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I'm taking the course Linear Algebra 1, and recently we've learned about matrix similarity. What is the motivation defining it? or, What are the uses/applications for this definition?


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Suppose $A$ is an $n \times n$-matrix (with real entries, say). Then $A$ defines a linear map $A : \mathbb{R}^n \to \mathbb{R}^n$ by $A(x) = Ax$. In fact, you can figure out the entries of $A$ by just applying $A$ to the standard basis vectors $e_1, \dots, e_n$ of $\mathbb{R}^n$: the $j$'th column of $A$ is simply $A(e_j)$.

Now sometimes (as in, very very often) we don't start with the matrix but with the linear map, and we could try to study this linear map by associating a matrix to it like above; by simply applying it to the standard basis vectors (what we get out is often called the transformation matrix).

Now we should ask ourselves what would happen, if we had used a different basis than the standard one (I should perhaps elaborate how this works but let's keep it rough and simple for now). In mathematics we often tend to prefer making as few unnatural choices as possible. Moreover, in linear algebra, vector spaces often do not come with preferred bases at all. Now as it turns out, had we chosen a different basis, we would get a different matrix for our map (perhaps not very surprising), but the two different matrices are similar!

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Great answer! I would just add that the converse is also true: if two matrices are similar, then they represent the same linear transformation. – Ittay Weiss Jan 8 '14 at 19:09
Good point indeed. – fuglede Jan 8 '14 at 19:35

Matrix similarity is all about change of basis.

An $F$-linear transformation from $F^n$ to $F^n$ has an existence independent of those numbers we fill squares with. However, to do computations, we pick a basis and then compute the numbers that belong in the square matrix.

But you can pick bases a lot of different ways. Shouldn't their matrices be related somehow? The answer is yes, of course: if you have a matrix $A$ for a transformation $T$ in one basis, and a matrix $B$ for $T$ in another basis, then $A=XBX^{-1}$ for some invertible matrix $X$.

What does this buy us? In many cases if you have a matrix, you can "move ot a different basis" to make the current matrix look simpler and easier to do algebra with. This is the spirit of the example GitGud already gave and won't repeat.

This is especially true with computing powers of matrices. It's hard to compute powers of matrices in general, but easy to compute powers of diagonal matrices. Thus if a matrix is diagonalizable (similar to a diagonal matrix), you can compute polynomial expressions of matrices very easily. Say, if $A=XDX^{-1}$ for a diagonal matrix, then the polynomial $A^{5000}=XD^{5000}X^{-1}$, and $D^{5000}$ is trivial to compute.

You can see it extends to polynomials too: $A^5+5A^3-2A=X(D^5+5D^3-2D)X^{-1}$.

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It's good to make up exercises for students.

Now seriously, a useful application is the following.

If $A=PJP^{-1}$, then $A^n=PJ^nP^{-1}$ (why?), for all $n\in \mathbb N$.

So if $J^n$ is simple to find, so is $A^n$ and it happens that there is always a matrix $J$ such that $J^n$ is easy to find, it's called a jordan matrix.

Sometimes it happens $J$ is diagonal and that makes matters trivial.

One reason why finding this powers is useful is to find $e^A\color{grey}{:=\sum \limits_{n=0}^\infty\left(\dfrac 1{n!}A^n\right)}$. This is useful to find solutions of systems of linear differential equations, which in turn have googol applications to the real world.

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Thanks for the info! – AndrePoole Jan 8 '14 at 18:56
@AndrePoole Don't accept my answer so fast, please. That my discourage people from giving other answers. Furthermore this is the kind of question which is common not too accept since it's a joint contribution from the community. It's OK to accept an answer if you want, though. – Git Gud Jan 8 '14 at 19:05
I don't quite agree that this is the motivation for defining matrix similarity. It is an application; an example of a very useful change of coordinates. – Ittay Weiss Jan 8 '14 at 19:12
@IttayWeiss Never meant to imply it was a motivation for it, I'll make this point clearer. – Git Gud Jan 8 '14 at 19:14
@gitgud don't mind my comment, I missed the part of the question where OP also asks for applications. Yours is certainly a (+1). – Ittay Weiss Jan 8 '14 at 19:26

Matrices are there in order to represent linear transformations. We really care about linear transformations, not so much about a rectangle of numbers. You probably know any matrix induces a linear transformation (given by multiplying the matrix by the vector). Now, in the other direction, given a vector space $V$ and a basis for it, if $T:V\to V$ is a linear transformation, then there exists a unique matrix $M$ such that $[T(v)]=M[v]$, where [x] denoted the vector of cooordinates of the vector $x$ in the given basis.

But, different bases for $V$ will generally give rise to different matrices. So, there is a bijection between $n\times n$ matrices (where $n$ is the dimension of $V$) and linear transformations $T:V\to V$. But the bijection depends on a choice of basis. Different bases may yield different bijections.

So, when do two matrices represent the same linear transformation but in two different bases? precisely when they are similar. Similarity of matrices is needed in order to counter the fact that we made an arbitrary choice of bases in order to obtain coordinates for the vectors. Linear transformations are coordinate free, but not so handy for computations. Matrices are handy for computations but different matrices may in fact be essentially the same in that they represent the same linear transformation, just in different coordinates.

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Let's view it from history. The website wrote the following paragraph:

"In 1826 Cauchy, in the context of quadratic forms in n variables, used the term 'tableau' for the matrix of coefficients. He found the eigenvalues and gave results on diagonalisation of a matrix in the context of converting a form to the sum of squares. Cauchy also introduced the idea of similar matrices (but not the term) and showed that if two matrices are similar they have the same characteristic equation. He also, again in the context of quadratic forms, proved that every real symmetric matrix is diagonalizable."

So I believe that the term "matrix similarlity" occurred suddenly in a case discussion. Then it became an independent topic and had links to the main topics of linear algebra.

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This is indeed an interesting answer, even though the question is old. It's always nice to look at things from historical perspective. – BigbearZzz Jul 11 at 5:20
A main reason I found an answer in historical reference is, I think the word "similar matrix" is a term just for convenience, or we can say it is an abbreviation ( of the existence of an invertible $Q$ so that $A=Q^{-1}BQ$ ). The fact that it enjoys transitivity and symmetry, reflexivity makes it useful. However, all above is not the same as the problem of diagonalizability problem, or Jordan form because the similarity is only to a special type of matrix?? I think the motivation of the problem is essential but these only occur in textbooks : Are two given matrices similar ? – user353343 Jul 11 at 5:31

As you may have learned, a matrix is a numerical description of a special kind of geometric transformation: a linear transformation.

Similar matrices describe the same transformation using different coordinate systems. This is useful because, in most applications, coordinate systems are figments of our imagination rather than things that exist naturally in the world, so we often switch back and forth between different coordinate systems when we want to look at a problem in different ways.

Here's how it works. Suppose you have a matrix $A$ that describes your favorite linear transformation in a certain coordinate system, and you want to find the matrix that describes the same transformation in a different coordinate system. A straightforward way to compute your transformation in the new coordinate system is to first apply the matrix $T$ that converts from the new system to the old system, then use $A$ to compute the trasformation in the old system, and finally apply $T^{-1}$ to convert back to the new system. In the conventional notation, matrices are applied from right to left, so the computation I just described is written $T^{-1}AT$. Using matrix multiplication, you can condense the whole computation down into a single matrix $A' = T^{-1}AT$. The matrix $A$ computes your transformation in the old system, the matrix $A'$ computes your transformation in the new system, and you can see explicitly that $A'$ and $A$ are similar.

It should be apparent from this explanation that you can also go the other direction: if two matrices $A$ and $A'$ are similar, that means $A' = T^{-1}AT$ for some matrix $T$, so you can think of $A$ and $A'$ as describing the same transformation in two different coordinate systems, which are related through the change-of-coordinates matrix $T$.

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