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I have been reviewing my Linear Algebra using Friedberg, Insel, and Spences' Linear Algebra 4th Edition and I found something curious in the exercises. In Section 2.2, exercise 16 states:

Let $V$ and $W$ be vector spaces such that $\dim(V) = \dim(W)$, and let $T: V \rightarrow W$ be linear. Show that there exist ordered bases $\beta$ and $\gamma$ for $V$ and $W$, respectively, such that $[T]_\beta^\gamma$ is a diagonal matrix.

My question is this: If I let $V = W$, then does this exercise say that every linear operator on a finite-dimensional vector space is diagonalizable? I know that this is not true, but I can't see where the issue is with the exercise statement. My gut feeling is that $\beta$ and $\gamma$ will always be distinct if $T$ is not diagonalizable, but I'm not sure if that's true.

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    $\begingroup$ That's right. You won't be able to choose a single basis for $V$ that makes $T$ diagonal, but you can find two separate bases, one for the domain, and one for the range, making it diagonal. $\endgroup$
    – Max
    Jun 7 '16 at 0:33
  • $\begingroup$ That's what I thought. I suppose I'll keep this question in mind until I get to the diagonalization chapter and try to prove it then. Thank you. $\endgroup$
    – Jeff L.
    Jun 7 '16 at 0:53
  • $\begingroup$ Actually I think a direct proof is not too hard. Suppose you are given $T$ and $\beta$. Now try to construct $\gamma$, by examining the image of $T$. $\endgroup$
    – Max
    Jun 7 '16 at 0:57
  • $\begingroup$ I think I might have a proof. Suppose $R(T) \neq V$ and that $\beta$ and $\gamma$ are ordered bases for $V$ such that $[T]_\beta^\gamma$ is diagonal. Then $\beta \neq \gamma$. Suppose that $\beta = \gamma$. Then the $j$th column of $[T]_\beta^\gamma$ is $[T(v_j)]_\beta$. Now $R(T) = span\{T(v_1), ..., T(v_n)\} \neq V$ so the vectors $T(v_1), ..., T(v_n)$, and hence the coordinate vectors with respect to $\beta$, must be linearly dependent. Then at least one $[T(v_j)]_\beta$ must be a linear combination of the others, which contradicts that $[T]_\beta^\gamma$ was diagonal. $\endgroup$
    – Jeff L.
    Jun 7 '16 at 1:49
  • $\begingroup$ Jeff, I'm afraid that's not quite true. Consider the $2 \times 2$ matrix with all zero entries except a $1$ in the top-left, viewed as a linear transformation with both bases chosen to be the standard one. This is diagonal, but its range is not everything. $\endgroup$
    – Max
    Jun 7 '16 at 2:43
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Let me try to clarify our discussion in the comments. Let $k$ be a field, and let $\beta=(e_1,e_2)$ be the standard ordered basis of $k^2$. The matrix

$$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=[T]_\beta^\beta $$

defines the linear transformation $T:k^2 \to k^2$ by declaring the relevant ordered bases. This matrix is not diagonalizable, but if we choose $\gamma=(e_1,e_1+e_2)$, then we have

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=[T]_\beta^\gamma .$$

However, $T$ is not diagonalizable. (Perhaps one can show this by elementary means, but all the ways I can think of demand we talk about eigenvalues and eigenvectors, so perhaps that is best put off until later.)

I can offer a hint to your exercise. Suppose you are given $\beta,T$. Let's construct $\gamma=(c_1,\dots,c_n)$ in three stages:

  1. $\gamma'=(c'_1,\dots,c'_k)$ will be a suitable basis of the range of $T$,
  2. The vectors $(c'_{k-1},\dots,c'_n)$ extend $\gamma'$ to a basis $\gamma''$ of $W$.
  3. In the approach I have in mind, $[T]_\beta^{\gamma''}$ has become a matrix with only ones and zeros. A suitable reordering of $\gamma''$ into $\gamma$ obtains $[T]_\beta^\gamma$ as a matrix with all zero entries except possibly ones on the diagonal.

Although there is probably a slicker way to do it.

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  • $\begingroup$ Sorry for the late reply, I was busy this week. I guess what I was trying to find out was that, for your example, if there was an elementary way to show that regardless of what basis I choose for $k^2$, $[T]_\beta^\beta$ will never be diagonal, but can be diagonal if we choose a distinct second basis $\gamma$. Since the $T$ you gave is not diagonalizable, there should be some way to show it. I had an idea, but I'm not sure if its a dead end. $\endgroup$
    – Jeff L.
    Jun 10 '16 at 22:30
  • $\begingroup$ Let $\beta=\{v_1, ..., v_n\}$ be a basis for $V$ and let $\gamma = \{w_1, ..., w_n\}$ be any other basis for which $[T]_\beta^\gamma$ is diagonal. Then, for each $j$, $[T(v_j)]_\gamma$ has all zeroes except possibly the $j$th entry since $[T]_\beta^\gamma$ is diagonal. But then $[T(v_j)]_\beta^\gamma = \lambda_j[w_j]_\gamma$ for some $\lambda_j$. That is, $T(v_j) = \lambda_jw_j$ for some $\lambda_j$. I want to say that in the case $\lambda_j \neq 1$, then $v_j \neq w_j$. $\endgroup$
    – Jeff L.
    Jun 10 '16 at 22:41
  • $\begingroup$ Actually that's not true. I think I fixed that, but it has a longer explanation. Should I use multiple comments to post it? I'm not really sure what the etiquette on mathstackexchange is like. I kinda wish we just had a blackboard in front of us. $\endgroup$
    – Jeff L.
    Jun 10 '16 at 23:25
  • $\begingroup$ I'm not sure. You could post as an answer, just to use the extended format. There is also a supposed math.se chat option. $\endgroup$
    – Max
    Jun 12 '16 at 16:56
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This amounts to Smith Normal Form. Given a (square) matrix $M,$ we can find separate invertible matrices $A,B$ such that $AMB$ is diagonal.

The more common notion called diagonalizing means $P^{-1}MP,$ far more restrictive.

The other notion is for real symmetric matrices, in which case $P^TMP$

The whole point of Smith is the independence of $A,B.$ If your coefficients are not in a field, rather a ring (PID), Smith becomes more intricate and more important. Over a field, I imagine you just take the first few (the rank of $M$) diagonal entries $1,$ the remainder all $0$

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