Diagonalization and Homomorphism

Question

Maybe this question could sound silly, but after carefully re-reading my linear algebra notes one particular detail catch my sight.

Let $V$ a $\Bbb K$-vector space, $\dim(V) = n$, $f$ an $\textbf{endomorphism}$ is said to be diagonalizable if exist a base $\mathcal{B}$ of $V$ s.t. $f$ representative matrix with respect to $\mathcal{B}$ is a diagonal matrix.

Till there nothing strange is the definition of diagonalization, but as far as I'm concerned there are multiple ways to define some application $g: V \to W$ with $\dim(V) = \dim(W)$, and the representative matrix of $g$ is a square matrix.

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The question is: we want $f$ to be an endomorphism because otherwise there isn't nothing relevant to say even tough algorithmically speaking we could make the same eigenvalue/eigenvector calculation over any square matrix, or since $V,W$ have the same dimension are isomorphic as vector spaces and we are basically working with endomorphism? My guess is that the second reasoning could makes sense since (using the little bit of understanding I have of category theory) let $F$ be a $W,V$ isomorphism and $g:V\to W$ $$\require{AMScd} \begin{CD} V @>{g}>> W \\ @V{id_{V}}VV @V{F}VV \\ V @>{Fg}>> V \end{CD}$$

This is a commutative diagram ad $Fg$ is an endomorphism. Hope for some clarification, thank you.

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Edit

After some responses, the possibility to choose different basis for $V,W$ lead to every homomorphism to have a diagonal form (this isn't the case considering only endomorphism). In the situation descibed above can we deduce some information (eigenvalues/eigenvectors,...) from $g$ that can be transported to $Fg$?

Answer 1

According to Wikipedia, when we speak of diagonalizable linear maps (or matrices) in linear algebra, we are always talking about endomorphisms in a finite-dimensional vector space over a field (or square matrices with coefficients in a field). However, the term diagonal matrix can be used to refer to non-square matrices or matrices over an arbitrary ring.

I think I know why, not all endomorphisms in a finite-dimensional vector space are diagonalizable in the form $P^{-1}AP$ where $P$ is some invertible matrix. But when we talk about homomorphisms from one vector space (free module) $X$ into another vector space (free module) $Y$ (the matrix can be square or non-square), we can change the matrix to $Q^{-1}AP$ by a change of basis, where $P,Q$ are invertible matrices over the ring or field. And over a field, any (square or non-square) matrix $A$ can be brought into the block form $$\begin{bmatrix} I&\\&0 \end{bmatrix}$$ with only $1$'s and $0$'s on the diagonal by the operation $Q^{-1}AP$, and over the ring of integers (I don't know if this holds in an arbitrary commutative ring), any (square or non-square) matrix $A$ can be brought into the diagonal form $$\begin{bmatrix} d_1&&&\\&d_2&&\\&&\ddots&\\&&&0 \end{bmatrix}$$ by the operation $Q^{-1}AP$. Therefore, if we allow non-square matrices (or allow homomorphisms with different domain and codomain even if they are of the same dimension), then every matrix is diagonalizable, and there's no need to talk about it anymore.

Answer 2

You are absolutely right, this can be adapted to any $g : V \rightarrow W$ with $\dim V = \dim W$.

However, instead of one basis $\mathcal{B}$ you would need two bases $\mathcal{B_V}$ and $\mathcal{B_W}$.

Answer 3

How are you defining an eigenvector in a way that makes sense if $V\neq W$? Typically we'd define it so that $f(v)=\lambda v$ for some scalar $\lambda$, but this can never hold if $f(v)\in W$. More or less this is just a type error: $f(v)$ is not in the right space to ever equal $\lambda v$.

If you have an isomorphism $g:V\rightarrow W$ you could change this definition to $f(v)=\lambda g(v)$, but this isn't much different than the endomorphism case since $f\circ g^{-1}$ is an endomorphism on $W$.

I think what you're getting at is that if $V$ and $W$ have the same (finite) dimension, then we can choose bases $B_v$ and $B_w$ for each space and $f:V\rightarrow W$ gives us a matrix representing the action of $f$ on these bases. We can use any technique we want to find eigenvalues of this matrix, so what do those represent?

An eigenvector of the matrix of $f$ (denote this matrix as $M_f(B_v,B_w)$ -- notice I'm including the bases, since the matrix depends on the bases!) is a vector $v\in\mathbb{K}^d$ of coefficients of the basis vectors such that $M_f(B_v,B_w)v = \lambda v$ for some $\lambda\in\mathbb{K}$. What does that mean? Well, we implicitly have two isomorphisms $\varphi_V:\mathbb{K}^d\rightarrow V$ and $\varphi_W:\mathbb{K}^d\rightarrow W$ defined by multiplying the coefficients of a vector in $\mathbb{K}^d$ by the corresponding basis vectors in $V$ (or $W$). Thus, the matrix is is really a map from $\mathbb{K}^d\rightarrow \mathbb{K}^d$ such that the following diagram commutes:

$$\require{AMScd} \begin{CD} \mathbb{K}^d @>{M_f(B_v,B_w)}>> \mathbb{K}^d \\ @V{\varphi_V}VV @V{\varphi_W}VV \\ V @>{f}>> W \end{CD}$$

We can see from this that $f(\varphi_V(v)) = \varphi_W(M_f(B_v,B_w)v)$. If $v$ was an eigenvector of $M_f(B_v,B_w)v$ such that $M_f(B_v,B_w)v=\lambda v$, this means

$$f(\varphi_V(v))=\varphi_W(\lambda v) = \lambda \varphi_W(v)$$

We can define an isomorphism $g:V\rightarrow W$ as $g=\varphi_W\circ\varphi_V^{-1}$, and then if we let $v'=\varphi_V(v)$, the above equation becomes the same as I previously defined for an arbitrary function with an isomorphism:

$$ f(v) = \lambda g(v)$$

So to summarize: you can't define eigenvalues without an isomorphism between $V$ and $W$, and composing a function with an isomorphism turns it into an endomorphism. Choosing bases for $V$ and $W$ implicitly defines an isomomorphism between them (by sending basis vectors in $V$ to the corresponding basis vectors in $W$). Representing a linear function $f:V\rightarrow W$ as a matrix in these bases gives you exactly the same eigenvalues as you would get from composing $f$ with this isomorphism and diagonalizing it as you would any other endomorphism.

The eigenvalues here are very dependent on the isomorphism you choose. Ricky's answer points out that diagonalization becomes almost trivial if you're free to choose the isomorphism.