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## Solution to Linear Algebra Done Wrong Exercise 1.1.3

True or false:

a) Every vector space contains a zero vector;
b) A vector space can have more than one zero vector;
c) An $m\times n$ matrix has $m$ rows and $n$ columns;
d) If $f$ and $g$ are polynomials of degree $n$, then $f+g$ is also a polynomial of degree $n$;
e) If $f$ and $g$ are polynomials of degree at most $n$, then $f+g$ is also a polynomial of degree at most $n$.

Solution:

a) Yes. Let $a,b$ be complex numbers, and $f(x),g(x)$ continuous functions on the interval $[0,1]$. Then $af(x)+bg(x)$ is also continuous on $[0,1]$. Therefore, the set is a vector space.

b) No. Let $f(x)\equiv 1$. Then $f(x)$ is a non-negative function on the interval $[0,1]$. However, its additive inverse $-f(x)\equiv -1$ which is a negative function on $[0,1]$ and does not belong the set. Therefore, this set is not a vector space.

c) No. The difference of two polynomials of degree $n$ may not be of degree $n$. For example, let $f(x)=x^n+1$ and $g(x)=x^n$, then
$$f(x)-g(x)=(x^n+1)-x^n=1.$$

d) Yes. Let
$$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$ $$g(x)=b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0,$$ then
$$f+g=(a_n+b_n)x^n+(a_{n-1}+b_{n-1})x^{n-1}+\cdots+(a_1+b_1)x+(a_0+b_0).$$ It is clear that

• if $a_n+b_n\ne 0$, then $f+g$ is of degree $n$;
• if $a_n+b_n\ne 0$, then $f+g$ is of degree less than $n$.

But the degree of $f+g$ is never greater than $n$.

### This Post Has 3 Comments

1. Hi, this is the solution to Question 1.2, not Question 1.3.

1. The solutions given here as solutions to Question 1.3 parts (a) and (b) do not fit the question, in fact the mistake is clearly that they are the solutions to 1.2 parts (a) and (b).

And Question 1.3 has parts (a) - (e), while the solutions here are for parts (a) - (D). The mistake is part (c) is missing.

So here the solutions to Question 1.3 parts (a) - (c) need to be revised.

I appreciate the work you're doing Linearity!