1. Let the column vectors a,b,c in R4 and the matrix A be as in the previous exercise:
a=[1; 2; 3; 4], b=[4; 3; 2; 1], c=2*a-3*b, A=[a b c]
Quickly repeat problems 4-6 of the previous exercise. We saw there that a,b,c are not linearly independent.
Show that the following are linearly independent:
(a) a,b hint: B=[a b], solve Bx=0 by Gauss elimination
(b) a,c hint: C=[a c], solve Cx=0 by Gauss elimination
(c) b,c hint: D=[b c], solve Dx=0 by Gauss elimination
Since you have already done Gauss elimination by hand several times, you can now save time by using the matlab function rref.
2. Basis, Ch 42.6-11. This means that we have found three bases for the space V=S(a,b,c), namely, a,b and a,c and b,c. Remember that the vector v=[1;1;1;1] belongs to V (Studio 6.2 exercise 4). Write v uniquely as a linear combination of
(a) a,b hint: solve Bx=v (use rref([B v])), then v=x1a+x2b
(b) a,c hint: solve Cy=v, then v=y1a+y2c
(c) b,c hint: solve Dz=v, then v=z1b+z2c
3. Linear function, Ch 42.16-17. Let f(x)=Ax with A as before. Show that
(a) f : R3 -> R4
(b) f is linear
4. Range R(A), Ch 42.11. The range (värderummet, värdemängden) of the linear function f(x)=Ax is
R(A)={y in R4 : y=Ax, x in R3}.
(a) Show that R(A) = V = S(a,b,c).
(b) Show that R(A) is a linear subspace of R4.
(c) Find more than one basis for R(A). (See exercise 2.)
Answer: (a) y = Ax = x1a+x2b+x3c . (b) y,v in R(A), a,b numbers => y=Ax, v=Au => ay+bv =A(ax+bu) => ay+bv in R(A).
5. Null space, Ch 42.12. The null space (nollrummet) of the linear function f(x)=Ax is
N(A)={x in R3 : Ax=0}.
(a) Show that N(A) is a linear subspace of R3.
(b) Find a basis for N(A).
Answer: (a) Solve Ax=0 by Gauss elimination. The solution is x=t[-2;3;1], where t is an arbitrary number. So N(A)=S(g) where g=[-2;3;1] and g is a basis for N(A), i.e., N(A)=S(g).
6. Solution of Ax=y, Ch 41.14. Solve Ax=y with
(a) y=[1;1;1;1]
(b) y=[1;1;1;0]
Answer: (a) x=[0.2;0.2;0]+t[-2;3;1]=xp+xh, where Axp=y and Axh=0, i.e., xh in N(A). Non-unique solution. This means that N(A) is not 0. (b) No solution. This means that y does not belong to R(A).
7. Let the matrices B, C, D be as before. We know that R(B)=R(C)=R(D)=R(A)=S(a,b,c)=V.
(a) Find N(B), N(C), N(D).
(b) Solve Bx=y with y=[1;1;1;1].
(c) Solve Cx=y with y=[1;1;1;1].
(d) Solve Dx=y with y=[1;1;1;1].
Answer: (a) N(B)=N(C)=N(D)=0 (this means the zero space consisting of only the zero vector). (b) x=[0.2;0.2], unique solution because N(B)=0.
8. Transpose of the linear function f(x)=Ax, Ch 42.20.
(a) Show that (AB)T=BTAT. Hint: use
formula (42.44).
(b) Show that (Ax,y)= (x,ATy).
Hint: remember that (x,y)=yTx
and use (a).
(c) Illustrate this by computing (y'*A)*x and (A'*y)'*x
for some A,x,y.
/stig