HW#1, solutions to selected problems

1) No. Each row operation can be undone by another row operation. (Switching two rows can be undone by switching the two rows again; multiplying a row by a nonzero number c can be undone by multiplying the row by 1/c; and adding c times the j^th row to the i^th row can be undone by adding (-c) times the j^th row to the i^th row.) If we could row reduce a nonzero matrix to the zero matrix, then we could work backwards to row reduce the zero matrix to a nonzero matrix. But every row operation sends the zero matrix to itself. (If you switch two zero rows, they're still zero; if you multiply a row of zeros by c it's still a row of zeros; and adding a multiple of zero to zero gives you zero.)

More generally we will see later that the rank of a matrix is well defined and unchanged by row operations.

2) First proof: suppose y and z are additive inverses of x. We want to show that y=z. We know that x+y=0 and x+z=0. So x+y=x+z. Now we want to subtract x from both sides to conclude that y=z. But what does it mean to subtract x from both sides? We have to add the additive inverse of x to both sides. Which additive inverse? Let's try adding y to both sides. So we have y+x+y=y+x+z. Since y+x=0, this gives 0+y=0+z, so y=z.

Slick proof: suppose y and z are additive inverses of x. Then y = y + 0 = y + (x + z) = (y+x)+z = 0 + z = z.

Remark: Of course the additive inverse of x equals (-1)x. To prove this, since we know that the additive inverse is unique, it is enough to show that (-1)x is an additive inverse of x, i.e. that x+(-1)x=0. But x+(-1)x=1x+(-1)x=(1+(-1))x=0x=0.

4) Section 1.2 problem 10. You have to check that the vector space operations are well defined, i.e. that the sum of two differentiable functions is differentiable, and that the product of a differentiable function by a scalar is differentiable. These are standard facts and you can just quote them. Then you need to check that the eight vector space axioms hold. For example, to prove (VS 1), we observe that if f and g are two differentiable functions, then for each real number x, we have (f+g)(x)=f(x)+g(x)=g(x)+f(x)=(g+f)(x). Here the first equality holds by the definition of addition of functions, the second equality holds because addition of real numbers is commutative, and the third equality holds again by the definition of addition of functions. Since this holds for each real number x, it follows that f+g=g+f. The verification of the other vector space axioms is similar; in a fairly straightforward manner, you reduce to standard algebraic properties of the real numbers.

5) First part: if y and z are two solutions to the system Ax=b, so that Ay=b and Az=b, then A(y-z)=Ay-Az=b-b=0, so y-z is a solution to the system Ax=0.

Second part: suppose Ax=0 has only the trivial solution. Suppose y and z are two solutions to Ax=b. Then by the first part, y-z is a solution to Ax=0. By our assumption, y-z is the trivial solution, i.e. y-z=0, so y=z.

Third part: suppose Ax=0 has a nontrivial solution z. Then cz is also a solution to Ax=0 for every scalar c, because A(cz)=c(Az)=c(0)=0. Since z is nonzero and our field F has infinitely many elements (because in this course we always take F to be the real numbers or the complex numbers), in this way we get infinitely many solutions cz to Ax=0. If y is a solution to Ax=b, then for each such c we get a different solution to Ax=b, namely y+cz, because A(y+cz)=Ay+A(cz)=b+0=b.

We will see later that the set of solutions to Ax=b, if it is nonempty, is an ``affine subspace'' parallel to the nullspace of A.