Definition 1: The matrices A and B are called similar if they represent the same linear transformation but in (possibly) different bases.

Definition 2: The matrices A and B are called similar if there is an invertible matrix P such that A = PBP^−1.

I'm supposed to show whether or not these two definitions are equivalent (saying the same thing). I know that they are but I am having trouble putting it into words. I'm also supposed to describe what it is the matrices P and P inverse do. At one point this was a homework question but I'm now just trying to understand the concepts. Any help is appreciated!

  • $\begingroup$ Do you remember what you learned about a change of basis matrix? $\endgroup$
    – saulspatz
    Mar 6, 2018 at 0:53

2 Answers 2


I'll give an analogy that will tell you what the role of $P$.

Imagine that Harry Potter has a spell that takes (only) black coloured broken chairs and fixes it , keeping it black after fixing . However, you have pink coloured chairs at home (I do so), and would like them to come out pink as well. Then, you do the following:

  • Paint your pink coloured chair black.

  • Ask Harry Potter to fix the broken black chair.

  • Take the fixed black chair and paint it pink.

Now, that is precisely what is happening here, with $P$ being the "paint black chairs pink" operation on chairs, and $B$ being Harry Potter. The first operation is the inverse of $P$. $A$, is then the procedure of fixing pink chairs.

Done with the analogy?

Now, remember the following very crucial statement : the matrix representation of a linear transformation depends on the basis with respect to which the matrix is written.

That is, when you try to figure out what $B$ does to a vector, you need to present the vector in the right form to $B$. So, for example, if $A$ is written with respect to the basis $\mathcal{A}$ and $B$ with respect to $\mathcal{B}$, then giving to $B$, a vector expressed in the basis $\mathcal{A}$, and expecting it to operate correctly, is an incorrect expectation. At least Harry Potter would refuse you if you gave him a pick chair : $B$ will complain very little, and give you the wrong answer!

So, what is the role of $P$? Note that $A$ is represented with respect to $\mathcal{A}$, so it expects only vectors written in the basis $\mathcal{A}$ to be fed to it. But if $A$ wants the help of $B$, then it cannot directly feed the vector that is coming to it, to $B$. Rather, it needs to repaint(or convert) the vector in the basis $\mathcal{B}$. Then, $B$ will see the repainted vector, do the right job on it, and return it, but in basis $\mathcal{B}$. So now, you repaint : get back the vector in basis $\mathcal{A}$.

Hence, $P$ is referred to as a change of basis matrix, for the obvious reasons I have mentioned earlier.

Apart from the fact that Harry takes black chairs and the procedure $A$ takes pink chairs, the sort of crucial thing between both of them is the fact that they fix the chair. That is, up to the job of painting/repainting the chair, which is, a reversible operation, they don't really differ in what they do. This is why the procedures are *similar" : they are doing essentially the same thing, but to different coloured chairs, and it is easy to go from one colour to the other.

Now, to your question : if $A,B$ are similar, then they are the same linear transformation in different bases, if and only if $A = PBP^{-1}$ for some invertible $P$.

What does "the same linear transformation" mean, without invoking change of basis? Note that it is not equality of matrices. What it means is the following : if you fix coefficients $c_i$, and $A$ changes $\sum c_ia_i$ to $\sum d_ia_i$, then $B$ changes $\sum c_ib_i$ to $\sum d_ib_i$.

That is, in some "coordinate free" sense, $A$ and $B$ are the same. I can explain it better, but wish not to.

So, we proceed with this notion.

(Allow me, from this point, to be a little more precise).

One direction : Let $A,B$ be the same linear transformation, but in different bases, say $\mathcal{A}$ and $\mathcal{B}$. Let $P$ be the matrix formed as follows : since $\mathcal{B}$ is a basis, we can express each element $a_i$ of $\mathcal{A}$ as $a_i = \sum_{j=1}^n c_{ij}b_j$, where $b_j$ are the elements of $\mathcal{B}$. Now, the matrix $P = c_{ij}$ does the job. I leave you to see this.

The other way is even simpler, and I leave you to do it. But please proceed only once you have understood what I have said before this section of the answer.


Say we have the transformation $y_a=Ax_a$ in basis $V_a$. Then $x=V_ax_a$ and $y=V_ay_a$ are the correponding vectors in the standard basis. We therefore have:


For $B$ to represent the same transformation in basis $V_b$ we need $V_aAV_a^{-1}=V_bBV_b^{-1}$. And therefore:


That matrix $P=(V_a^{-1}V_b)$ is obviously invertible iff $V_a$ and $V_b$ are invertible, that is, if their columns form indeed a basis.


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