MS-C1342 Linear algebra, V/2023 Noferini / Puska
Linear algebra
Exercise sheet 2 / Model solutions
1. (a) Show that k · k1 and k · k∞ satisfy the three conditions in the definition of a norm. (See
Definition 2.1 in the Lecture notes.)
(b) Let x = [1, 2, 3]T ∈ R3 . Calculate kxk1 , kxk2 and kxk∞ .
Solution.
(a) For the 1-norm, kxk1 , being the sum of the absolute values of the components of x, is
clearly never negative and zero if and only if x = 0. Moreover,
n
X n
X
kαxk1 = |αxi | = |α| |xi | = |α|kxk1 .
i=1 i=1
Lastly,
n
X n
X
kx + yk1 = |xi + yi | ≤ |xi | + |yi | = kxk1 + kyk1 .
i=1 i=1
For the ∞-norm, kxk∞ , being the maximum of the absolute values of the components
of x, is clearly never negative and zero if and only if x = 0. Moreover,
kαxk∞ = max |αxi | = |α| max |xi | = |α|kxk∞ .
i=1,...,n i=1,...,n
Lastly,
kx+yk∞ = max |xi +yi | ≤ max (|xi |+|yi |) ≤ max |xi |+ max |yi | = kxk∞ +kyk∞ .
i=1,...,n i=1,...,n i=1,...,n i=1,...,n
(b) We have
kxk1 = |1| + |2| + |3| = 6,
√ √
kxk2 = 12 + 22 + 32 = 14,
kxk∞ = max{|1|, |2|, |3|} = 3.
2. Let W1 , W2 ⊂ Rn be subspaces with bases
{q 1 , . . . , q k } and {v 1 , . . . , v m }, respectively.
Denote Q = q 1 , . . . , q k and V = v 1 , . . . , v m .
(a) Show that W1 + W2 = R( Q V ), where we define W1 + W2 as the vector subspace
{x + y | x ∈ W1 , y ∈ W2 } of Rn .
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MS-C1342 Linear algebra, V/2023 Noferini / Puska
(b) Let T : Rk+m → Rk be such that
(T x)i = xi
for i = 1, . . . , k.
Show that W1 ∩ W2 = {QT x | x ∈ N ( Q V )}.
Solution.
(a) By definition, the range of a matrix is the linear span of its columns. The sum W1 + W2
is equal to the span of W1 ∪ W2 . This proves the equality.
(b) If y ∈ W1 ∩ W2 then y = Qa = V b for some vectors a and b. If we set
a
x= ,
−b
then we get [Q V ]x = 0, if and only if x ∈ N ([Q V ]). Clearly, T x = a and so
QT x = y. This shows that W1 ∩ W2 ⊆ N ([Q V ]).
Conversely, let x ∈ N ([Q V ]) and write
Tx
x= ,
y
for some y ∈ Rm . Then we have that QT x = −V y, and hence QT x ∈ W1 ∩ W2 .
This proves N ([Q V ]) ⊆ W1 ∩ W2 and concludes the proof.
3. Let A ∈ Rm×n .
(a) Show that y T x = 0 for any x ∈ N (A) and y ∈ R(AT ). (Hint: y = AT z for some
z ∈ Rm .)
(b) Let x ∈ Rn be such that y T x = 0 for any y ∈ R(AT ). Show that x ∈ N (A). (Hint:
Choose y = AT Ax.)
Solution.
(a) By definition of R(AT ) there exists z ∈ Rm such that y = AT z. Hence,
T
y T x = AT z x.
Using the calculation rules for the transpose (AB)T = (B T AT ) and (AT )T = A, one
gets
T T
AT z x = z T AT x = z T Ax.
By noticing that x ∈ N (A), so that Ax = 0, the proof is complete.
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MS-C1342 Linear algebra, V/2023 Noferini / Puska
(b) Choosing y = AT Ax gives
T
y T x = AT Ax x.
Using calculation rules of the transpose,
T
AT Ax x = xT AT Ax = kAxk22 .
Hence, by assumption, kAxk22 = 0. By the properties of a norm, this implies that
Ax = 0, thus x ∈ N (A).
4. Use Matlab to visualize the set
S := {x ∈ R2 | kxk∗ = 1},
for ∗ = 1, 2 or ∞. You may modify the function plot_norm.m found at the MyCourses
page. Return both the script that you wrote and a printout of the resulting figure.
When kxk2 = 1, are kxk1 ja kxk∞ larger or smaller that one? Which four vectors x ∈ R2
satisfy
kxk1 = kxk2 = kxk∞ = 1?
Justify your answer based on the figure that you draw.
Solution: The challenge in the visualization of the set S is to find a parametric presentation
for it. Any x ∈ R2 can be written as
sin θ
x = rv θ , where v θ = , r ∈ R, r ≥ 0 and θ ∈ [0, 2π].
cos θ
Now, one has
S = {rv θ | r ∈ R, r ≥ 0, θ ∈ [0, 2π] and krv θ k∗ = 1}.
By the properties of a norm, rkv θ k∗ = 1. This implies that r = 1/kv θ k∗ and thus
vθ
S= | θ ∈ [0, 2π] .
kv θ k
The set S is now easy to draw. Here follows the code.
N = 1000;
t = linspace(0,2*pi,N);
x = cos(t);
y = sin(t);
figure;
for p=[1,2,Inf]
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MS-C1342 Linear algebra, V/2023 Noferini / Puska
for i=1:N
v = [x(i) ; y(i)];
rho = 1./norm(v,p);
xplot(i) = rho*x(i);
yplot(i) = rho*y(i);
end
hold on;
plot(xplot,yplot);
end
By the figure one gets kxk∞ ≤ kxk2 ≤ kxk1 , where the equality holds for x = ±e1 and
x = ±e2 .
1 1-norm
2-norm
Inf-norm
-1
-1 0 1
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