I'm just really really wondering....is there such a thing? I mean, in my lecture notes, it's defined as, for a vector space $V$, and it's subspace or submodule $U$, to be the mapping $f:V \to V/U$, sending $v$ to $v+U$.

The thing is, I can find ZERO resource or reference on this massive massive web of information called the internet. Nothing hits with "canonical mapping". All that his is canonical "basis" or canonical "Jordan Form" which is not what I want. Does it have a different name, that most people are familiar with? Is it my uni being weird and using a very very obscure name to it?

I am asked to find a matrix to represent a canonical map but I have no idea how to tackle it. Is this "canonical" mappping thing a linear transformation?? It seems like it but I'm not sure....And I don't know how to find this representation matrix for the mapping.

Any advice? Or source, recommended information I can refer to?

  • 1
    $\begingroup$ It's not limited to vector spaces, which is why you may be having a difficult time searching for it. This is an instance of a more general map called the "natural projection" (sometimes also called the "canonical projection") which sends an element to it's equivalence class. $\endgroup$ – EuYu May 10 '15 at 21:19

Suppose $U$ is a subspace of a finite-dimensional vector-space $V$. There is an equivalence relation $\sim$ on $V$ defined by $\vec v\sim\vec w$ if and only if $\vec v-\vec w\in W$. Let $V/U$ be the set of equivalence classes of $V$ under $\sim$ and define addition and scalar-multiplication in $V/U$ by the formulas \begin{align*} [\vec v]+[\vec w]&=[\vec v+\vec w] & \lambda\cdot[\vec v]=[\lambda\cdot\vec v] \end{align*} One checks that this operation is well-defined and endows $V/U$ with the structure of a vector-space called the quotient space. In fact, if we extend a basis $\{\vec u_1,\dotsc,\vec u_k\}$ of $U$ to a basis $\{\vec u_1,\dotsc,\vec u_k,\vec v_1,\dotsc,\vec v_n\}$ of $V$ one checks that $\{[\vec v_1],\dotsc,[\vec v_n]\}$ is a basis for $V/U$. That is, $$ \dim V/U=\dim V-\dim U $$

Now, the map you are interested in is the so-called quotient-map $$ \pi:V\to V/U $$ defined by $\pi(\vec v)=[\vec v]$. One checks that this map is a well-defined linear map. One also checks that $\pi$ is surjective with kernel described by $$ \ker\pi=\DeclareMathOperator{Span}{Span}\Span\{\vec u_1,\dotsc,\vec u_k\} $$ This map is considered canonical because it is exactly the quotient map one defines in set-theory.

For a concrete example, let $V=\Bbb R^4$ and $U$ be the line through the origin of $V$ and the point $(1,1,1,1)$. Then $$ \vec u_1=\begin{bmatrix}1\\1\\1\\1\end{bmatrix} $$ is a basis for $U$ and putting \begin{align*} \vec v_1&=\begin{bmatrix}1\\0\\0\\0\end{bmatrix} & \vec v_2&=\begin{bmatrix}0\\1\\0\\0\end{bmatrix} & \vec v_3&=\begin{bmatrix}0\\0\\1\\0\end{bmatrix} \end{align*} gives a basis $\{\vec u_1,\vec v_1,\vec v_2,\vec v_3\}$ for $V$. In particular, note that $\dim V=4$, that $\dim U=1$, and that $\dim V/U=3$. Can you give a description of the quotient map $\pi:V\to V/U$?

  • $\begingroup$ Hey Brian, Thanks a lot for answering :) Along with Neal's idea, I think I've got a better understanding of this what-seemed-obscure quotient mapping now. For the example...It goes through the origin so $U$ is a subspace of $V$ and so the map takes an element $v \in V$ and takes it to some $w+U \in V/U$ which can be expressed as a linear combination of $v_1, v_2, v_3$? $\endgroup$ – Melba1993 May 11 '15 at 14:32

When "canonical" is used like this, it usually means "obvious."[*] There are potentially many maps $f:V \to V/U$, but the obvious one, the one you think of first, takes $v$ to the equivalence class that contains it. (Note that your professor actually included the definition of the map, $v\mapsto v+U$.)

If you're not sure whether it's a linear transformation, check it! You know the definitions, so you can apply them. Then find the matrix of the map.

Remember that if you have a linear mapping $M:V\to W$ of vector spaces $V$ and $W$, to assign it a matrix you must first choose bases $\{v_1,\ldots,v_n\}$ of $V$ and $\{w_1,\ldots,w_m\}$ of $W$. Writing vectors as column vectors, the matrix of $M$ is defined so that

the $i^{th}$ column comprises the coordinates of $Mv_i$ with respect to the $\{w_1,\ldots,w_m\}$.

So in this case, you have chosen a basis $\{b_1,b_2,b_3,b_4\}$ of $V$ and a basis of $\{q_1,q_2\}$ of $V/U$. Can you see what the images of the $b_i$ are under the projection, in terms of the $q_j$?

More specifically, here's how I'd recommend approaching the problem. In linear algebra (and everything that depends on linear algebra, which is basically everything else), cleverly choosing a basis is 75% of the problem. So instead of going with the standard basis of $\mathbb{R}^4$, let's use a basis $\{c_1, c_2, c_3, c_4\}$ where $U = \operatorname{span}(c_1,c_2)$.

The projection map is a surjection (you should verify this) so the image of a basis is a spanning set. In fact, this is how we can choose a nice basis for $V/U$: let's push the $c_i$s through the projection and whittle that set down to a basis.

Ah-hah! Our earlier cleverness paid off. The vectors $c_1,c_2$ span $U$, hence span the kernel of the projection map. Thus the images of $c_3,c_4$ span $V/U$, and so we have a useful basis of $V/U$: $\{ c_3 + U, c_4 + U\}$.

Now let's find a matrix for the projection in terms of these vectors. Where does $c_1$ go? Zero. So the first column of the matrix is all zeros. Same for $c_2$. What about $c_3$? It goes to $c_3 + U$, so its column in the matrix is $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Likewise for $c_4$.

So the matrix with respect to these bases is $$ \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. $$

If you insist on using the standard basis for $\Bbb R ^4$, the numbers will be more complicated, but you should be fine as long as you hold onto the general principle that coordinates of the matrix are coordinates of the image of each basis vector.

[*] In different contexts, there are many "canonical X"s, but I don't know if the word "canonical" itself has a commonly accepted precise definition. That seems like a question for the category theorists.

  • $\begingroup$ Hi Neal, Thanks for taking your time to answer! I've checked the axioms and it does seem like a linear transformation...But I still can't apply it to solve specific problems :/. Say I have $V=R^4$ and $U$, its subspace. And I picked a basis $b_1=(1 1 0 0)^T,b_2=(0 0 1 1)^T$ for U. I've further extended this basis to a basis of V by adding $b_3=(1 -1 0 0)^T,b_4=(0 0 1 -1)^T$ to my set. Then $b_3+U,b_4+U$ is a basis for $V/U$(I think). However I can't find the method to finding "Matrix for the canonical mapping wrp to the standard basis and $b_3+U,b_4+U$. It's apparently a 2x4 matrix... $\endgroup$ – Melba1993 May 10 '15 at 21:41
  • $\begingroup$ @Melba1993 No problem. It's always fun to review this stuff. I've edited my answer --- does this help? $\endgroup$ – Neal May 10 '15 at 23:59
  • $\begingroup$ I really appreciate your help Neal. So, given the hint, I thought I should add some more context to the question; I am actually given $U={v \in R^4 : v_1=v_2 , v_3=v_4}$ as a subspace of $V=R^4$. So the bases are as stated above, $b_1,b_2,b_3,b_4$. Now, I've worked out that for some $v \in V$, $v= \frac{1}{2}(v_1+v_2)b_1+\frac{1}{2}(v_1-v_2)b_3+\frac{1}{2}(v_3+v_4)b_2+\frac{1}{2}(v_2-v_4)b_4$. And the mapping sends this to $v+U$ i.e. to $\frac{1}{2}(v_1-v_2)b_3+\frac{1}{2}(v_3+v_4)b_4+U$. But I can't figure out how to construct a matrix from here.... $\endgroup$ – Melba1993 May 11 '15 at 13:51
  • $\begingroup$ I mean, on my LHS I have $v$ in terms of for basis so that gives me a 4 by 4 matrix i.e the columns in $b_1,b_2,b_3,b_4$ but the RHS is...a matrix with $U$ in it..? Like, $(b_3+U, b_4+U)(\frac{1}{2}(v_1-v_2) ; \frac{1}{2}(v_3+v_4)$? Even if so, $(b_3+U, b_4+U)$ isn't a square matrix so, how can I find its inverse, to multiply it along with the LHS to find my final matrix?? I am sorry to ask much but it would be more than great if you could guide me further....thanks $\endgroup$ – Melba1993 May 11 '15 at 13:58
  • $\begingroup$ @Melba1993 I think you're thinking too hard. I edited my answer. A minor point - the elements of $V/U$ are cosets which can be represented as $v + U$, so when you write out an expression for vectors in $V/U$ using this representation, it will have lots of "$+U$"s in it. $\endgroup$ – Neal May 11 '15 at 14:23

Let $X$ be the solution space of the ODE $\ddot x+ x=0$, and let $Y$ be the solution space of the ODE $\ddot y-y=0$. Then both $X$ and $Y$ are two-dimensional vector spaces (over ${\mathbb R}$, say). Therefore $X$ and $Y$ are isomorphic. But there is no "canonical" bijective linear map $f: \>X\to Y$ that is the prime candidate for such an isomorphism. In order to define any particular isomorphism you have to choose a basis $(e_1,e_2)$ in $X$ and to declare what the images $\bar e_i:=f(e_i)\in Y$ shall be.

Contrasting this, a canonical map between spaces $X$ and $Y$ arises if it can be defined from the present situation without introducing new data chosen in some arbitrary way.

In your case the "situation" consists of a vector space $V$ and a subspace $U$, and nothing else. (Of course there are certain data necessary to define this subspace $U$ precisely.) Given $V$ and $U$ there is a completely abstract construction that defines a new vector space $V/U$. Unlike the vectors $u\in U$ the elements of $V/U$ are not certain special vectors of $V$, but are entities $\hat v$ which are addressed in the form $v+U$, whereby the $v\in V$ is not uniquely determined by $\hat v$. It is then shown that the map $$f:\quad v\to V/U,\qquad v\mapsto v+U$$ is (a) well defined, and is (b) a linear map from the vector space $V$ the vector space $V/U$. Note that in order to define this $f$ we did neither have to choose a basis in $V$, nor the images of the chosen basis vectors.

But if you want to do computations with this $f$ you need a basis of the abstract space $V/U$. The way to go is to replace $V/U$ by an isomorphic copy living as a bona fide subspace in $V$. Any complement of $U$ in $V$ will do for this purpose. I won't go into this here.


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