## Notes on diagonalizable linear operators

Okay, I have a few things in my mind about this topic and I don’t want to lose them tomorrow or the day after. Therefore, I created this note. Hope it is useful to other people as well. Background of this note are as follows:

- Suppose is an matrix and it is diagonalizable. Therefore, by definition, there is an invertible matrix and a diagonal matrix . However, I always forget the relationship between and , whether it is or . Generally, those two are no different, but we can choose such that the column vectors of matrix are the eigenvectors of matrix , hence we have to know precisely which one it is, otherwise it gets wrong. Until now, when I forget about this, I always derive it and it takes some of my precious time (haha) but now after I have this, I can just open the Internet and look for this note.
- Point 1 above talks about diagonalizability of a matrix. But generally, we have a linear operator acting on a general vector field instead. A linear operator is called diagonalizable if there is a basis B such that the matrix representation of this linear operator is a diagonal matrix. It seems for me, diagonalizability of a linear operator and that of a matrix are talking about two different things. One is talking about how to find a basis, while the other is talking about how to find an invertible matrix. However, these two concepts turn out to be the same.

We begin with the following theorem.

*Theorem 1 (Coordinate-change matrix)*

*Let be a standard basis and be another basis for an -dimensional linear space . Then there is an invertible matrix such that for every *

*. (1)*

*Matrix is called the transition or coordinate-change matrix from the basis to the basis . Note that and are the coordinates of vector with respect to basis and respectively. In fact, the column vectors of matrix are the coordinates of the basis vectors of with respect to the standard bases (i.e. if then ). *

We won’t prove the theorem as it can be found in any linear algebra textbook. However, using this theorem we can relate the diagonalizability of a linear operator and that of a matrix.Suppose is a diagonalizable linear operator on an -dimensional linear space . Take then let be . Suppose is a standard basis in and is a basis in that consists of independent eigenvectors of .

Then we have the following.

(i) , where is a representation matrix of with respect to standard basis .

(ii) , where is a representation matrix of with respect to basis . We also know that is a diagonal matrix.

We want to show that using the theorem 1, or . From equation (1) we get:

.

Hence, we have . While from point (ii) we have . As a conclusion, we have,

.

We have derived that in fact, is the right one. We also derived that the invertible matrix has such a close relationship with the basis as the column vectors of are the basis vectors of , which are the eigenvectors of the linear operator . This conclude my note.

## The solvability condition of A x = b when A is not a square matrix

In an elementary linear algebra course, we learned the solvability condition of the linear system

where is an matrix and are vectors.

To mention a few, the followings are the solvability conditions (equivalent conditions) such that the linear system above has a solution:

- determinant of is not zero
- rank of is full
- all eigenvalues of are non zero
- the vector columns (also rows) are independent
- has an inverse
- et cetera

However, I would like to discuss the case when either (#) is not a square matrix or

(##) the rank of is not full (< $latex n$).

There is actually a condition such that the linear system has a solution whenever we have either (#) or (##). The solvability condition is

.

The thing is this condition is not applicable to most problems I have ever had. For instances, I bumped into the following problem.

Let be the following

, and . I want to seek a condition such that the linear system is solvable for . I want to find a condition for and such that there is a solution of .

The statement did not help me at that time and I look for another implication for this statement that is applicable enough to my problem.

**Fundamental Theorem of Orthogonality **(see [1])

Let be any matrices. The row space of is orthogonal to the null space of .

Proof: Suppose is a vector in the nullspace. Then . This equation can be considered as rows of multiplying :

= = .

We can clearly see that a dot product of row 1 and is zero, a dot product of row 2 and is zero, and so on. This means that the dot product of every row of and is zero, which means every row of is orthogonal to , which completes the proof.

**Corrolary 1**

**The column space of is orthogonal to the left null space of .**

Proof: The proof is easy, we just need to consider the transpose of the matrix .

Going back to our original problem, when the vector is in the column space of , the vector will be orthogonal to any vector in the left null space of .

Thus, we have the following solvability condition:

**Solvability condition of for any matrices
**The linear system has a solution if the dot product of and is zero, where is the base vector of the left null space of .

Remark, the above solvability condition is closely related to the linear version of Fredholm solvability condition.

Reference

[1] Strang, G., “Linear Algebra and Its Applications,” Thomson Brooks/Cole

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