Linear-Algebra-Done-Right · Unit 22 · Video 4 · Interactive Practice

Polar Decomposition: T = S√(T*T)

IKey Formulas

Formula	Name	Meaning
$z = \dfrac{z}{\|z\|}\sqrt{\bar{z}z}$	Polar form of $z \ne 0$	Direction on the unit circle times a length
$T = S\sqrt{T^*T}$ , $S$ unitary	Polar decomposition	Every $T \in \mathcal{L}(V)$ , over $\mathbf{R}$ or $\mathbf{C}$
$\sqrt{T^*T}\,v = \sum_{k=1}^m s_k\langle v, e_k\rangle e_k$	The positive factor	Stretches along $e_k$ by $s_k$
$Sv = \sum_{k=1}^n \langle v, e_k\rangle f_k$	The unitary factor	Sends each $e_k$ to $f_k$

Key Insight: When $\mathbf{F} = \mathbf{C}$ , $S$ and $\sqrt{T^*T}$ have diagonal matrices with respect to one common orthonormal basis exactly when $T$ is normal.

IIThe Complex-Number Picture

Multiplication by $z$ on $\mathbf{C}$ factors as multiplication by $z/|z|$ after multiplication by $|z| = \sqrt{\bar{z}z}$ .

IIIFirst Stretch, Then Move Rigidly

$\sqrt{T^*T}$ stretches the unit circle along $e_1, e_2$ ; then $S$ carries that ellipse onto $T$ 's image without changing any length.

Operator Stage 0.00

IVQuiz Questions

Problem 1 · An Operator on One Dimension

Given: $T \in \mathcal{L}(\mathbf{C})$ defined by $Tw = (-5 + 12i)w$ — find the number $c \ge 0$ with $\sqrt{T^*T}\,w = cw$ .

✅ Correct!

T^*

is multiplication by

\overline{z} = -5 - 12i

, so

T^*T

is multiplication by

|z|^2 = 169

and

\sqrt{T^*T}

13

;

S

is multiplication by

\tfrac{-5 + 12i}{13}

❌ Not quite.

169 = \bar{z}z

is what

T^*T

multiplies by. Take its nonnegative square root.

❌ Not quite.

|-5| + |12| = 17

is not the absolute value:

|z| = \sqrt{25 + 144}

❌ Not quite. That is

\sqrt{z^2}

. The adjoint conjugates:

T^*T

multiplies by

\bar{z}z

, not

z^2

Show solution

With $\langle w, u\rangle = w\bar{u}$ on $\mathbf{C}$ and $z = -5 + 12i$ : $\langle zw, u\rangle = zw\bar{u} = \langle w, \bar{z}u\rangle$ , so $T^*u = \bar{z}u$ . Then

T^*Tw = \bar{z}z\,w = (25 + 144)w = 169w,

and the positive square root is multiplication by $13$ . The polar decomposition is $Tw = \tfrac{-5 + 12i}{13}\cdot 13\,w$ , with $S$ = multiplication by $\tfrac{-5 + 12i}{13}$ , which has absolute value $1$ .

Problem 2 · The Positive Factor

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x, y) = (y, -3x)$ — find $\sqrt{T^*T}$ .

✅ Correct!

T^*T(x, y) = (9x, y)

, and

(3x, y)

is the positive operator whose square is that.

❌ Not quite.

(x, 3y)

\sqrt{TT^*}

: you composed in the other order. The polar decomposition uses

T^*T

, the adjoint applied after

T

❌ Not quite.

(9x, y)

T^*T

itself. Its positive square root takes the square root of each diagonal entry:

(3x, y)

❌ Not quite. Compute

T^*

from

\langle T(x, y), (a, b)\rangle

, then

T^*T

Show solution

$\langle T(x, y), (a, b)\rangle = ya - 3xb = \langle (x, y), (-3b, a)\rangle$ , so $T^*(a, b) = (-3b, a)$ . Then

T^*T(x, y) = T^*(y, -3x) = (9x, y).

The operator $R(x, y) = (3x, y)$ has a diagonal matrix with nonnegative entries with respect to the standard (orthonormal) basis, so it is positive, and $R^2(x, y) = (9x, y)$ . By uniqueness of positive square roots, $\sqrt{T^*T}(x, y) = (3x, y)$ .

Problem 3 · The Unitary Factor

Given: the same $T(x, y) = (y, -3x)$ — find the unitary $S \in \mathcal{L}(\mathbf{R}^2)$ with $T = S\sqrt{T^*T}$ .

✅ Correct!

S(3x, y) = (y, -3x) = T(x, y)

, and

S

is a rotation by

-90\degree

, so it preserves lengths.

❌ Not quite. This rotation goes the other way:

S(3x, y) = (-y, 3x) = -T(x, y)

❌ Not quite. Require

S(3x, y) = (y, -3x)

for all

x, y

, and substitute

a = 3x

b = y

Show solution

We need $S\big(\sqrt{T^*T}(x, y)\big) = T(x, y)$ , i.e. $S(3x, y) = (y, -3x)$ for all $x, y$ . Writing $a = 3x$ and $b = y$ (which range over all of $\mathbf{R}^2$ ):

S(a, b) = (b, -a).

$\|S(a, b)\|^2 = b^2 + a^2 = \|(a, b)\|^2$ , so $S$ is unitary. Since $T$ is invertible, $\sqrt{T^*T}$ is invertible and this $S$ is the only one.

Problem 4 · A Shared Diagonalizing Basis

Given: four operators on $\mathbf{C}^2$ with the usual inner product — which one has a polar decomposition $T = S\sqrt{T^*T}$ in which $S$ and $\sqrt{T^*T}$ have diagonal matrices with respect to the same orthonormal basis?

✅ Correct! Its matrix

\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}

satisfies

TT^* = T^*T = 2I

, so

T

is normal, and only normal operators have such a pair of factors.

❌ Not quite. Check normality: here

TT^* \ne T^*T

, and if both factors were diagonal in one orthonormal basis,

T

would be too, making it normal.

❌ Not quite. Distinct eigenvalues

1, 3

make

T

diagonalizable, but not with an orthonormal basis: its eigenvectors

(1, 0)

and

(1, 1)

are not orthogonal, so

T

is not normal.

Show solution

If $S$ and $\sqrt{T^*T}$ are both diagonal with respect to one orthonormal basis, so is their product $T$ , and an operator with a diagonal matrix with respect to an orthonormal basis is normal (complex spectral theorem). Conversely, every normal $T$ has such a decomposition. So the answer is the normal operator.

For (c), $\mathcal{M}(T) = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ and $\mathcal{M}(T)\mathcal{M}(T)^* = \mathcal{M}(T)^*\mathcal{M}(T) = 2I$ , so $T$ is normal. For (a), $\mathcal{M}(T)\mathcal{M}(T)^* = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \ne \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = \mathcal{M}(T)^*\mathcal{M}(T)$ ; for (b), $TT^*(z_1, z_2) = (z_1, 0)$ but $T^*T(z_1, z_2) = (0, z_2)$ ; for (d), the $(1, 1)$ entries of $\mathcal{M}(T)\mathcal{M}(T)^*$ and $\mathcal{M}(T)^*\mathcal{M}(T)$ are $5$ and $1$ .

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