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I've been studying some linear algebra for a test and I've stumbled across this particular issue.

I'm given the transformation matrix $M_{BB'}$, corresponding to some $f \colon \mathbb{R}^n \to \mathbb{R}^m$ with respect to the (ordered) bases $B=(B_1, \dotsc, B_n)$ and $B'=(B'_1, \dotsc, B'_n)$. As I understand, if I have a vector $v$ expressed with respect to the basis $B$, then I can find $f(v)$ by calculating $M_{BB'}v$, obtaining a result expressed in the basis $B'$. Also, if I want to find $f(v)$ for any $v$ expressed in the canonical basis $E$, I can calculate $C_{B'E}M_{BB'}C_{EB}$, where $C_{B'E}$ and $C_{EB}$ are the matrices for the change of basis from $B'$ to $E$ and from $E$ to $B$, respectively, and apply it to the vector $v$, obtaining $f(v)$ with respect to the canonical basis.

As far as finding the image of a particular vector goes, I don't think much else can be said. Nonetheless, when I'm asked to find the image of $f$, given $M_{EE} = C_{B'E}M_{BB'}C_{EB}$, I reason that, since $C_{EB}$ is invertible, instead of computing $C_{B'E}M_{BB'}C_{EB}v$ for a generic $v$, I can assume that one and only one $w = C_{EB}^{-1}v$ exists for each $v$, so it suffices to find the image of $C_{B'E}M_{BB'}v$, i.e., to find the image of $M_{BB'}$, find a basis for this space and transform the basis vectors to the canonical basis.

For the null space of $f$, I'm trying to find all $v$ such that $M_{EE}v = C_{B'E}M_{BB'}C_{EB}v = 0$. And since $C_{B'E}$ is invertible, all I need to solve is: $M_{BB'}C_{EB}v = 0$.

Now, I've tried these "simplified" methods and I get different answers, probably due to a mistake in the calculation, but my questions are the following. Are these methods valid? Is there anything I'm missing? Are these the most efficient ways to compute $Im f$ and $Ker f$? (I'm not really worried about working with large matrices or computational complexity, but I would simply like not to perform unnecessary computations.) Are there other procedures for these goals that I'm not considering? Also, is there any significance as to the "disappearance" of one of the two change of basis matrix in each equation?

Any help or insight into this matter is greatly appreciated.

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In your third paragraph, "$M_{BB'} = C_{B'E}M_{BB'}C_{EB}$" is incorrect; the left hand side should be $M_{EE}$. Same in the next paragraph. – Arturo Magidin Jan 31 '11 at 3:20
Oops. You're right. Thanks. – Abel Jan 31 '11 at 3:28
up vote 3 down vote accepted

Think of the matrices $C_{EB}$ as "dictionaries" or "translators". $C_{EB}$ is a translator which, if you talk to it in the $E$-language (describing vectors in terms of the basis $E$) will "reply" by telling you the same thing, but in $B$-language (in terms of the basis $B$). Likewise, $C_{B'E}$ is the translator that takes something spoken in $B'$-language, and translates it into $E$-language.

What does $C_{B'E}M_{BB'}C_{EB}$ give, then? You "speak" to it in $E$-language; $C_{EB}$ translates it into $B$ language. $M_{BB'}$ will apply $M$ and give the answer in $B'$-language; and finally $C_{B'E}$ will translate this into $E$-language. So $C_{B'E}M_{BB;}C_{EB} = M_{EE}$, where $M_{EE}$ is the matrix of the same linear transformation as $M_{BB'}$, but relative to the standard basis.

Similarly, what is $C_{B'E}M_{BB'}$? It is the matrix $M_{BE}$: you give it a vector described in $B$-language, and it will answer with its image under $M$, but described in $E$-language.

The key to the manipulations/computations you are doing, then, is to keep straight in what language you are speaking and in what language you are getting the answers. I suspect that the reason you are getting "different answers" is that you are obtaining the 'same' answers, but in different languages.

If you try to find the nullspace of $f$ by looking for all vectors $\mathbf{v}$ for which $M_{BB'}\mathbf{v}=\mathbf{0}$, what you will find is the nullspace of $f$, described in terms of $B$. So if $B=[\mathbf{x}_1,\ldots,\mathbf{x}_n]$, and you get, say, that $\mathbf{v}=(1,0,0,\ldots,0)^T$ is one of the vectors in the nullspace, what you really have is that the vector whose coordinates relative to $B$ are $\mathbf{v}$ is in the nullspace; that is, $\mathbf{v}$ "really" describes the vector $\mathbf{x}_1$.

By contrast, if you try to find the nullspace of $f$ by trying to find all vectors $\mathbf{w}$ for which $C_{B'E}M_{BB'}C_{EB}\mathbf{w}=\mathbf{0}$ you will find the vectors in the nullspace described in terms of $E$, not in terms of $B$. You can "translate" form one to the other by using $C_{EB}$ (which translates from $E$-speak to $B$-speak), or $C_{BE} = (C_{EB})^{-1}$ (which translates from $B$-speak to $E$-speak).

(You get the same answer if you use $M_{BB'}C_{EB}$ as if you use $C_{B'E}M_{BB'}C_{EB}$ because you are still dealing with vectors in the domain described in $E$-speak; the fact that you are translating the results into $B'$-speak at the end is immaterial, since the only vector that is described as $\mathbf{0}$ in any speak is $\mathbf{0}$).

For the image, you want to keep track of the same things: if you try to find the image using the image of $M_{BB'}$, you will get the answer described in $B'$-speak; but if you use $C_{B'E}M_{BB'}$, you'll get the answer described in $E$-speak. You can translate from one to the other by using $C_{B'E}$ and $C_{EB'}=(C_{EB'})^{-1}$.

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@Nick: If you are still having trouble, you can give the explicit matrices/bases/computations. Then we can discuss specifics rather than generalities. – Arturo Magidin Jan 31 '11 at 3:37
I guess I'm having a little trouble treating change of basis matrices and linear transformation matrices differently, because after all, every matrix describes a linear transformation (when considering the standard basis). I may be too fixed on the idea of the standard basis. Your post described exactly what I reasoned about the problem. So it is possible that some mistakes occurred during the computations. I'll take a look again and see how that works out. Thanks. – Abel Jan 31 '11 at 3:50
@Nick: You have to be careful: yes, every matrix describes a linear transformation relative to the standard basis, but also every matrix describes a linear transformation relative to whatever bases you want to specify ahead of time. You need to know what "language" the matrix expects to hear, and what "language" the matrix is replying in, or else you'll get gibberish. If you have a word, it may be a word in both Spanish and English, yet mean different things (e.g., "simpatico" means "funny, amusing" in Spanish, but it means "in synch" in English). You need to know what language is being used – Arturo Magidin Jan 31 '11 at 4:04

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