and

3

; a Jordan block has one number all down its diagonal.

❌ That one is a Jordan matrix. Split it along the diagonal: each block has a single number on its diagonal,

1

's directly above it, and every entry outside the blocks is

0

(

1 \times 1

blocks and repeated eigenvalues are allowed).

Show solution

(a) blocks $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ and $(2)$ . (b) blocks $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $(0)$ . (d) blocks $(5)$ and $\begin{pmatrix} 5 & 1 \\ 0 & 5 \end{pmatrix}$ .

In (c) both entries above the diagonal are $1$ , so all three rows lie in one block, and its diagonal $2, 3, 3$ is not constant. Treating the $1$ in position $(1, 2)$ as lying between blocks is not allowed either, since entries outside the blocks are $0$ . So (c) is not a Jordan matrix.

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