Linear-Algebra-Done-Right · Unit 22 · Video 2 · Interactive Practice

Computing the Norm of T: Three Formulas and Examples

IKey Formulas

Formula	Name	When to use it
$\\|T\\| = s_1 = \sqrt{\text{largest eigenvalue of } T^*T}$	Formula (a)	Computing a number from a matrix
$\\|T\\| = \max\{\\|Tv\\| : v \in V,\ \\|v\\| = 1\}$	Formula (b)	Only unit vectors need checking
$\\|T\\| = \min\{c \ge 0 : \\|Tv\\| \le c\\|v\\| \text{ for all } v \in V\}$	Formula (c)	Proofs: $\\|Tv\\| \le \\|T\\|\,\\|v\\|$
$\\|T^*\\| = \\|T\\|$	Norm of the adjoint	Cauchy–Schwarz and $(T^)^ = T$

Key Insight: If $V$ has an orthonormal basis of eigenvectors of $T$ with eigenvalues $\lambda_1, \dots, \lambda_n$ , then $\|T\| = \max\{|\lambda_1|, \dots, |\lambda_n|\}$ ; without such a basis, the eigenvalues can badly underestimate $\|T\|$ .

IIApproximating the Largest Eigenvalue of $T^*T$

Applying $T^*T$ and rescaling to length one pulls any starting vector toward $\pm e_1$ , so $\|Tx\|$ climbs to $s_1$ .

Map Step j 0

III $T$ and $T^*$ Stretch by the Same Maximum

The unit circle's images under $T$ and $T^*$ are differently tilted ellipses with the same longest semi-axis $s_1$ .

Map Angle θ of v 20°

IVQuiz Questions

Problem 1 · An Orthonormal Eigenbasis

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ and an orthonormal basis $e_1, e_2, e_3$ of $\mathbf{C}^3$ with $Te_1 = 3e_1$ , $Te_2 = -4e_2$ , $Te_3 = 2i\,e_3$ — find $\|T\|$ .

✅ Correct!

\|T\| = \max\{|3|, |-4|, |2i|\} = 4

, reached at

e_2

❌ Not quite.

3

is the largest eigenvalue as a signed number, but the stretch along

e_2

|-4| = 4

❌ Not quite.

\sqrt{9 + 16 + 4}

\|Tv\|

for

v = e_1 + e_2 + e_3

, which has length

\sqrt{3}

, not

1

❌ Not quite. Compare the absolute values of the three eigenvalues.

Show solution

Write $v = a_1e_1 + a_2e_2 + a_3e_3$ . Then $Tv = 3a_1e_1 - 4a_2e_2 + 2i\,a_3e_3$ , and since the basis is orthonormal,

\|Tv\|^2 = 9|a_1|^2 + 16|a_2|^2 + 4|a_3|^2 \le 16\big(|a_1|^2 + |a_2|^2 + |a_3|^2\big) = 16\|v\|^2.

So $\|Tv\| \le 4\|v\|$ , and $\|Te_2\| = 4$ with $\|e_2\| = 1$ . By formula (c), $\|T\| = 4$ .

Problem 2 · When Eigenvalues Mislead

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ defined by $T(x, y) = (x + 4y, y)$ — find $\|T\|$ .

✅ Correct! The largest eigenvalue of

T^*T

9 + 4\sqrt{5} = (2 + \sqrt{5})^2

❌ Not quite.

1

is the only eigenvalue of

T

, but

T

has no orthonormal eigenbasis, and

\|T(0, 1)\| = \|(4, 1)\| = \sqrt{17} > 1

already.

❌ Not quite. The largest matrix entry is not the norm:

\|T(0, 1)\| = \sqrt{17} > 4

❌ Not quite.

\sqrt{1 + 16 + 1}

is the square root of the sum of the squared entries, which is

\sqrt{s_1^2 + s_2^2}

, larger than

s_1

Show solution

With respect to the standard basis, $\mathcal{M}(T) = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}$ , so

\mathcal{M}(T^*T) = \begin{pmatrix} 1 & 0 \\ 4 & 1 \end{pmatrix}\begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 4 & 17 \end{pmatrix}.

Its eigenvalues solve $\lambda^2 - 18\lambda + 1 = 0$ , so $\lambda = 9 \pm 4\sqrt{5}$ . The largest is $9 + 4\sqrt{5} = (2 + \sqrt{5})^2$ , hence $\|T\| = s_1 = 2 + \sqrt{5} \approx 4.24$ .

Problem 3 · The Norm of an Adjoint

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

✅ Correct!

\|T^*\| = \|T\| = \sqrt{3}

, the square root of the largest eigenvalue

3

T^*T

❌ Not quite.

2

is the square root of the sum of the squared entries,

\sqrt{s_1^2 + s_2^2} = \sqrt{3 + 1}

❌ Not quite.

3

is the largest eigenvalue of

T^*T

; the norm is its square root.

❌ Not quite.

\|T^*\| = \|T\|

, so compute

\mathcal{M}(T^*T)

and its largest eigenvalue.

Show solution

A linear map and its adjoint have the same norm, so it suffices to find $\|T\|$ . With respect to the standard bases,

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

The eigenvalues of $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ are $3$ (eigenvector $(1, 1)$ ) and $1$ (eigenvector $(1, -1)$ ). So $s_1 = \sqrt{3}$ and $\|T^*\| = \|T\| = \sqrt{3}$ . Check: $T(1, 1)/\sqrt{2} = (1, 1, 2)/\sqrt{2}$ has length $\sqrt{6}/\sqrt{2} = \sqrt{3}$ .

Problem 4 · The Norm of an Inverse

Given: an invertible $T \in \mathcal{L}(\mathbf{R}^3)$ whose singular values are $6$ , $2$ , $\tfrac14$ — find $\|T^{-1}\|$ .

✅ Correct! The singular values of

T^{-1}

are

4

\tfrac12

\tfrac16

, and the largest is

4

❌ Not quite.

\tfrac16 = 1/\|T\|

is the smallest singular value of

T^{-1}

. The largest comes from the smallest singular value of

T

❌ Not quite. The singular values of

T^{-1}

are the reciprocals of those of

T

Show solution

The singular values of $T^{-1}$ are the reciprocals of the singular values of $T$ : $\tfrac16$ , $\tfrac12$ , $4$ . By formula (a), $\|T^{-1}\|$ is the largest of these:

\|T^{-1}\| = \frac{1}{1/4} = 4.

So a small singular value of $T$ makes $T^{-1}$ large: $T$ shrinks some unit vector to length $\tfrac14$ , and $T^{-1}$ must stretch it back by a factor of $4$ .

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Computing the Norm of T: Three Formulas and Examples

IKey Formulas

IIApproximating the Largest Eigenvalue of T∗TT^*TT∗T

IIITTT and T∗T^*T∗ Stretch by the Same Maximum

IVQuiz Questions

IIApproximating the Largest Eigenvalue of $T^*T$

III $T$ and $T^*$ Stretch by the Same Maximum