MATH 271: Linear Algebra and Partial Differential Equations
Lecture 1: Heat Exchange, Modes, and the Spectral Viewpoint
Tsogtgerel Gantumur
Tuesday, September 1, 2026
A recurring idea in this course is that a linear problem may become much simpler after we find
the right directions in which to describe it. In finite dimensions these directions are eigenvectors.
Later, for differential equations, they will become eigenfunctions.
We begin with a concrete physical problem rather than with an abstract matrix.
1. Two identical bodies exchanging heat
Consider two identical bodies with temperatures u1 (t) and u2 (t). Let C > 0 be the heat capacity of
each body. Assume that the heat current from body 1 to body 2 is proportional to the temperature
difference:
heat current from 1 to 2 = k(u1 − u2 ),
k > 0.
The same current leaves one body and enters the other. Conservation of energy therefore gives
Cu′1 = −k(u1 − u2 ),
Cu′2 = k(u1 − u2 ).
Writing
α=
we obtain
k
> 0,
C
u′1 = α(u2 − u1 ),
u′2 = α(u1 − u2 ).
Two physical facts are already visible:
• the hotter body cools while the colder body warms;
• heat is only exchanged internally, so the total heat is conserved.
2. Solving first as scalar equations
Add the two equations in (1):
(u1 + u2 )′ = 0.
1
(1)
Hence
u1 (t) + u2 (t) = u1 (0) + u2 (0).
The average temperature
m=
u1 + u2
2
is constant.
Now subtract the equations:
(u1 − u2 )′ = −2α(u1 − u2 ).
If
d=
u1 − u2
,
2
then
d′ = −2αd,
d(t) = d(0)e−2αt .
Therefore
u1 (t) = m + d(0)e−2αt ,
u2 (t) = m − d(0)e−2αt .
Thus the evolution splits into two independent pieces:
• the common-temperature part stays fixed;
• the temperature-difference part decays exponentially.
In particular, we have
u1 (t), u2 (t) −→
u1 (0) + u2 (0)
2
(t → ∞).
3. The same calculation in matrix form
Set
u1 (t)
.
u(t) =
u2 (t)
Then (1) is
′
u = Lu,
−1 1
L=α
.
1 −1
The two directions that appeared naturally in the scalar calculation are
1
1
q0 =
,
q1 =
.
1
−1
A direct calculation gives
Lq1 = −2αq1 .
Lq0 = 0q0 ,
Thus q0 and q1 are eigenvectors, with eigenvalues
λ1 = −2α.
λ0 = 0,
2
Writing
u(0) = c0 q0 + c1 q1 ,
we obtain
u(t) = c0 q0 + c1 e−2αt q1 .
This is exactly the scalar solution, written in a language that extends immediately to larger systems.
The combinations u1 + u2 and u1 − u2 were therefore not accidental: they are the coordinates of
the state along two eigenvector directions.
Eigenvectors are modes. Suppose
Lq = λq
and look for a solution of u′ = Lu in the form
u(t) = a(t)q.
Then
a′ q = aLq = λaq,
so
a′ = λa,
a(t) = a(0)eλt .
An eigenvector is therefore a direction that evolves by itself. We call such a direction a mode; the
eigenvalue tells us how that mode evolves.
4. Three bodies
For three identical bodies, let kij = kji ≥ 0 denote the exchange rate between bodies i and j (after
dividing by the common heat capacity). The symmetry kij = kji expresses that the heat transferred
across a given contact is the same heat leaving one body and entering the other; the conductance
belongs to the pair, not to a direction of transfer.
The equations are
u′1 = k12 (u2 − u1 ) + k13 (u3 − u1 ),
u′2 = k12 (u1 − u2 ) + k23 (u3 − u2 ),
u′3 = k13 (u1 − u3 ) + k23 (u2 − u3 ).
In matrix form,
u′ = Lu,
with
Notice that


−(k12 + k13 )
k12
k13
.
k12
−(k12 + k23 )
k23
L=
k13
k23
−(k13 + k23 )
 
1
L 1 = 0.
1
Thus the common-temperature direction is again a zero mode.
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The same strategy as before is now clear: rather than solving the three coupled scalar equations
directly, look for special directions q satisfying
Lq = λq.
Along each such direction the vector problem again reduces to one scalar equation.
5. The general N -dimensional recipe
Consider a linear system
u′ (t) = Au(t),
u(t) ∈ RN .
Suppose that A has a basis of eigenvectors
q1 , . . . , q N ,
Aqj = λj qj .
Write the initial state in this basis:
u(0) = c1 q1 + · · · + cN qN .
Then each mode evolves independently:
u(t) =
N
X
cj eλj t qj .
j=1
If

λ1

Λ=
0
Q = q1 · · · qN ,
0
..


,
.
λN
then
AQ = QΛ.
If Q is invertible, this is equivalent to
A = QΛQ−1 .
With modal coordinates
a = Q−1 u,
the coupled vector equation becomes
a′ = Λa,
that is,
a′j = λj aj
(j = 1, . . . , N ).
So diagonalization converts a coupled linear system into independent scalar equations.
6. The spectral theorem
For an arbitrary matrix there need not be enough independent eigenvectors to form a basis. Real
symmetric matrices are much better behaved.
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Spectral theorem for real symmetric matrices. If A = AT is a real symmetric N ×N matrix,
then all eigenvalues of A are real and there is an orthonormal basis of RN consisting of eigenvectors
of A.
Equivalently, there is an orthogonal matrix Q and a real diagonal matrix Λ such that
A = QΛQT ,
QT Q = I.
The columns of Q are eigenvectors of A, and the corresponding diagonal entries of Λ are their
eigenvalues. Indeed,
AQ = QΛ
means column by column that
Aqj = λj qj .
This is why the spectral theorem is precisely a theorem about eigenvectors and eigenvalues.
The collection of eigenvalues is called the spectrum. In this course we will repeatedly use the same
pattern:
find spectral modes −→ expand the state in those modes −→ solve scalar equations.
For matrices the modes are eigenvectors. Later, for differential operators, they will be eigenfunctions.
Practice
1. For the two-body system, verify directly that
1
,
q0 =
1
1
q1 =
−1
are eigenvectors of L, and find the corresponding eigenvalues.
2. Let u1 (0) = 10 and u2 (0) = 2. Solve the two-body heat-exchange system explicitly and
determine the limiting temperature.
3. Show directly from the differential equations that u1 + u2 is constant in time.
4. Suppose Aqj = λj qj for a basis q1 , . . . , qN . Starting from
u(0) =
N
X
cj qj ,
j=1
verify directly that
u(t) =
N
X
j=1
solves u′ = Au.
5
cj eλj t qj
5. For the three-body heat-exchange matrix above, verify that
 
1
1
1
is a zero eigenvector and explain its physical meaning.
6. Different heat capacities. Suppose N bodies have positive heat capacities C1 , . . . , CN ,
and symmetric pairwise conductances kij = kji ≥ 0. Let


C1
0


..
C=
,
.
0
CN
and let K be the symmetric exchange matrix defined by
X
(Ku)i =
kij (uj − ui ).
j̸=i
Then the temperature equation is
Cu′ = Ku,
or
u′ = C −1 Ku.
The matrix C −1 K need not be symmetric.
Since each Ci > 0, define
√
C1

1/2
C
=
0
0
..
.
√
 √
1/ C1

−1/2
C
=
0


,
CN
0
..
.
√
1/ CN


.
Thus
C 1/2 C 1/2 = C,
C −1/2 C −1/2 = C −1 ,
C 1/2 C −1 = C −1/2 .
Now introduce the new variable
v = C 1/2 u.
Since C is constant in time,
v ′ = C 1/2 u′ .
Show that
v ′ = C −1/2 KC −1/2 v,
and verify that the matrix in parentheses is symmetric.
Thus unequal heat capacities do not destroy the symmetric spectral structure; they merely
hide it in the original temperature coordinates.
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