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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 $n$.