I am going to start the topic eigen values and eigen vectors which is in our present course.I am a mathematics major and want to get the complete essence of the topic.One of my seniors told me that I should look into properties of polynomials first before starting this topic.Should I have a deep understanding of polynomials for understanding eigenvalues or basic knowledge(what we studied in theory of equations) is sufficient.Since,we have yet not studied abstract algebra,we do not know much about principal ideal domains or polynomial rings.
So my question is exactly how much knowledge from polynomials is required as a prerequisite for the eigenvalue course and exactly what topics in polynomial should I go through necessarily.Do I need to study minimal and annihilating polynomials also(they are not included in our course)?
No, you don't need to understand principle ideal domains, nor even know what a principle ideal is to study eigenvalues and eigenvectors. What you need to know mostly is just how to compute roots of polynomials, either by factoring or completing the square or by the quadratic formula. For example, given a matrix $$ A = \begin{pmatrix} -2&-4&2\\-2&1&2\\4&2&5 \end{pmatrix} $$ We compute the characteristic polynomial \begin{align} p(\lambda) &= \det(A-\lambda I)\\ &= \det\begin{pmatrix} -2-\lambda&-4&2\\-2&1-\lambda&2\\4&2&5-\lambda \end{pmatrix}\\ &= (2-\lambda)((1-\lambda)(5-\lambda)-2\cdot 2) + 4(-2(5-\lambda)-2\cdot 4)+2(-2\cdot 2 - 4(1-\lambda))\\ &= -\lambda^3 +4\lambda^2 +27\lambda-90.\\ \end{align} By the rational roots theorem, any rational number $\frac pq$ that is a root of $p$ must satisfy $p\vert -90$ and $q\vert -1$. It follows immediately that $q=1$, as $p$ is monic. Noting that $3$ divides both the constant coefficient and the coefficient of $\lambda$, we compute $p(3) = (-3)^3 + 4(3)^2 + 27(3) - 90 = 0$. It follows that $3$ is a root of $p$, so $p(\lambda) = (3-\lambda)(\lambda^2 -\lambda -30)$ Since $-6\cdot 5=30$ and $-6+5=-1$ we see that $-5$ and $6$ are the other eigenvalues.
An important result to know is the Cayley-Hamilton theorem, which states that if $p$ is the characteristic polynomial of $A$, then $p(A)=0$. Indeed we may compute \begin{align} p(A) &= -A^3 + 4A^2 +27A - 90\\ &= \left( \begin{array}{ccc} 64 & 76 & -46 \\ 14 & 11 & -86 \\ -140 & -38 & -193 \\ \end{array} \right)+\left( \begin{array}{ccc} -54 & -108 & 54 \\ -54 & 27 & 54 \\ 108 & 54 & 135 \\ \end{array} \right) -\begin{pmatrix}90&0&0\\0&90&0\\0&0&90 \end{pmatrix}\\ &= \begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}, \end{align} the zero matrix. But the characteristic polynomial need not be the simplest polynomial that satisfies $p(A)=0$. Take for example $$ B=\begin{pmatrix} 4&0&-3\\4&-2&-2\\4&0&-4. \end{pmatrix} $$ Then taking the vector $e_1=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and successively applying powers of $A$, we have $$ e_1 = \begin{pmatrix}1\\0\\0\end{pmatrix},\quad Be_1 = \begin{pmatrix}4\\4\\4\end{pmatrix},\quad B^2 e_1 = B\begin{pmatrix}4\\4\\4\end{pmatrix} = \begin{pmatrix}4\\0\\0\end{pmatrix} = 4e_1 $$ This gives the linear dependency $(B^2-4I)e_1=0$, so the minimal polynomial $m(\lambda)$ such that $m(B)=0$ has $\lambda^2-4$ as a factor. Similarly, successively applying powers of $B$ to $e_2=\begin{pmatrix}0\\1\\0\end{pmatrix}$ yields $$ e_2 = \begin{pmatrix}0\\1\\0\end{pmatrix},\quad Be_2 = \begin{pmatrix}0\\-2\\0\end{pmatrix} = -2e_2, $$ so $Be_2 + 2e_2=0$, and hence $\lambda + 2$ divides $m(\lambda)$. Since the three vectors $e_1, Ae_1, e_2$ span $\mathbb R$, we find $$ m(\lambda) = \mathrm{lcm}(\lambda^2-4,\lambda+2) = \lambda^2-4, $$ and indeed we compute $$ m(B) = B^2 - 4I = 0. $$ This is most of what you need to know about polynomials. Do note that the minimal polynomial $m(\lambda)$ divides the characteristic polynomial $p(\lambda)$.