Linear-Algebra-Done-Right · Unit 21 · Video 1 · Interactive Practice

Singular Values: Square Roots of the Eigenvalues of T*T

IKey Formulas

Formula	Name	What it says
$(T^T)^ = T^T$ and $\langle T^Tv, v\rangle = \langle Tv, Tv\rangle = \\|Tv\\|^2 \ge 0$	$T^*T$ is positive	For every $T \in \mathcal{L}(V, W)$ , $T^*T$ is a positive operator on $V$ , even when $W \ne V$
$\operatorname{null} T^T = \operatorname{null} T$ , $\ \operatorname{range} T^T = \operatorname{range} T^*$	Null space and range of $T^*T$	$T^Tv = 0 \Rightarrow \\|Tv\\|^2 = \langle T^Tv, v\rangle = 0$ ; the range follows from $\operatorname{range} S^* = (\operatorname{null} S)^\perp$ with $S = T^*T$
$\dim \operatorname{range} T = \dim \operatorname{range} T^* = \dim \operatorname{range} T^*T$	Equal ranks	From $\operatorname{range} T = (\operatorname{null} T^)^\perp$ and the fundamental theorem of linear maps applied to $T^ \in \mathcal{L}(W, V)$
$s = \sqrt{\lambda}$ , $\lambda$ an eigenvalue of $T^T$ , repeated $\dim E(\lambda, T^T)$ times, $s_1 \ge s_2 \ge \cdots$	Singular values	$E(\lambda, T^T) = \operatorname{null}(T^T - \lambda I)$ ; by the spectral theorem these dimensions add up to $\dim V$ , so $T$ has exactly $\dim V$ singular values

Key Insight: For $T(z_1, z_2, z_3, z_4) = (0,\ 3z_1,\ 2z_2,\ -3z_4)$ the eigenvalues are $-3$ and $0$ , but the singular values are $3$ , $3$ , $2$ , $0$ : they record the factor $2$ , and they are not the numbers $|\lambda|$ .

IIEach Entry of $\mathcal{M}(T^*T)$ Is a Dot Product of Columns

With orthonormal bases and real entries, entry $(j, k)$ of $\mathcal{M}(T^*T) = \mathcal{M}(T)^{\mathrm{t}}\mathcal{M}(T)$ is $c_j \cdot c_k$ .

Map

IIISingular Values Record What Eigenvalues Miss

For $T(z) = (0,\ 3z_1,\ cz_2,\ -3z_4)$ the eigenvalues stay $-3, 0$ for every $c$ , while $|c|$ appears among the singular values.

c 2

IV $T^*T$ and $T$ Kill the Same Vectors

For the map $\mathbf{F}^4 \to \mathbf{F}^3$ of the second example, $\|Tz\|^2 = \langle T^*Tz, z\rangle$ , so $Tz = 0$ exactly when $T^*Tz = 0$ .

z₁ 1 z₂ −1 z₃ 2 z₄ 0

💡 Here $\operatorname{null} T = \operatorname{span}\big((1,-1,0,0),\ e_3\big) = E(0, T^*T)$ , which is why two of the four singular values $5, \sqrt{2}, 0, 0$ are zero.

VQuiz Questions

Problem 1 · Singular Values of a Coordinate Map

Given: $T \in \mathcal{L}(\mathbf{F}^3)$ , $T(z_1, z_2, z_3) = (2z_3,\ 0,\ -z_2)$ , with the Euclidean inner product. Find the singular values of $T$ .

✅ Correct!

T^*T(z) = (0,\ z_2,\ 4z_3)

has eigenvalues

4, 1, 0

, and their square roots in decreasing order are

2, 1, 0

❌ Not quite.

4, 1, 0

are the eigenvalues of

T^*T

. The singular values are their nonnegative square roots.

❌ Not quite.

0

is the only eigenvalue of

T

(so

0, 0, 0

lists it with multiplicity), but singular values come from

T^*T

, not from

T

❌ Not quite. Singular values are nonnegative square roots, so the coefficient

-1

contributes

\sqrt{(-1)^2} = 1

, and the list is written in decreasing order.

Show solution

Read off the images of the standard basis: $Te_1 = 0$ , $Te_2 = -e_3$ , $Te_3 = 2e_1$ . These are the columns of $\mathcal{M}(T)$ :

\mathcal{M}(T) = \begin{pmatrix} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & -1 & 0 \end{pmatrix}, \qquad \mathcal{M}(T^*) = \mathcal{M}(T)^{\mathrm{t}} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 2 & 0 & 0 \end{pmatrix}.

So $T^*(w_1, w_2, w_3) = (0,\ -w_3,\ 2w_1)$ and

T^*T(z_1, z_2, z_3) = T^*(2z_3,\ 0,\ -z_2) = \big(0,\ -(-z_2),\ 2\cdot 2z_3\big) = (0,\ z_2,\ 4z_3).

Each $e_k$ is an eigenvector of $T^*T$ : $E(4, T^*T) = \operatorname{span}(e_3)$ , $E(1, T^*T) = \operatorname{span}(e_2)$ , $E(0, T^*T) = \operatorname{span}(e_1)$ , each of dimension $1$ . Taking square roots, the singular values are $2, 1, 0$ : three of them, matching $\dim \mathbf{F}^3$ .

For comparison, $Tz = \lambda z$ with $\lambda \ne 0$ gives $z_2 = 0$ , then $z_3 = 0$ , then $z_1 = 0$ , so $0$ is the only eigenvalue of $T$ .

Problem 2 · Not the Absolute Values of the Eigenvalues

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (x + y,\ y)$ , with the Euclidean inner product. Its only eigenvalue is $1$ . Find the singular values of $T$ .

✅ Correct!

T^*T

has eigenvalues

\tfrac{3 \pm \sqrt{5}}{2}

, and their square roots are

\tfrac{1 + \sqrt{5}}{2}

and

\tfrac{\sqrt{5} - 1}{2}

, far from the eigenvalue

1

T

❌ Not quite. That is

|\lambda|

for the eigenvalue

1

listed twice. Singular values are not the absolute values of the eigenvalues; compute the eigenvalues of

T^*T

❌ Close, but… these are the eigenvalues of

T^*T

. The singular values are their nonnegative square roots.

❌ Not quite.

\sqrt{2}

and

1

are the lengths of the columns

(1, 0)

and

(1, 1)

, that is, the square roots of the diagonal entries of

\mathcal{M}(T^*T)

. Its off-diagonal entry

1

also matters, because

\mathcal{M}(T^*T)

is not diagonal.

Show solution

The columns of $\mathcal{M}(T)$ are $c_1 = Te_1 = (1, 0)$ and $c_2 = Te_2 = (1, 1)$ , so

\mathcal{M}(T^*T) = \mathcal{M}(T)^{\mathrm{t}}\mathcal{M}(T) = \begin{pmatrix} c_1\cdot c_1 & c_1\cdot c_2 \\ c_2 \cdot c_1 & c_2\cdot c_2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}.

Its characteristic polynomial is $\lambda^2 - 3\lambda + 1$ , so its eigenvalues are $\lambda = \tfrac{3 \pm \sqrt{5}}{2}$ , both positive, each with a one-dimensional eigenspace.

Since $\big(\tfrac{1 + \sqrt{5}}{2}\big)^2 = \tfrac{6 + 2\sqrt{5}}{4} = \tfrac{3 + \sqrt{5}}{2}$ and $\big(\tfrac{\sqrt{5} - 1}{2}\big)^2 = \tfrac{6 - 2\sqrt{5}}{4} = \tfrac{3 - \sqrt{5}}{2}$ , the singular values are

s_1 = \tfrac{1 + \sqrt{5}}{2} \approx 1.618, \qquad s_2 = \tfrac{\sqrt{5} - 1}{2} \approx 0.618.

The eigenvalue $1$ of $T$ (with $|\lambda| = 1$ ) appears nowhere in this list.

Problem 3 · A Map From F³ to F²

Given: $T \in \mathcal{L}(\mathbf{F}^3, \mathbf{F}^2)$ with matrix $\mathcal{M}(T) = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \end{pmatrix}$ in the standard bases. Find how many singular values $T$ has, and list them.

How many singular values?

Which list?

✅ Correct!

T

has

\dim \mathbf{F}^3 = 3

singular values,

2, \sqrt{2}, 0

; the zero comes from

\operatorname{null} T = \operatorname{span}\big((1, 0, -1)\big)

❌ Not quite.

2 = \dim \mathbf{F}^2

, the dimension of the target.

T^*T

is an operator on the domain

\mathbf{F}^3

, so there are

\dim \mathbf{F}^3

singular values.

❌ Not quite. The eigenspace dimensions of

T^*T \in \mathcal{L}(V)

add up to

\dim V

, and here

V = \mathbf{F}^3

❌ Not quite.

4, 2, 0

are the eigenvalues of

T^*T

; take their nonnegative square roots.

❌ Not quite.

\mathcal{M}(T^*T)

is not diagonal: columns

1

and

3

\mathcal{M}(T)

are equal, so

c_1 \cdot c_3 = 1

. Find the eigenvalues of the

3 \times 3

matrix, and keep the zero in the list.

Show solution

The columns are $c_1 = (1, 0)$ , $c_2 = (0, 2)$ , $c_3 = (1, 0)$ . Entry $(j, k)$ of $\mathcal{M}(T^*T) = \mathcal{M}(T)^{\mathrm{t}}\mathcal{M}(T)$ is $c_j \cdot c_k$ :

\mathcal{M}(T^*T) = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 4 & 0 \\ 1 & 0 & 1 \end{pmatrix}.

Eigenvectors: $e_2 \mapsto 4e_2$ ; $(1, 0, 1) \mapsto (2, 0, 2) = 2(1, 0, 1)$ ; $(1, 0, -1) \mapsto (0, 0, 0)$ . These three vectors are orthogonal, so $E(4) = \operatorname{span}(e_2)$ , $E(2) = \operatorname{span}\big((1, 0, 1)\big)$ and $E(0) = \operatorname{span}\big((1, 0, -1)\big)$ , each of dimension $1$ , and the dimensions add up to $3 = \dim \mathbf{F}^3$ .

Taking square roots in decreasing order, the singular values are $2, \sqrt{2}, 0$ . The zero is predicted by $E(0, T^*T) = \operatorname{null} T$ : indeed $T(1, 0, -1) = (1 - 1,\ 0) = (0, 0)$ .

Problem 4 · Counting Zeros Without Computing

Given: $T \in \mathcal{L}(\mathbf{F}^5, \mathbf{F}^3)$ is surjective. Determine how many singular values $T$ has and how many of them equal $0$ .

✅ Correct! There are

\dim \mathbf{F}^5 = 5

singular values, and the zeros fill

E(0, T^*T) = \operatorname{null} T

, which has dimension

5 - 3 = 2

❌ Not quite. The count is the dimension of the domain,

5

, not of the target

\mathbf{F}^3

❌ Not quite.

T

maps a

5

-dimensional space onto a

3

-dimensional one, so it cannot be injective, and

\operatorname{null} T = E(0, T^*T)

is nonzero.

❌ Not quite. The number of zeros is

\dim \operatorname{null} T

, not

\dim \operatorname{range} T = 3

. The fundamental theorem gives

\dim \operatorname{null} T = 5 - 3

Show solution

Count. $T^*T$ is a self-adjoint operator on $\mathbf{F}^5$ , so by the spectral theorem the dimensions of its eigenspaces add up to $5$ : $T$ has $5$ singular values.

Zeros. The singular value $0$ is repeated $\dim E(0, T^*T)$ times, and $E(0, T^*T) = \operatorname{null} T^*T = \operatorname{null} T$ by part (b). Since $T$ is surjective, $\dim \operatorname{range} T = 3$ , and the fundamental theorem of linear maps gives

\dim \operatorname{null} T = \dim \mathbf{F}^5 - \dim \operatorname{range} T = 5 - 3 = 2.

So exactly $2$ singular values are $0$ and the other $3$ are positive, matching part (d): $\dim \operatorname{range} T^*T = \dim \operatorname{range} T = 3$ .

Solved: 0 / 4

Singular Values: Square Roots of the Eigenvalues of T*T

IKey Formulas

IIEach Entry of M(T∗T)\mathcal{M}(T^*T)M(T∗T) Is a Dot Product of Columns

IIISingular Values Record What Eigenvalues Miss

IVT∗TT^*TT∗T and TTT Kill the Same Vectors

VQuiz Questions

IIEach Entry of $\mathcal{M}(T^*T)$ Is a Dot Product of Columns

IV $T^*T$ and $T$ Kill the Same Vectors