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When we study diagonalization, we only deal with matrices associated to Endomorphisms $\phi:E\longrightarrow E$ where $E$ is a finite dimensional $K$-vector space. My question is why don't we diagonalize matrices associated to a linear map $\phi: E\longrightarrow F$ where $E$ is an $n$-dimensional $K$-vector space and $F$ is an $m$-dimensional $K$-vector space? Thank you for your help!!

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You can choose bases $\mathfrak{E}$ of $E$ and $\mathfrak{F}$ of $F$ so that the matrix of $\phi$ with respect to them is of the block form $$ \begin{bmatrix} I & 0\\ 0 & 0 \end{bmatrix} $$ where $I$ is an appropriate identity matrix. In other words the matrix has all zeroes, except for a (possibly incomplete) diagonal of ones. This is of course as diagonal a matrix as they come.

This is very easy to see. Choose a basis of $\ker(\phi)$, and extend it to a basis of $E$; this is $\mathfrak{E}$. Now consider the non-zero images under $\phi$ of the vectors of this basis. They will form a basis for the image of $\phi$. Extend this to a basis of $F$; this is $\mathfrak{F}$.

PS I am addressing a question OP asked in a comment below. The reason we can do this is that we are assuming $E$ and $F$ to be two unrelated vector spaces, for which we can choose independently two bases. In the case of $\phi : E \to E$, the idea is to choose one basis for $E$ (that is, the same basis for the domain and the codomain) to obtain a matrix of a convenient form. This is of course more limiting than being able to choose different bases for the domain and the codomain.

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  • $\begingroup$ Are you saying that when $dim(E)=dim(F)=n$ then we can always find basis $B_E$ of $E$ and $B_F$ of $F$ for which the matrix associated to the linear map $\phi:E\longrightarrow F$ is a square diagonal matrix? $\endgroup$ – palio Sep 11 '13 at 16:38
  • $\begingroup$ @palio, please see the PS to my answer. But, yes, the answer to your question is yes. $\endgroup$ – Andreas Caranti Sep 11 '13 at 16:40
  • $\begingroup$ Dear Andreas, your PS is exactly what i want to understand : If we know a process how to find a diagonal square matrix for any linear map $\phi:E\rightarrow F$ where $E$ and $F$ are vector spaces with same dimension, why do we extensively study the case of : 1) $E=F$ and 2) take the same basis for the domain and codomain? $\endgroup$ – palio Sep 11 '13 at 17:00
  • $\begingroup$ @palio, there are many reasons, one of them is that choosing the same basis allows for composition, i.e. iteration. For instance, you can compute more easily the $1000$-th power of the matrix $\begin{bmatrix}0&1\\-2&3\\\end{bmatrix}$ by putting it into diagonal form $\begin{bmatrix}2&0\\0&1\\\end{bmatrix}$ first. If you choose two different bases, you may get a slightly nicer form, but composition fails. $\endgroup$ – Andreas Caranti Sep 11 '13 at 18:21
  • $\begingroup$ I thought a linear map $f:E\rightarrow E$ does not depend on a choice of basis for the domain an codomain when an explicit formula for $f$ is given, i mean when for each element $x\in E$ the corresponding image $f(x)\in E$ is known and then composition is always defined, the choice of basis is relevant when we get from maps to matrices, in that case we have to choose the two basis to write the matrix, and in the setting of matrices composition is not defined since for the matrix $M^2$ the first $M$ correspond to a $f$ from basis $B_1$ to $B_2$ while the second $M$ from $B_2$ to $B_1$. $\endgroup$ – palio Sep 11 '13 at 18:41

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