In Lipschutz's (Schaum's Outline) Linear Algebra, 6e, the attached motivation is provided for the study of eigenvalues. I'm struggling to fully understand the statement (essentially, to prove it), and I'm hoping someone can help me. I've spent the last 25 or so minutes trying to digest it as it's obviously the basis (no pun intended) for trying to find the eigenvalues of a linear transformation.
To clarify notation: $A$'s elements come from some field $K$ (and it can thus be thought of as a matrix representation of some linear transformation $T$ in terms of the standard basis of $K^n$).
Now my attempt at a proof looks something like the below. I am trying to come at the (admittedly simple) proof in the spirit of Axler in LADR (emphasis on the meaning of the underlying transformations as opposed to getting bogged down in the matrix multiplication):
Suppose $A$ is similar to $D$. Then, by definition, $D$=$P^{-1}AP$, so $PD=AP$. The LHS of the equality yields a matrix where the jth of $P$ has been multiplied by the lone entry at $[D_{j,j}]$. The RHS is the transformation of $A$ on $P$ and (considered column-wise in terms of P) yields the $Au_j$. So that direction seems clear to me. For the reverse, I can't see why a basis existing such that the above (picture) is true means A is diagonalizable. Is it as simple as saying that if such a basis exists, then consider encoding them column-wise in a matrix $P$ and then you get $PD=AP$?
Ultimately, I'm hoping someone can elucidate (with a short discussion) exactly what is going with this statement by Lipschutz. Thanks!
Let $B$ be any $n\times n$ matrix. We treat the matrix $B$ as a linear transformation $B:K^n\longrightarrow K^n$ defined by $B(e_i):=Be_i$ for all $1\leq i\leq n$, where $e_i$ is the column vector $(0,...,1,...,0)^t$ (ie, $1$ occurs in the $i$th position) and the product $Ae_i$ is just multiplication of matrices. Consider the following statements:
(i) There exists an invertible matrix $P$ such that $P^{-1}AP=diag (a_1,...,a_n)$, for some $a_i\in K$.
(ii) There exists a basis $\{u_1,..,u_n\}$ of $K^n$ such that $A(u_i)=k_iu_i$ ($k_i\in K, 1\leq i\leq n$)
We claim that (i) $\Leftrightarrow$ (ii)
Proof of claim : First assume that (i) is true. Then $P^{-1}AP=diag (a_1,...,a_n)\Rightarrow (P^{-1}AP)(e_i)=diag(a_1,...,a_n)(e_i)=a_ie_i$ for all $i=1,...,n$ $\Rightarrow APe_i=a_iPe_i$ for all $i=1,...,n$ (this being obtained by premultiplication by $P$). Note that since $P$ is invertible, the set $\{Pe_1,...,Pe_n\}$ is a basis of $K^n$ (since $\{e_1,..,e_n\}$ is a basis of $K^n$). So taking $Pe_i=u_i$ we see that (ii) holds, where the role of $k_i$ is played by $a_i$ for each $i$.
Conversely, now assume that (ii) holds. In order to establish (i) we need to cook up a matrix (or equivalently a linear transformation of $K^n$ onto itself) $P$. We do this as follows:
Consider the linear transformation $T:K^n\longrightarrow K^n$ defined by $T(e_i)=u_i$ and extended linearly over $K$. This makes sense because the $e_i$'s and $u_i$'s form bases of $K^n$. NOw with respect to the stanadard basis $\{e_1,...,e_n\}$ of $K^n$, the map $T$ can be represented by a matrix $P$(say) such that $T(e_i)=Pe_i$ for all $1\leq i\leq n$. This $P$ is invertible matrix since the map $T$ is an invertible linear transformation. Therefore, $APe_i=Au_i=k_iu_i=Pk_ie_i\Rightarrow P^{-1}APe_i=k_ie_i=diag (k_1,...,k_n)e_i$ for all $1\leq i\leq n$. Since $P^{-1}AP$ and $diag(k_1,...,k_n)$ agree (as a function) on each basis element $e_i$, they denote the same function and hence $P^{-1}AP=diag(k_1,...,k_n)$ as matrices. Thus we see that (i) holds.
This shows that any one of the statements (i) or (ii) can be used as a definition of diagonalizability of a matrix.
I hope that this clarifies the situation.
P.S. The Schaum series book is perhaps okay for obtaining a healthy supply of problems to work out. However I would suggest any learner to read Hoffman and Kunze's Linear Algebra or Michael Artin's Algebra for a much better exposition.