By the definition following Example 2.2, a matrix is diagonalizable if it represents a transformation with the property that there is some basis such that is a diagonal matrix— the starting and ending bases must be equal. But Theorem 2.6 says only that there are and such that we can change to a representation and get a diagonal matrix. We have no reason to suspect that we could pick the two and so that they are equal.
Must matrix equivalent matrices have matrix equivalent transposes?
Yes. Row rank equals column rank, so the rank of the transpose equals the rank of the matrix. Same-sized matrices with equal ranks are matrix equivalent.
Show that matrix-equivalence is an equivalence relation.
For reflexivity, to show that any matrix is matrix equivalent to itself, take and to be identity matrices. For symmetry, if then (inverses exist because and are nonsingular). Finally, for transitivity, assume that and that . Then substitution gives . A product of nonsingular matrices is nonsingular (we've shown that the product of invertible matrices is invertible; in fact, we've shown how to calculate the inverse) and so is therefore matrix equivalent to .
This exercise is recommended for all readers.
Show that a zero matrix is alone in its matrix equivalence class. Are there other matrices like that?
By Theorem 2.6, a zero matrix is alone in its class because it is the only of rank zero. No other matrix is alone in its class; any nonzero scalar product of a matrix has the same rank as that matrix.
What are the matrix equivalence classes of matrices of transformations on ? ?
There are two matrix-equivalence classes of matrices— those of rank zero and those of rank one. The matrices fall into four matrix equivalence classes.
How many matrix equivalence classes are there?
For matrices there are classes for each possible rank: where is the minimum of and there are classes for the matrices of rank , , ..., . That's classes. (Of course, totaling over all sizes of matrices we get infinitely many classes.)
Are matrix equivalence classes closed under scalar multiplication? Addition?
They are closed under nonzero scalar multiplication, since a nonzero scalar multiple of a matrix has the same rank as does the matrix. They are not closed under addition, for instance, has rank zero.
Let represented by with respect to .
Find in this specific case.
Describe in the general case where .
and thus the answer is this.
As a quick check, we can take a vector at random
while the calculation with respect to
yields the same result.
and, as in the first item of this question
so, writing for the matrix whose columns are the basis vectors, we have that .
Let have bases and and suppose that has the basis . Where , find the formula that computes from .
Repeat the prior question with one basis for and two bases for .
The adapted form of the arrow diagram is this.
Since there is no need to change bases in (or we can say that the change of basis matrix is the identity), we have where .
Here, this is the arrow diagram.
We have that where .
If two matrices are matrix-equivalent and invertible, must their inverses be matrix-equivalent?
If two matrices have matrix-equivalent inverses, must the two be matrix-equivalent?
If two matrices are square and matrix-equivalent, must their squares be matrix-equivalent?
If two matrices are square and have matrix-equivalent squares, must they be matrix-equivalent?
Here is the arrow diagram, and a version of that diagram for inverse functions.
Yes, the inverses of the matrices represent the inverses of the maps. That is, we can move from the lower right to the lower left by moving up, then left, then down. In other words, where (and invertible) and are invertible then .
Yes; this is the prior part repeated in different terms.
No, we need another assumption: if represents with respect to the same starting as ending bases , for some then represents . As a specific example, these two matrices are both rank one and so they are matrix equivalent
but the squares are not matrix equivalent— the square of the first has rank one while the square of the second has rank zero.
No. These two are not matrix equivalent but have matrix equivalent squares.
This exercise is recommended for all readers.
Square matrices are similar if they represent the same transformation, but each with respect to the same ending as starting basis. That is, is similar to .
Prove that similar matrices are matrix equivalent.
Show that similarity is an equivalence relation.
Show that if is similar to then is similar to , the cubes are similar, etc. Contrast with the prior exercise.
Prove that there are matrix equivalent matrices that are not similar.
The definition is suggested by the appropriate arrow diagram.
Call matrices similar if there is a nonsingular matrix such that .
Take to be and take to be .
This is as in Problem 10. Reflexivity is obvious: . Symmetry is also easy: implies that (multiply the first equation from the right by and from the left by ). For transitivity, assume that and that . Then and we are finished on noting that is an invertible matrix with inverse .
Assume that . For the squares: . Higher powers follow by induction.
These two are matrix equivalent but their squares are not matrix equivalent.
By the prior item, matrix similarity and matrix equivalence are thus different.