Real n-by-n Matrices... Let $M_n(\mathbb{R})$ denote the vector space of real $n\times n$ matrics, and let $A \in M_n(\mathbb{R})$.
Part (a) of this question says: Suppose $B \in M_n(\mathbb{R})$ such that $AB = I_n$ (the $n \times n$ identity matrix. If $C \in M_n(\mathbb{R})$ such that $CA = 0$, then prove $C = 0$.
I have already proven part (a).
Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + \dots + t_mA^m = 0$ for some $t_0, \dots, t_m \in \mathbb{R}$ with $t_m \neq 0$. Also, suppose that $AB = I_n$ for some $B \in M_n(\mathbb{R})$.Prove that $t_0 \neq 0$. (Hint: Use the result from part (a)).
My idea is to use induction on $m$. That is, suppose
$$t_0I = 0.$$
But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as
\begin{align*}
t_0I &= 0
\\
t_0AB &= 0
\\
t_0CAB &= 0
\\
t_0C &= 0.
\end{align*}
But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".
I am studying for my linear algebra comp in a few weeks so any help is appreciated!
 A: Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + \cdots  + t_m A^m = 0
$$
If $t_0=0$ you have,
$$
t_1 A + \cdots  + t_m A^m = 0
$$
Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + \cdots t_n ' A^n = 0
$$
where $n<m$. 
Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.
I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.
A: Hint: assume $t_0=0$ then
$$
t_1A+\ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.
A: Here is another hint/solution:


*

*assume it to equal $0$,

*multiply your equation by $A$, and factor by $A$,

*using part (a), get a contradiction. 

