This question is really several short general questions to clear up some confusions. We'll start with where I'm at:

An endomorphism $\phi$ is a map from a vector space $V$ to itself. After choosing a basis $\mathcal{B}$ of $V$, one may determine a matrix to represent such an endomorphism with respect to that basis, say $[\phi]_{\mathcal{B}}$.

Question 1) If given a matrix without a basis specified, could you deduce a unique endomorphism it corresponds to? (my lean is no)

Question 2) In a similarity transform, say $A=SDS^{-1}$ where $D$ is diagonal, $S$ is the change of basis matrix from one basis to another. My question is, since $D$ is diagonal does that mean the matrix $D$ is the same endomorphism as $A$ with respect to the standard basis in $\mathbb{R^{n}}$. Or are we unable to determine which bases are involved if only given the matrices.

Question 3) Given a matrix related to an endomorphism, is it possible to determine the basis used to represent the endomorphism. (Lean yes)

Overarching Question) I am trying to understand what happens under similarity transforms. I understand we input a vector, a change of basis is applied to it, the endomorphism is applied with respect to the new basis, and then it is changed back to the old basis, but my confusion relates to the construction of the similarity matrix. If a matrix is diagonalizable, then $S$ turns out to be the eigenvectors arranged in a prescribed order. Why is this! Why do the eigenvectors for the endomorphism $\phi$ with respect to one basis act as a change of basis matrix, and what basis to they go to? This is really part of question 2. Does this send the vectors to the standard basis? Or some other basis that just happens to diagonalize the endomorphism.


  • $\begingroup$ (1) Consider the identity matrix. (3) Consider the zero matrix. $\endgroup$ – anon Jun 21 '12 at 2:38
  • $\begingroup$ @toypalme: sometimes an endomorphism has the same matrix with respect to any matrix, this is the case with the identity and the zero matrix, so (1) sometimes you can determine the endomorphism form the matrix, and (3) sometimes you cannot determine the basis, because all bases work. $\endgroup$ – Jack Schmidt Jun 21 '12 at 2:53
  • $\begingroup$ @JackSchmidt Thank you, so what I was looking for, some systematic way to find a specific basis a matrix is with respect to and the endomorphism may not pan out. $\endgroup$ – toypajme Jun 21 '12 at 2:55
  • $\begingroup$ An invertible endomorphism is called an automorphism. Replacing $\phi:V\to V$ with $\theta \phi \theta^{-1}$ for an automorphism $\theta$ takes an endomorphism to a very similar, but often different, endomorphism. If two endomorphisms have the same matrix (with respect to different bases) then they are "very similar" in this way. The same endomorphism with respect to different bases have similar matrices. These ideas of similarity and "very similar" are closely related, and exactly the obstacle to endomorphisms having specific bases. $\endgroup$ – Jack Schmidt Jun 21 '12 at 2:56
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    $\begingroup$ I'm familiar with the SVD decomposition. Thats a nice way to look at it, and is exactly the sort of insight I'm looking for. So intrinsically, if I diagonalize $A=S^{-1}DS$, one way to look at this is that $A$ represents some endomorphism w.r.t. some basis. Writing $SAS^{-1}=D$ shows that the eigenvectors of $D$ are the standard basis, but it shows the eigenvectors of $A$ are simply the entries of $S$. Since $S$ is also the change of basis matrix, does this show the basis that $A$ is with respect to is the basis found in the columns of $S$. And $D$ represents the endo in the standard basis? $\endgroup$ – toypajme Jun 21 '12 at 3:39

Qiaochu's answer is a great treatment but I think #3 was dismissed a little hastily.

As I read it, #1 and #3 are not the same: much more is given for #3, namely you get both the matrix and a transformation and the fact that they are related.

So my reading of #3 is:

If we are told that a matrix $B$ is a matrix for transformation $f$ in terms of an unknown basis $\beta$, can the elements of $\beta$ be recovered explicitly?

Fix any basis you like, call it $\alpha$, and express $f$ as a matrix $A$ corresponding to that basis. Since we know they correspond to the same transformation, we know $A$ and $B$ are similar via a nonsingular matrix $X$. This matrix $X$ is a change of basis matrix, which converts between the coefficients in terms of mystery basis $\beta$ and those of our known basis $\alpha$.

Either $X$ or its inverse will convert from $\beta$ to $\alpha$ depending on how you set it up, so let's just pick the one which goes this direction and relabel it $X$. Let's also assume we have been using matrix multiplication on the left of column vectors.

To find $\beta_1$, for example, compute $Xe_1$ where $e_1$ is the column unit vector with $1$ in the first spot. (That is the representation of $\beta_1$ in basis $\beta$.) The output is the coefficients of $\beta_1$ represented in terms of $\alpha$, so you may just arrange these coefficients in front of your known basis elements and add everything up to get $\beta_1$. The $\beta_i$ are recovered using the other unit vectors $e_i$.

I suppose there could be some practical difficulty, but theoretically this seems sound. I cannot tell if the last paragraph is addressing this or not.

Added: Jack Schmidt pointed out to me the leak in my idea: $X$ is not uniquely determined, because, for example, $B$ could be self-similar by a nonidentity matrix $Y$ and then $A=X^{-1}BX=(YX)^{-1}B(YX)\dots$ so #3 is still an underdetermined problem in general. It's still more determined than #1, but not as determined as I wanted it to be.

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  • $\begingroup$ I also think this does answer #3 and it sheds light on the overarching. Given any matrix, we can just fix the basis as the standard basis and that gives rise to an endomorphism. The matrix may be a different endomorphism w.r.t. another basis, but I think in the end they are equivalent. Call the endo. $\phi$ and let it be diagonalizable. We then have, $[\phi]_{B}=S^{-1}[\phi]_{B\prime}S_{B\prime\leftarrow B}$. If we multiply this by a vector $[x]_{B}$ we see that: $S_{B\prime\leftarrow B}[x]_{B}=[x]_{B\prime}$ and so $x$ is converted to the new basis, then the map is applied in the new basis. $\endgroup$ – toypajme Jun 21 '12 at 13:14
  • $\begingroup$ From this it makes it clear that then $B^{\prime}$ is the basis which diagonalizes $\phi$. To find out what $\phi$ is, the basis $B^{\prime}$ can be written in vectors relative to $B^{\prime}$ so the first basis vector $v_{1}$ is $v_{1}=[(1,0,0,0\cdots)^{T}]_{B^{\prime}}$ From this we can recover it in the standard basis by writing: $S^{-1}v_{1}=[v_{1}]_{B}$. This would give the basis that diagonalizes the endomorphism, which in this case ends up being the eigenvectors. $\endgroup$ – toypajme Jun 21 '12 at 13:18
  • $\begingroup$ Perhaps I should have used normal matrices since I think unitary equivalence makes this easier to see. If you have a normal matrix $A=UDU^*$ then if you consider an eigenvector of $\phi$ defined as fixing the std basis, $v_{1}$, $[v_{1}]_{B}$ is the first column of $U$, and $UDU^{*}[v_{1}]_{B}=UD[(1,0,0,0,...)^{T}]_{eigen}$, and so immediately it is clear that the basis "eigen" diagonalizes the endomorphism $\phi$ given from considering $A$ with respect to the standard basis, and the eigenvectors are the basis for the diagonal form. $\endgroup$ – toypajme Jun 21 '12 at 13:29
  • $\begingroup$ #3 does not have a unique answer in general, only if there is a $D$ and its diagonal elements are distinct. In particular, the theoretical difficulty (which is also practical) is that "X" is not uniquely defined. $\endgroup$ – Jack Schmidt Jun 21 '12 at 17:41
  • $\begingroup$ (Actually it never has a unique answer over a field of size larger than 2, but for a stupid reason: Any nonzero multiple of a basis gives exactly the same matrix for every endomorphism.) $\endgroup$ – Jack Schmidt Jun 21 '12 at 17:44

1) No. Any choice of basis determines an endomorphism that the matrix corresponds to and in general different choices of basis will give different endomorphisms. (They are, of course, all similar.)

2) No. Consider the case that $A$ is not diagonal. Two matrices have the same entries if and only if they represent the same linear transformation with respect to a fixed basis.

3) No. This is the same question as 1).

4) To see this concretely, write out the condition $AS = SD$ for $D$ a diagonal matrix explicitly. To see this abstractly, let $T$ be a linear transformation with respect to a basis $e_i$, and suppose it has a basis $v_i$ of eigenvectors with eigenvalues $\lambda_i$. Then $T$ is diagonal with respect to the basis $v_i$. If $S$ denotes the linear transformation which sends $v_i$ to $e_i$, then $$STS^{-1}(e_i) = ST v_i = S \lambda_i v_i = \lambda_i S v_i = \lambda_i e_i$$

so $STS^{-1}$ is diagonal with respect to the basis $e_i$. Now, the above is a statement about linear transformations which is independent of basis. Writing everything above in terms of the basis $e_i$ gives you a corresponding statement about matrices, and in that statement $S^{-1}$ is more or less by definition the matrix whose columns are the entries of $v_i$ (with respect to the basis $e_i$).

I have often thought that elementary linear algebra would be less confusing if it were made explicit that when changing bases one is really working with two vector spaces; first the original vector space $V$ one cares about and second the concrete vector space $\mathbb{C}^n$ where $n = \dim V$ with its distinguished basis. A basis for $V$ is then equivalent to the choice of an isomorphism $f : \mathbb{C}^n \to V$ and changing bases corresponds to changing the choice of this map. Crucially, there are two natural ways to do this: either precompose with an automorphism $\mathbb{C}^n \to \mathbb{C}^n$ or postcompose with an automorphism $V \to V$. The two give the same result, but the notion of sameness here is itself dependent on the choice of $f$.

In category theory, one says that the finite-dimensional vector space (over a fixed field) of a given dimension is unique up to isomorphism, but not unique up to unique isomorphism, and so when identifying different vector spaces one must keep track of the identifications one is using or else risk getting hopelessly lost.

In other words, when dealing with objects that are isomorphic but for which the isomorphism is not unique, it is better to behave as if they are different objects even if they are in some sense "the same."

It might help to think of the same linear transformation with respect to two different bases as operating on two different data types. That is, with respect to a given basis $\mathcal{B}$, the corresponding matrix should perhaps be thought of as a function which accepts and spits out "$\mathcal{B}$-type" vectors, and with respect to a different basis $\mathcal{C}$ accepts and spits out $\mathcal{C}$-type vectors. These operations are compatible but to specify the compatibility you need to typecast from $\mathcal{B}$-type vectors to $\mathcal{C}$-type vectors and back again, and this is exactly what the change-of-basis matrix is supposed to do.

The change-of-basis matrix therefore has different input and output types.

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  • $\begingroup$ First thank you for your response. I have some confusion with "seeing it abstractly." You are talking about a linear transformation with respect to a basis, and then fixing a basis and turning it into a matrix equation? My confusion is, if you have an endomorphism and write the matrix representing the endomorphism with respect to its eigenvectors, will this be diagonal if the matrix is diagonalizable? I think that is what is coming from this, that the eigenvectors for some endomorphisms form a "natural" basis in the sense that its corresponding matrix is diagonal. $\endgroup$ – toypajme Jun 21 '12 at 3:53
  • $\begingroup$ @toy: yes, that is equivalent to being diagonalizable. $\endgroup$ – Qiaochu Yuan Jun 21 '12 at 3:54
  • $\begingroup$ But this slightly confuses me because I am not sure how to think about the eigenvectors in this abstract setting without a basis besides an element in the kernel of $(A-\lambda I)$. Whenever I have seen endomorphisms written out, and the exercise was to convert it to a matrix with respect to a certain basis, it appeared to me that it was already given with respect to the standard basis.. let me explain: $\phi(x,y)=(x+y,x-2y)$ is one example. Is this not given with respect to any basis? $\endgroup$ – toypajme Jun 21 '12 at 3:54
  • $\begingroup$ @toypajme: that linear transformation is given with respect to the basis $(1, 0)$ and $(0, 1)$. What do you mean by "besides an element in the kernel of $A - \lambda I$?" $\endgroup$ – Qiaochu Yuan Jun 21 '12 at 3:59
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    $\begingroup$ @toypajme: it sounds like you have never seen an endomorphism specified without a choice of basis at all, so here's one: consider the action of differentiation $\frac{d}{dx}$ on the space of polynomials of degree at most $d$. $\endgroup$ – Qiaochu Yuan Jun 21 '12 at 4:00


Let $n=\dim V$, $A$ a given matrix $n \times n$ with entries in commutatif field $\mathbb K$ and let us considere the map : $f_A : \mathbb K^n \to \mathbb K^n $ such that , if $X=(x_1,...,x_n) \in \mathbb K^n$ then $f_A(X)=A \;^tX$ we can proof that $A$ is the matrix of $f_A$ on canonical basis of $\mathbb K^n$ , thus $B_0=(e_i)_{1 \leq i \leq n}$ where $e_i=(\delta_{ij})_{1 \leq j \leq n}$ and $\delta_{ij}=0$ if $i \neq j$ and $\delta_{ij}=1$ if $i=j$

If $B$ is some basis of $V$ there exists a unique isomorphisme $\phi : V \to \mathbb K^n$ sush that $\phi(B)=B_0$ then we have : $g_{\phi}= \phi^{-1} \circ f_A \circ \phi$ is the unique map from $V$ to $V$ represented bay $A$ on the basis $B$.

Conversly all linear map from $V$ to $V$ represented by $A$ on some basis $B$ of $V$ has the for $ g_{\phi}$ where $\phi$ is some isomorphisme from $V$ to $\mathbb K^n$. Precisely $\phi$ is the unique isomorphism from $V$to $\mathbb K^n$ who trnsforms the basis $B$ of $V$ to the canonical basis $B_0$ of $\mathbb K^n$.

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