I am working on some image filters and I have come across this issue. I can express every point $u$ of the image (the space for all that matters) in two distinct packets (bases for the space) $\{x_i\}, \{y_i\}$. That is for all $u$,
$$u=\sum u_i x_i, \ u=\sum v_i y_i$$ where $u_i,v_i$ are scalars.

Applying the filter amounts to considering two linear maps on the space, call them $B$ and $C$. I managed to express these on the two distinct bases and they have the same matrix $a_{ij}$: $$ Bx_i=\sum_j a_{ij}x_j, \ Cy_i=\sum_j a_{ij}y_j $$

My question is, how are $B$ and $C$ related in this context?
Am I justified to say that they are equal? Is there a theorem I can invoke?

Thanks a lot.

  • 2
    $\begingroup$ They are doing the same type of thing but to different vectors, for example, rotate the space along different axes. Such linear operators are called similar, i.e. there is a transformation map $T$ such that $B=T^{-1}CT$. $\endgroup$
    – Conifold
    Jan 18, 2021 at 8:56
  • 1
    $\begingroup$ No they're not equal. They're similar though. $\endgroup$ Jan 18, 2021 at 8:56
  • $\begingroup$ Understood. Put this down as an answer, I will accept it. $\endgroup$ Jan 18, 2021 at 9:14

1 Answer 1


Edit: the discussion is relevant only when the domain and the range of the linear map are the same, otherwise we cannot talk about similarity, regards to Christoph for the remark.

In general case - no, they are not equal. Let us declare a linear transformation $f: R^2 \to R^2$ and $g: R^2 \to R^2$. Let us fix 2 bases in $R^2: B_{1} = \{(1, -1)^T, (2, 3)^T\}$ and $B_{2} = \{(4, 6)^T, (3, 1)^T\}$ (they constitute a basis, because they are linearly independent and there are 2 of them, which is $dim(R^2)$).

Let us say that linear transformations $f$ and $g$ have the same matrix with respect to bases $B_{1}$ and $B_{2}$ respectively:

$[f]_{B_{1}} = [g]_{B_{2}}= \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$

We would need the following property on which you can read more in the Change of basis topic (or here) while discussing linear maps. This is a corollary of a more general theorem for the linear maps which have the same domain and range:

Assume that $f : R^n \to R^n$ is a linear map and let $B=\{v_{1},...,v_{n}\}$ and $C=\{v' _{1},...,v'_{n}\}$ be bases of $R^n$. Let $g : R^n \to R^n$ be the uniquely determined linear map for which $g(v_{j}) = v'_{j}$ holds for every $1 ≤ j ≤ n.$ Then $[f]_{C} = [g]^{−1}_{B}[f]_{B}[g]_{B}$. Likewise $[f]_{B} = [g]_{B}[f]_{C}[g]_{B}^{-1}$.

We will show that the linear maps defined above have different matrices with respect to the standard basis $\{(1,0)^T, (0, 1)^T\}$ in $R^2$.

  1. $[f]$

We need matrix of linear map $h$ which maps standard basis in $R^2$ to the basis $B_{1}$

$[h] = \begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$

$[h]^{-1} = \frac{1}{5}*\begin{bmatrix}3 & -2\\1 & 1\end{bmatrix}$

$[f] = [h][f]_{B_{1}}[h]^{-1} = \frac{1}{5}*\begin{bmatrix}7 & -3\\8 & -7\end{bmatrix}$

  1. $[g]$

As in the first case, we need matrix of linear map $h$ which maps standard basis in $R^2$ to the basis $B_{2}$

$[h] = \begin{bmatrix}4 & 3\\6 & 1\end{bmatrix}$

$[h]^{-1} = \frac{1}{14}*\begin{bmatrix}-1 & 3\\6 & -4\end{bmatrix}$

$[g] = [h][g]_{B_{2}}[h]^{-1} = \frac{1}{2}*\begin{bmatrix}3 & -1\\5 & -3\end{bmatrix}$

As we can see, matrices of linear maps $f$ and $g$ w.r.t. the standard basis are not equal, that is, these maps are not the same.

  • 2
    $\begingroup$ There is no notion of similarity when domain and codomain aren't equal, so this assumption is not "for simplicity" but an essential assumption for the concept of similarity. $\endgroup$
    – Christoph
    Jan 18, 2021 at 9:42
  • $\begingroup$ A critical remark, thank you. $\endgroup$
    – junumboxo
    Jan 18, 2021 at 9:43

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