A similarity transformation is
Where are square matrices. The goal of similarity transformation is to ﬁnd a matrix which has a simpler form than so that we can use in place of to ease some computational work. Lets set our goal in having be a diagonal matrix (a general diagonal form is called block diagonal or Jordan form, but here we are just looking at the case of being a diagonal matrix).
The question becomes: Given , can we ﬁnd such that is diagonal?
The standard method to show the above is via an algebraic method, which we show ﬁrst. Next we show a geometric method that explains similarity transformation geometrically.
Starting with
Our goal is to ﬁnd a real matrix such that it is diagonal. From the above, by pre multiplying each side by we obtain
Now, since our goal is to make diagonal, let us select the eigenvalues of to be the diagonal of . Now we write the above in expanded form as follows
The above can be written as separate equations by looking at the column view of matrix multiplication
and
all the way to the n column of
Each of the above equations is in the form
and this is the key idea.
The above shows that if we take the diagonal elements of to be the eigenvalues of then the matrix is the (right) eigenvectors of
Hence, if we want the matrix to be diagonal and real, this means the matrix itself must have some conditions put on it. Speciﬁcally, eigenvalues must be all distinct and real. This happens when is real and symmetric. Therefore, we have the general result that given a matrix which has distinct and real eigenvalues, only then can we ﬁnd a similar matrix to it which is real and diagonal, and these diagonal values are the eigenvalues of .
It turns out that this derivation is more complicated than the algebraic approach. It involves a change of basis matrix (which will be our matrix) and the representation of linear transformation as a matrix under some basis (which will be our matrix). Let us start from the beginning.
Given the vector in , it can have many diﬀerent representations (or coordinates) depending on which basis are used. There are an inﬁnite number of basis that span , hence there are inﬁnite representations for the same vector. Consider for example . The standard basis vectors are , but another basis are , and yet another are , and so on.
In any linearly independent vectors can be used as basis. Let be the vector representation in w.r.t. basis , and let be the vector representation w.r.t. basis .
An important problem in linear algebra is to obtain given and the change of basis matrix .
This requires ﬁnding a matrix representation for the change of basis. Once the matrix is found, we can write
 (1) 
Where is the change of basis matrix which when applied to returns the coordinates of the vector w.r.t. basis
From (1) we see that given , then to obtain we write
 (1A) 
Another important problem is that of linear transformation , where now we apply some transformation on the whole space and we want to ﬁnd what happens to the space coordinates (all the position vectors) due to this transformation.
Consider for example which is a rotation in by some angle . Here, every position vector in the space is aﬀected by this rotation but the basis remain ﬁxed, and we want to ﬁnd the new coordinates of each point in space.
Let be the position vector before applying the linear transformation, and let be the new position vector. Notice that both vectors are written with respect to the same basis . Similar to change basis, we want a way to ﬁnd from , hence we need a way to represent this linear transformation by a matrix, say with respect to the basis and so we could write the following
 (2) 
Assume the basis is changed to . Let the representation of the same linear transformation w.r.t. basis be called , so we write
 (2A) 
Hence when the basis changes from to , representation changes from to
Now assume we are given some linear transformation and its representation w.r.t. basis , how could we ﬁnd representation w.r.t. to new basis ?
From (2) we have
But from (1) , hence the above becomes
Premultiplying from left the above equation by we obtain
But since , then above reduces to
But from (2A) , hence the above becomes
Therefore
 (3) 
Notice the diﬀerence between change of basis and linear transformation. In change of basis, the vector remained ﬁxed in space, but the basis changed, and we want to ﬁnd the coordinates of the vector w.r.t the new basis. With linear transformation, the vector itself is changed from to , but both vectors are expressed w.r.t. the same basis.
Equation (3) allows us to obtain a new matrix representation of the linear transformation by expressing the same w.r.t. to diﬀerent basis. Hence if we are given a representation of which is not the most optimal representation, we can, by change of basis, obtain a diﬀerent representation of the same by using (3). The most optimal matrix representation for linear transformation is a diagonal matrix.
Equation (3) is called a similarity transformation. To make it easier to compare (3) above with what we wrote in the previous section when we derived the above using an algebraic approach, we let , hence (3) is
The matrix is similar to . (i.e. is similar to . Both matrices represent the same linear transformation applied on the space. We will show below some examples of how to ﬁnd given and . But ﬁrst we show how to obtain matrix representation of
Given a vector and some linear transformation (or a linear operator) that acts on the vector transforming this vector into another vector according to some prescribed manner . Examples of such linear transformation are rotation, elongation, compression, and reﬂection, and any combination of these operations, but it can not include the translation of the vector, since translation is not linear.
The question to answer here is how to write down a representation of this linear transformation? It turns out that can be represented by a matrix of size , and the actual numbers that go into the matrix, or the shape of the matrix, will depend on which basis in we choose to represent .
We would like to pick some basis so that the ﬁnal shape of the matrix is the most simple shape.
Let us pick a set of basis that span hence
Now

And since is linear we can rewrite the above as
 (1) 
We see from above that the new transformed vector has the same coordinates as if we view as the new basis.
Now itself is an application of the linear transformation but now it is being done on each basis vector . Hence it will cause each speciﬁc basis vector to transform in some manner to a new vector. Whatever this new vector is, we must be able to represent it in terms of the same basis vectors therefore, we write
Where is the coordinate of the vector
And we do the same for
and so on until
Or
Now plug the above into (1) and obtain
Hence, if we take the as common factors, we have
Hence, since from (1), then by comparing coeﬃcients of each basis vector we obtain
Or in matrix form
Hence we ﬁnally obtain that
Let us call the matrix that represents under basis as
We see from the above that the column of contain the coordinates of the vector . This gives us a quick way to construct : Apply to each basis vector , and take the resulting vector and place it in the column of .
We now see that will have a diﬀerent numerical values if the basis used to span are diﬀerent from the ones used to obtain the above
Let use pick some new basis, say . Let the new representation of now be the matrix , then the column of is found from applying on the new basis
Where now is the coordinates of the the vector which will be diﬀerent from since in one case we used the basis set and in the other case we used the basis set . Hence we see that will numerically be diﬀerent depending on the basis used to span the space even though the linear transformation itself did not change.
Now we show how to determine , the matrix which when applied to a vector will result in the vector
Given a vector , it is represented w.r.t. basis as
and w.r.t. basis as
But the vector itself is invariant under any change of basis, hence
 (1) 
Now write each basis in terms of the basis , hence we have
Substitute (2) into RHS of (1) we obtain
Factor out the basis from the RHS, we obtain
hence
Now comparing coeﬃcients of each of the basis we obtain the following result
or in Matrix form, we write
 (3) 
The above gives a relation between the coordinates of the vector w.r.t. basis (these are the coordinates) to the coordinates of the same vector w.r.t. to basis (these are the coordinates ). The mapping between these coordinates is the matrix shown above which we call the matrix. Since the above matrix returns the coordinates of the vector w.r.t. when it is multiplied by the coordinates of the vector w.r.t. basis , we write it as to make it clear from which basis to which basis is the conversion taking place.
Looking at (2) and (3) above, we see that column in is the coordinates of the w.r.t. basis .
Hence the above gives a method to generate the change of basis matrix . Simply represent each basis in w.r.t. to basis . Take the result of each such representation and write it as a column in . We now show few examples to illustrate this.
Example showing how to ﬁnd change of basis matrix
Given the basis ,
and new basis , ﬁnd the change of basis matrix
Column 1 of is the coordinates of w.r.t. basis . i.e.
 (4) 
and column 2 of is the coordinates of w.r.t. basis . i.e.
 (5) 
But (4) is
Hence
and , solving, we obtain
and (5) is
Hence and , solving we obtain , , hence
Now we can use the above change of basis matrix to ﬁnd coordinates of the vector under given its coordinates under . For example, given , then
The ﬁrst step is to ﬁnd
The ﬁrst column of is found by writing w.r.t. basis , and the second column of is found by writing w.r.t. basis ,
Hence
or and
And
Hence and
Hence
Now we need to ﬁnd the representation of w.r.t. basis
The ﬁrst column of is the new coordinates of the basis after applying onto it. but
and the second column of is the new coordinates of the basis after applying onto it. but
Hence
Therefore
Hence we see that the linear transformation is represented as w.r.t. basis , while it is represented as w.r.t. basis . Therefore, it will be better to perform all calculations involving this linear transformation under basis instead of basis
Example 2
Consider , and let be a rotation of space by . Let the ﬁrst basis and let the second basis be .
The ﬁrst step is to ﬁnd
The ﬁrst column of is found by writing w.r.t. basis , and the second column of is found by writing w.r.t. basis ,
Hence
or and and
Hence and
Hence
Now we need to ﬁnd the representation of w.r.t. basis . The ﬁrst column of is the new coordinates of the basis after applying onto it. but
and the second column of is the new coordinates of the basis after applying onto it. but
Hence
Therefore
Hence we see that the linear transformation is represented as w.r.t. basis , while it is represented as w.r.t. basis . Therefore the change of basis selected above did not result in making the representation of any simpler (in fact there was no change in the representation). This means we need to ﬁnd a more direct method of ﬁnding the basis under which has the simplest representation. Clearly we can't just keep trying diﬀerent basis to ﬁnd if has simpler representation under the new basis.
Our goal of simpler representation of is that of a diagonal matrix or as close as possible to being diagonal matrix (i.e. Jordan form). Given and an inﬁnite number of basis that we could select to represent under in the hope we can ﬁnd a new representation such that it is simpler than , we now ask, how does one go about ﬁnding such basis?
It turns out the if we select the eigenvectors of as the columns of , this will result in being diagonal or as close to being diagonal as possible (block diagonal or Jordan form).
Let us apply this to the second example in the previous section. In that example, we had
The eigenvectors of are and ,
Therefore
Which is diagonal representation.
Hence we see that the linear transformation is represented as w.r.t. basis
To obtain such that is real and diagonal requires that be real and symmetric. The eigenvalues of goes into the diagonal of and the eigenvectors of go into the columns of This is an algebraic view.
Geometrically, is viewed as the matrix representation under some basis of a linear transformation . And is the matrix representation of the same linear transformation but now under a new basis and is the matrix that represents the change of basis from to .
The question then immediately arise: If must be real and symmetric (for to be real and diagonal), what does this mean in terms of the linear transformation and change of basis matrix ? This clearly mean that, under some basis , not every linear transformation represented by can have a similar matrix which is real and diagonal. Only those linear transformations which result in real and symmetric representation can have a similar matrix which is real and diagonal. This is shown in the previous examples, where we saw that deﬁned as rotation by under the standard basis resulted in and since this is not symmetric, hence we will not be able to factor this matrix using similarity transformation to a diagonal matrix, no matter which change of basis we try to represent under. The question then, under a given basis what is the class of linear transformation which leads to a matrix representation that is symmetric? One such example was shown above which is the projection into the line at . This question needs more time to look into.
We now write to represent a diagonal matrix, hence the similarity transformation above can be written as
or
Using similarity transformation, we found that for a real and symmetric matrix we are able to decompose it as where is diagonal and contains the eigenvalues of on the diagonal, and contains the right eigenvectors of in its columns.
Right eigenvectors are the standard eigenvectors we have been talking about all the time. When is an eigenvalues of then is a right eigenvector when we write
However, is a left eigenvector of when we have
Given any arbitrary matrix it can be factored into 3 matrices as follows where is a unitary matrix ( or ), and is also unitary matrix.
These are the steps to do SVD
where the matrix was for example.
The following diagram help illustrate the SVD process described above.
As a conclusion, the following diagram shows the diﬀerence between similarity transformation factorization of a matrix and SVD.
Trying to understand SVD from geometrical point view requires more time, and this is left for future research into this subject.