Linear-Algebra-Done-Right · Unit 26 · Video 1 · Interactive Practice

Every Invertible Complex Operator Has a Square Root

IKey Formulas

Formula	Name	Hypotheses
$R^2 = T$	$R$ is a square root of $T$	$R, T \in \mathcal{L}(V)$
$\bigl(I + \tfrac12 N - \tfrac18 N^2 + \tfrac1{16} N^3 + \cdots + a_{m-1} N^{m-1}\bigr)^2 = I + N$	Identity plus nilpotent	$N \in \mathcal{L}(V)$ , $N^m = 0$ ; $\mathbf{F} = \mathbf{R}$ or $\mathbf{C}$
$T\|_{G(\lambda_k, T)} = \lambda_k\bigl(I + T_k/\lambda_k\bigr)$ , $\ R_k = \sqrt{\lambda_k}\,\bigl(I + T_k/\lambda_k\bigr)^{1/2}$	A square root on each generalized eigenspace	$\mathbf{F} = \mathbf{C}$ , $T$ invertible, $\lambda_k \neq 0$ an eigenvalue, $T_k = (T - \lambda_k I)\|_{G(\lambda_k, T)}$ nilpotent
$Rv = R_1 u_1 + \cdots + R_m u_m$ for $v = u_1 + \cdots + u_m$ , $\ u_k \in G(\lambda_k, T)$	Gluing the pieces: $R^2 = T$	same; $(I + T_k/\lambda_k)^{1/2}$ is the lemma's root

Key Insight: Nilpotence turns the Taylor series of $\sqrt{1+x}$ into a finite sum, and invertibility lets $T$ be written on every generalized eigenspace as a nonzero scalar times the identity plus a nilpotent.

IISolving for the Coefficients

Each coefficient $a_k$ enters the square only through $2a_k$ , so matching the coefficient of $N^k$ determines it.

Coefficients solved (k) 3

IIIA Square Root of Every Complex Number

Taking $\sqrt r$ of the length and half the angle gives a $w$ with $w^2 = z$ ; the root of each $\lambda_k$ comes from here.

💡 Over $\mathbf{R}$ the step fails at a negative eigenvalue: no real $c$ has $c^2 = -1$ .

IVQuiz Questions

Problem 1 · A Root When $N^3 = 0$

Given: $N \in \mathcal{L}(V)$ with $N^3 = 0$ — find an operator whose square is $I + N$ .

✅ Correct!

\bigl(I + \tfrac12 N - \tfrac18 N^2\bigr)^2 = I + N + \bigl(\tfrac14 - \tfrac14\bigr)N^2 + (\text{multiples of } N^3, N^4) = I + N

❌ One term short.

\bigl(I + \tfrac12 N\bigr)^2 = I + N + \tfrac14 N^2

, and

N^2

need not be

0

when only

N^3 = 0

❌ Check the sign of $a_2$ . The coefficient of

N^2

in the square is

2a_2 + a_1^2 = 2a_2 + \tfrac14

, which must be

0

❌ The $N^2$ term is $2a_2 + a_1^2$ , not $a_2 + a_1^2$ . With

a_2 = -\tfrac14

the square has

N^2

-coefficient

-\tfrac12 + \tfrac14 = -\tfrac14 \neq 0

Show solution

Try $R = I + a_1 N + a_2 N^2$ . Since powers of $N$ commute and $N^3 = N^4 = 0$ ,

R^2 = I + 2a_1 N + (2a_2 + a_1^2) N^2.

Matching with $I + N$ : $2a_1 = 1$ , so $a_1 = \tfrac12$ ; then $2a_2 + \tfrac14 = 0$ , so $a_2 = -\tfrac18$ . Hence $R = I + \tfrac12 N - \tfrac18 N^2$ .

Problem 2 · The Next Coefficient

Given: $N^5 = 0$ and the lemma's square root $I + a_1 N + a_2 N^2 + a_3 N^3 + a_4 N^4$ with $a_1 = \tfrac12$ , $a_2 = -\tfrac18$ , $a_3 = \tfrac1{16}$ — find $a_4$ .

✅ Correct!

2a_4 + 2a_1a_3 + a_2^2 = 2a_4 + \tfrac1{16} + \tfrac1{64} = 0

gives

a_4 = -\tfrac5{128}

❌ A product is missing.

N^4

arises as

N \cdot N^3

N^3 \cdot N

and

N^2 \cdot N^2

, so the term

a_2^2 = \tfrac1{64}

belongs in the equation.

❌ The new coefficient appears as $2a_4$ .

-\tfrac5{64}

is the value of

2a_4

, not of

a_4

❌ Check the sign. The known terms

\tfrac1{16} + \tfrac1{64}

are positive, so

2a_4

must be negative.

Show solution

The coefficient of $N^4$ in $(I + a_1 N + a_2 N^2 + a_3 N^3 + a_4 N^4)^2$ collects the products whose exponents add to $4$ :

2a_4 + 2a_1 a_3 + a_2^2 = 2a_4 + 2 \cdot \tfrac12 \cdot \tfrac1{16} + \tfrac1{64} = 2a_4 + \tfrac5{64}.

Setting it equal to $0$ gives $a_4 = -\tfrac5{128}$ .

Problem 3 · Factor Out the Eigenvalue

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ defined by $T(w, z) = (9w + 6z,\ 9z)$ — find a square root $R$ of $T$ .

✅ Correct!

R^2(w, z) = R(3w + z,\ 3z) = (9w + 3z + 3z,\ 9z) = (9w + 6z,\ 9z) = T(w, z)

❌ Square roots are not taken entry by entry.

R^2(w, z) = (9w + 6\sqrt6\, z,\ 9z)

, because the off-diagonal entry is hit twice by the factor

3

❌ Halve the nilpotent part after factoring out $9$ . With

T = 9(I + N)

N(w, z) = (\tfrac23 z, 0)

, the root is

3(I + \tfrac12 N)

; this option squares to

(9w + 18z,\ 9z)

❌ Not quite.

R^2(w, z) = (9w + 12z,\ 9z)

: the

z

-term is doubled once more by the factor

3

on the diagonal.

Show solution

The only eigenvalue is $9$ , and $T = 9I + M$ with $M(w, z) = (6z, 0)$ , $M^2 = 0$ . Factor out $9$ :

T = 9\bigl(I + N\bigr), \qquad N = \tfrac19 M, \quad N(w, z) = \bigl(\tfrac23 z,\ 0\bigr), \quad N^2 = 0.

Since $N^2 = 0$ , the lemma's root is $I + \tfrac12 N$ , and $\sqrt9 = 3$ :

R = 3\bigl(I + \tfrac12 N\bigr) = 3I + \tfrac32 N, \qquad R(w, z) = (3w + z,\ 3z).

Check: $R^2(w, z) = (9w + 3z + 3z,\ 9z) = (9w + 6z,\ 9z)$ . ( $-R$ is a square root too.)

Problem 4 · Which Has No Square Root?

Given: four operators on $\mathbf{C}^2$ — find the one that has no square root.

✅ Correct! If

R^2 = T

then

R^4 = T^2 = 0

, so

R

is nilpotent on a

2

-dimensional space and

R^2 = 0

, contradicting

T \neq 0

❌ That one is invertible. Over

\mathbf{C}

R(w, z) = (iw,\ iz)

satisfies

R^2 = -I

❌ The zero operator is its own square root. Invertibility is sufficient, not necessary.

❌ That one is invertible. It is

I + N

with

N(w, z) = (z, 0)

N^2 = 0

, so

R = I + \tfrac12 N

R(w, z) = (w + \tfrac12 z,\ z)

, works.

Show solution

(b) and (d) are invertible operators on a complex vector space, so they have square roots: $iI$ for (b), and $I + \tfrac12 N$ with $N(w, z) = (z, 0)$ for (d). (c) is the zero operator, and $0^2 = 0$ .

For (a), $T \neq 0$ but $T^2(w, z) = T(z, 0) = (0, 0)$ . Suppose $R^2 = T$ . Then $R^4 = T^2 = 0$ , so $R$ is nilpotent. A nilpotent operator on a space of dimension $2$ satisfies $R^2 = 0$ . Then $T = R^2 = 0$ , a contradiction. So (a) has no square root.

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