There is an interesting problem I found on an old algebra prelim and it has to do with linear algebra. I want to know how I should approach it. The question is let $A,B \in M_n(F)$, where $F$ is a field and $A \sim B$, i.e. $A,B$ are similar matrices. Let $\lambda \in F$ be a scalar. Prove or disprove that $$ \text{dim Nul}(A- \lambda I_n) = \text{dim Nul}(B- \lambda I_n)\text{,}$$ i.e., the dimension of the nullspace of $A – \lambda I_n$ is equal to the dimension of the nullspace of $B- \lambda I_n$.
Here are my questions. Is the dimension of the nullspace of a matrix also referred to as the nullity of said matrix? Second, is the statement even true to begin with, that is, can it actually be proved? If it can be proved, could we just use the fact that similar matrices have the same characteristic polynomial, and hence, the same eigenvalues to come to a conclusion? Could the definition of what it means for a matrix to be similar to another, i.e. there exists an invertible $n \times n$ matrix $P$ such that $B= P^{-1}AP$, be useful here as well (I would think so!)?
Best Answer
Yes nullity is same as dimension of null space.
Observe that if $A$ and $B$ are similar then $\exists \, P$ such that $A=PBP^{-1}$. Using this we can also say that $A-\lambda I=PBP^{-1}-\lambda I = P(B-\lambda I)P^{-1}$. This means both $A-\lambda I$ and $B-\lambda I$ are also similar. So to prove what you have asked it is enough to show that the nullity of similar matrices is same. So here is an outline to show that if $A$ and $B$ are similar then there nullities will be same.
Let us assume there exists an invertible matrix $P$ such that $B=PAP^{-1}$.
Let $S=\{x_1,x_2, \ldots, x_k\}$ be a basis of the null space $N(B)$, then $Bx_i=0$ for all $i \in \{1,2,3, \ldots, k\}$. Observe that \begin{align*} Bx_i & = 0 \\ PAP^{-1}x_i & = 0\\ A(P^{-1}x_i) & = 0 \end{align*} Thus the set $\{P^{-1}x_1, P^{-1}x_2, \ldots, P^{-1}x_k\} \subseteq N(A)$. You can easily show that this set is linearly independent. Now if we can show that $\text{span}\{P^{-1}x_1, P^{-1}x_2, \ldots, P^{-1}x_k\}=N(A)$, then we are done.
Let $y \in N(A)$, then $Ay=P^{-1}BPy=0$. This means $Py \in N(B)$. Thus $Py=c_1x_1+c_2x_2+ \dotsb + c_kx_k$ for some scalars $c_i$. Now multiply both sides by $P^{-1}$.