Tensor analysis — scalars, vectors, rank-2 tensors: Worked examples

Before anything, a reminder of the two objects we keep using, in plain words:

Definition Index notation, in plain words (read this before the formulas)

A rank-2 tensor's entries carry two labels. We write $T_{ij}$ (or $T^{ij}$ , or the mixed $T^{i}_{j}$ ) where each of $i, j$ runs over $1, 2$ in 2D. Think of $i$ = row, $j$ = column, so $T_{11}$ is top-left, $T_{12}$ top-right, and so on.

A lower index (like $T_{ij}$ ) is called covariant; an upper index (like $T^{ij}$ ) is contravariant. In the Cartesian, rotation-only world of this page they have the same numbers — the distinction only bites in curved/non-orthonormal coordinates (see Ex 6). We keep both notations only because the parent note uses them.
The Einstein summation convention means: whenever the same index appears once up and once down in a product, sum over it. So $A^{ij} V_{j}$ silently means $\sum_{j} A^{ij} V_{j}$ . No $Σ$ symbol is written; the repeated index is the instruction to add.
Example: $tr T = T^{1}_{1} + T^{2}_{2}$ is just "add the two diagonal entries."

The scenario matrix

Everything a rank-2 tensor problem can throw at a first learner sorts into these cells. Each column is a distinct kind of difficulty; the examples below are labelled with the cell(s) they cover.

Cell	Case class	What's tricky	Covered by
A	Symmetric tensor, "nice" rotation ( $4 5^{\circ}$ )	diagonalising, principal axes	Ex 1
B	Symmetric tensor, "ugly" angle ( $3 0^{\circ}$ )	messy $cos, sin$ , invariants still hold	Ex 2
C	Antisymmetric tensor	off-diagonal only, what rotation does to it	Ex 3
D	Degenerate input: zero / isotropic tensor	is $0$ or $λ I$ a tensor?	Ex 4
E	Sign cases: negative & mixed-sign entries	tension vs compression, sign of trace/det	Ex 5
F	Limiting / blow-up coordinate (polar $r \to 0$ )	metric degenerates, $det g \to 0$	Ex 6
G	Real-world word problem	translate physics → tensor	Ex 7
H	Exam twist: quotient theorem / "is it a tensor?"	prove without brute force	Ex 8

The invariants we sanity-check on almost every example are the three numbers a rotation can never change:

Example 1 — Cell A: symmetric tensor, clean $4 5^{\circ}$

Worked example Rotate

$T = (3113)$ MPa by $θ = 4 5^{\circ}$ Forecast first: the two $1$ 's are the shear. Guess: after a smart rotation, can the shear vanish? If so, what fills the diagonal?

Step 1 — Write $R$ for $4 5^{\circ}$ .

=\frac{1}{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}.$$ *Why this step?* $\cos45^\circ=\sin45^\circ=\tfrac{1}{\sqrt2}$; plug into the rotation template. **Step 2 — Apply the rank-2 rule $\tilde T=RTR^{\top}$.** $$RT=\frac{1}{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix}3&1\\1&3\end{pmatrix} =\frac{1}{\sqrt2}\begin{pmatrix}2&-2\\4&4\end{pmatrix}.$$ *Why this step?* Each index of $T$ eats one rotation factor; matrix multiplication is exactly that summation over the shared index. **Step 3 — Multiply by $R^{\top}$.** $$\tilde T=\frac{1}{\sqrt2}\begin{pmatrix}2&-2\\4&4\end{pmatrix}\cdot\frac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix} =\begin{pmatrix}2&0\\0&4\end{pmatrix}.$$ *Why this step?* The second $R^{\top}$ rotates the *other* slot. Off-diagonals died → these are the ==principal axes==. (The eigenvalues $2,4$ now sit on the diagonal; which one lands top-left depends on the rotation direction — here $45^\circ$ counter-clockwise puts $2$ first.) ![[deepdives/dd-maths-4.10.07-d3-s01.png]] **Figure s01.** The purple ellipse is the "stress ellipse" of $T$; its two axes point along the principal directions found in Step 3. The magenta arrow marks the axis with eigenvalue $4$, the orange arrow the axis with eigenvalue $2$. Read it as: rotating the frame to line up with these arrows is exactly what kills the off-diagonal shear. **Verify:** $\operatorname{tr}T=3+3=6$ and $\operatorname{tr}\tilde T=2+4=6$ ✔. $\det T=3\cdot3-1\cdot1=8$, $\det\tilde T=2\cdot4=8$ ✔. Eigenvalues of $T$: solve $(3-\lambda)^2-1=0\Rightarrow\lambda=4,2$ — exactly the new diagonal. Units MPa throughout ✔.

Example 2 — Cell B: symmetric tensor, ugly $3 0^{\circ}$

Worked example Same

$T = (3113)$ , now $θ = 3 0^{\circ}$ Forecast: at $3 0^{\circ}$ we are not at the principal axes (those were at $4 5^{\circ}$ ), so the off-diagonal should be nonzero but the trace should still be $6$ . Guess the new shear sign.

Step 1 — Exact numbers for $3 0^{\circ}$ . Keep them exact: $cos 3 0^{\circ} = \frac{3}{2}$ , $sin 3 0^{\circ} = \frac{1}{2}$ . $R = (\frac{3}{2} \frac{1}{2} - \frac{1}{2} \frac{3}{2}) .$ Why this step? We keep surds ( $3$ ) rather than decimals so the invariants come out as clean integers at the end — "ugly angle" means irrational entries, not sloppier arithmetic.

Step 2 — Use the closed form for a symmetric $2 \times 2$ under rotation. For $T = (a b b a)$ , $\tilde{T}_{11} = a + b sin 2 θ, \tilde{T}_{22} = a - b sin 2 θ, \tilde{T}_{12} = b cos 2 θ .$ Why this formula holds (derivation): multiply out $\tilde{T} = R T R^{⊤}$ with $R = (c s - s c)$ (writing $c = cos θ, s = sin θ$ ) and $a$ on both diagonal entries of $T$ . The top-left entry becomes $\tilde{T}_{11} = a c^{2} + 2 b cs + a s^{2} = a (c^{2} + s^{2}) + b (2 cs) = a + b sin 2 θ$ , using $c^{2} + s^{2} = 1$ and the double-angle identity $2 cs = sin 2 θ$ . Likewise $\tilde{T}_{12} = b (c^{2} - s^{2}) = b cos 2 θ$ (double-angle $c^{2} - s^{2} = cos 2 θ$ ) and $\tilde{T}_{22} = a - b sin 2 θ$ . That is why only $2 θ$ appears — a rotation of the axes by $θ$ rotates the stress pattern by $2 θ$ . With $a = 3, b = 1, 2 θ = 6 0^{\circ}$ : $sin 6 0^{\circ} = \frac{3}{2}$ , $cos 6 0^{\circ} = \frac{1}{2}$ .

Step 3 — Plug in the exact values.

\approx\begin{pmatrix}3.8660&0.5\\0.5&2.1340\end{pmatrix}\text{ MPa}.$$ *Why this step?* The exact form makes the trace obviously $6$; the decimals are only for reading off sizes. The off-diagonal $\tfrac12$ is the leftover shear because $30^\circ$ isn't the principal angle. **Verify (exact):** trace $=(3+\tfrac{\sqrt3}{2})+(3-\tfrac{\sqrt3}{2})=6$ ✔ (surds cancel). $\det=(3+\tfrac{\sqrt3}{2})(3-\tfrac{\sqrt3}{2})-(\tfrac12)^2=9-\tfrac34-\tfrac14=8$ ✔. Off-diagonal $\tfrac12\neq0$ confirms we are *not* on the principal axes — consistent with Ex 1.

Example 3 — Cell C: antisymmetric tensor

Worked example Rotate the antisymmetric

$A = (0 - 2 20)$ by any $θ$ (use $6 0^{\circ}$ ) Forecast: antisymmetric means $A_{ij} = - A_{j i}$ . A guess worth testing: does an antisymmetric tensor stay antisymmetric after rotation, and does the single number $2$ survive?

Step 1 — Recall the split. Any tensor $= \frac{1}{2} (T + T^{⊤}) + \frac{1}{2} (T - T^{⊤})$ : symmetric + antisymmetric. $A$ is purely the antisymmetric part. Why this step? Symmetry type is coordinate-independent, so rotation cannot leak $A$ into a symmetric piece — a strong prediction to test.

Step 2 — Show $\tilde{A}$ is antisymmetric first. With $R$ orthogonal, $(R A R^{⊤})^{⊤} = R A^{⊤} R^{⊤} = R (- A) R^{⊤} = - R A R^{⊤}$ , so $\tilde{A} = - \tilde{A}^{⊤}$ . Why this step? Proving the structure first tells us $\tilde{A} = (0 - c c 0)$ for a single number $c$ ; we only need to find that one $c$ .

Step 3 — Compute $c$ explicitly. Write $R = (cos θ sin θ - sin θ cos θ)$ and multiply out $\tilde{A} = R A R^{⊤}$ . The top-right entry is

=\underbrace{(\cos\theta)(A_{12})(\cos\theta)}_{\text{from }R_{1\cdot},\,R^{\top}_{\cdot2}} +\underbrace{(-\sin\theta)(A_{21})(-\sin\theta)}_{} =A_{12}\cos^2\theta - A_{21}\sin^2\theta.$$ Because $A_{21}=-A_{12}$, this becomes $A_{12}\cos^2\theta+A_{12}\sin^2\theta=A_{12}(\cos^2\theta+\sin^2\theta)$. Now the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$ collapses it to $c=A_{12}=2$. $$\tilde A=\begin{pmatrix}0&2\\-2&0\end{pmatrix}.$$ *Why this step?* Writing the entry term-by-term shows *exactly where* $\cos^2+\sin^2$ comes from — it is the antisymmetry $A_{21}=-A_{12}$ turning a minus into a plus. The lone component is a rotational **invariant** (the 2D "curl"/spin). **Verify:** trace of any antisymmetric $2\times2$ is $0$: $0+0=0$ before and after ✔. $\det A=0\cdot0-2\cdot(-2)=4$; $\det\tilde A=4$ ✔. Antisymmetry preserved ✔.

Example 4 — Cell D: degenerate inputs (zero & isotropic)

Worked example Is the zero tensor

$0$ a tensor? Is $λ I$ frame-independent? Forecast: guess whether $(0000)$ and $(5005)$ keep their form in every frame.

Step 1 — Zero tensor. $\tilde{0} = R 0 R^{⊤} = 0$ for every $R$ . Why this step? Multiplying anything by the zero matrix gives zero; the transformation law is satisfied trivially, so $0$ is a valid (rank-2) tensor.

Step 2 — Isotropic tensor $T = λ I$ . $\tilde{T} = R (λ I) R^{⊤} = λ R R^{⊤} = λ I .$ Why this step? $R R^{⊤} = I$ by orthonormality; the scalar $λ$ passes through. So a pure "pressure" state (equal normal stress, no shear) looks the same on every rotated axis.

Figure s02. The purple circle is the stress "ellipse" of an isotropic tensor

λ I

— a perfect circle because both eigenvalues are equal. The magenta pair of arrows is one frame; the orange pairs are rotated frames. Notice every frame reads off the same radius in every direction: that is what "same components in every frame" looks like.

Verify: with $λ = 5$ : trace $= 10$ both frames ✔; $det = 25$ both frames ✔; eigenvalues both $5$ (repeated) ✔. Zero tensor: trace $0$ , det $0$ , invariant ✔.

Example 5 — Cell E: sign cases (compression, mixed sign)

Worked example

$T = (- 4 0 06)$ MPa — one axis in compression, one in tension. Rotate $4 5^{\circ}$ . Forecast: negative on one axis (compression), positive on the other (tension). Guess: does the $4 5^{\circ}$ frame create shear, and what is the sign of the trace/determinant?

Step 1 — Read the signs physically. $- 4$ = pushing in (compression), $+ 6$ = pulling out (tension). Trace $= - 4 + 6 = 2$ (net tension). $det = - 24 < 0$ : opposite signs $\Rightarrow$ a saddle stress state. Why "saddle"? When the two eigenvalues have opposite signs (equivalently $det T < 0$ ), the material is being pulled apart along one principal axis while squeezed along the other — geometrically the stress "surface" rises in one direction and falls in the perpendicular one, exactly the shape of a Pringle/horse-saddle. It is the tensor signature of a mixed-sign state.

Step 2 — Rotate by $4 5^{\circ}$ using the closed form (now $b = 0$ , but $a_{1} \neq = a_{2}$ ): for diagonal $(p 0 0 q)$ , $\tilde{T} = \frac{1}{2} (p + q p - q p - q p + q)$ . With $p = - 4, q = 6$ : $\tilde{T} = (1 - 5 - 5 1)$ MPa. Why this step? Rotating away from the principal axes creates shear — the exact opposite of Ex 1. The $4 5^{\circ}$ direction feels the maximum shear $\frac{∣ q - p ∣}{2} = 5$ MPa (sign depends on rotation sense; magnitude $5$ is what matters physically).

Verify: trace $= 1 + 1 = 2$ ✔ (matches). $det = 1 - 25 = - 24$ ✔ (sign preserved, still a saddle). Eigenvalues of $\tilde{T}$ : $1 \pm 5 = {6, - 4}$ — same as original ✔.

Example 6 — Cell F: limiting / blow-up coordinate (polar $r \to 0$ )

Worked example The polar metric

$g_{ij} = (10 0 r^{2})$ as $r \to 0$ Forecast: the metric encodes distance $d s^{2} = d r^{2} + r^{2} d θ^{2}$ . Guess what goes wrong at the origin $r = 0$ — is it a real hole in space, or just the coordinates failing?

Step 1 — Determinant of the metric. $det g = 1 \cdot r^{2} - 0 = r^{2}$ . Why this step? $det g$ is the area element $r d r d θ$ ; if $det g \to 0$ the coordinate patch is degenerating (a whole circle of $θ$ collapses to one point).

Step 2 — Take the limit $r \to 0$ . $det g \to 0$ , so $g$ becomes non-invertible: the inverse metric $g^{ij}$ defined just above no longer exists, and you cannot raise the $θ$ index. The angle $θ$ is undefined at the origin. Here the upper-vs-lower distinction from the opening definitions finally bites: $V_{i} = g_{ij} V^{j}$ needs an invertible $g$ to run backwards. Why this step? A metric must be invertible to define $g^{ij}$ ; failure here is a coordinate singularity, not a curvature singularity — the flat plane is perfectly smooth at its centre.

Figure s03. The magenta curve is

det g = r^{2}

plotted against radius

r

. It is a happy positive parabola everywhere except at the violet dot

r = 0

, where it touches zero. That single touch-point is the coordinate singularity: the polar chart "pinches shut" at the origin — a full circle of

θ

values all crowd into one point — even though the flat plane underneath is perfectly smooth there.

Verify: at $r = 2$ , $det g = 4 > 0$ (fine); at $r = 0$ , $det g = 0$ (degenerate) ✔. Compare Cartesian $g = δ_{ij}$ : $det = 1$ everywhere — the plane itself has no problem, confirming the singularity lives in the polar chart, matching the metric tensor story.

Example 7 — Cell G: real-world word problem

Worked example Moment-of-inertia tensor of a thin square plate

A thin flat plate has principal moments $I_{x} = 2, I_{y} = 8 kg\cdotm^{2}$ about its centre, no product of inertia. Engineer mounts it rotated $3 0^{\circ}$ . What is the product of inertia $I_{x y}$ felt along the new axes? Forecast: rotating away from principal axes should switch on a product of inertia. Guess its rough size.

Step 1 — Model as a rank-2 tensor. $I = (I_{x} 0 0 I_{y}) = (2008)$ . Why this step? Inertia obeys $\tilde{I} = R I R^{⊤}$ — it is a rank-2 tensor (same machinery as the stress cases D/E), so the closed forms derived in Ex 2 apply verbatim.

Step 2 — Off-diagonal formula for a diagonal tensor. From the Ex 2 derivation with $b = 0$ on the original but distinct diagonal entries, the general diagonal-tensor result is $\tilde{I}_{x y} = \frac{1}{2} (I_{x} - I_{y}) sin 2 θ$ . With $2 θ = 6 0^{\circ}$ , $sin 6 0^{\circ} = \frac{3}{2}$ : $\tilde{I}_{x y} = \frac{1}{2} (2 - 8) \cdot \frac{3}{2} = - \frac{3 3}{2} \approx - 2.598 kg\cdotm^{2}$ , i.e. magnitude $2.598$ . Why this step? This is exactly the shear-generation formula from Ex 5 in physics clothing.

Step 3 — New diagonal. $\tilde{I}_{xx} = \frac{I _{x} + I _{y}}{2} - \frac{I _{y} - I _{x}}{2} cos 2 θ = 5 - 3 (0.5) = 3.5$ , and $\tilde{I}_{y y} = 5 + 1.5 = 6.5 kg\cdotm^{2}$ . Why this step? Same closed form as Ex 2 with the average $\frac{I _{x} + I _{y}}{2}$ playing the role of $a$ ; the state splits into the average $\pm$ a rotated swing.

Verify: trace $= 3.5 + 6.5 = 10 = 2 + 8$ ✔ (invariant). $det = 3.5 \cdot 6.5 - 2.59 8^{2} = 22.75 - 6.75 = 16 = 2 \cdot 8$ ✔. Product of inertia magnitude $\approx 2.6 kg\cdotm^{2}$ is nonzero as forecast; units kg·m² throughout ✔.

Example 8 — Cell H: exam twist (quotient theorem)

Worked example Prove

$A^{ij}$ is a tensor given only that $A^{ij} V_{j}$ is a vector for every covector $V_{j}$ Forecast: you are not given how $A$ transforms. Guess whether the behaviour of the output forces the tensor law on $A$ .

Step 1 — Name what we know. For every $V_{j}$ , the quantity $U^{i} = A^{ij} V_{j}$ (summed over $j$ by the Einstein convention) transforms as a contravariant vector: $\tilde{U}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} U^{k}$ . Why this step? The summation convention means $A^{ij} V_{j}$ sums over $j$ ; the hypothesis is a statement about this contraction.

Step 2 — Write both sides in the new frame. $\tilde{U}^{i} = \tilde{A}^{ij} \tilde{V}_{j}$ and also $\tilde{U}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k l} V_{l}$ . Using the covector rule $V_{l} = \frac{\partial x ~ ^{m}}{\partial x ^{l}} \tilde{V}_{m}$ , substitute and equate. Why this step? We force the two expressions for the same physical vector $\tilde{U}^{i}$ to agree.

Step 3 — Since $\tilde{V}_{j}$ is arbitrary, coefficients match. $\tilde{A}^{ij} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ~ ^{j}}{\partial x ^{l}} A^{k l} .$ Why this step? "True for every $V$ " kills the arbitrary factor and leaves precisely the contravariant rank-2 transformation law — so $A^{ij}$ is a tensor. QED.

Verify (numeric sanity): take $A = (3113)$ and the covector $V = (1, 0)$ , so $U = A V = (3, 1)$ . Now rotate by $9 0^{\circ}$ with $R = (01 - 1 0)$ . Transforming the output directly gives $\tilde{U} = R U = (- 1, 3)$ . Transforming the pieces instead: $\tilde{A} = R A R^{⊤} = (3 - 1 - 1 3)$ and $\tilde{V} = R V = (0, 1)$ , so $\tilde{A} \tilde{V} = (- 1, 3)$ . The two routes give the same $\tilde{U} = (- 1, 3)$ ✔ — exactly what the quotient theorem promises: transforming before or after the contraction agrees, so $A$ must be a tensor.

Recall Self-test

Why does $tr T$ survive any rotation? ::: Because $R T R^{⊤}$ is a similarity transform and trace is invariant under similarity (the Jacobians contract to $δ$ ). At a $4 5^{\circ}$ rotation of a diagonal tensor, what is the induced shear magnitude? ::: $\frac{1}{2} ∣ q - p ∣$ , the maximum shear — off-axis rotations create off-diagonals. Is $det g \to 0$ in polar as $r \to 0$ a real singularity? ::: No — it's a coordinate singularity; the flat plane is smooth, only the polar chart degenerates.