# Linear Algebra: #13 Determinant

**Deﬁnition**

A mapping det : M(n × n, F) → F is called a determinant function if it satisfies the following three conditions.

- det(I
_{n}) = 1, where I_{n}is the identity matrix. - If A ∈ M(n×n, F) is changed to the matrix A' by multiplying all the elements
in a single row with the scalar a ∈ F, then det(A') = a · det(A). (This is our
row operation S
_{i}(a).) - If A'
is obtained from A by adding one row to a different row, then det(A') =
det(A). (This is our row operation S
_{ij}(1).)

**Simple consequences of this definition**

Let A ∈ M(n × n, F) be an arbitrary n × n matrix, and let us say that A is transformed into the new matrix A' by an elementary row operation. Then we have:

- If A' is obtained by multiplying row i by the scalar a ∈ F, then det(A') = a · det(A). This is completely obvious! It is just part of the definition of “determinants”.
- Therefore, if A' is obtained from A by multiplying a row with −1 then we have det(A' ) = −det(A).
- Also, it follows that a matrix containing a row consisting of zeros must have zero as its determinant.
- If A has two identical rows, then its determinant must also be zero. For can we multiply one of these rows with −1, then add it to the other row, obtaining a matrix with a zero row.
- If A
′
is obtained by exchanging rows i and j, then det(A') = −det(A). This
is a bit more difficult to see. Let us say that A =
**(u**, . . . ,_{1}**u**, . . . ,_{i}**u**, . . . ,_{j}**u**), where_{n}**u**is the k-th row of the matrix, for each k. Then we can write_{k}

(This is the elementary row operation S_{ij}.) - If A'
is obtained from A by an elementary row operation of the form S
_{ij}(c), then det(A') = det(A). For we have:

Therefore we see that each elementary row operation has a well-defined effect on the determinant of the matrix. This gives us the following algorithm for calculating the determinant of an arbitrary matrix in M(n × n, F).

**How to ﬁnd the determinant of a matrix**

Given: An arbitrary matrix A ∈ M(n × n, F).

Find: det(A).

Method:

- Using elementary row operations, transform A into a matrix in step form, keeping track of the changes in the determinant at each stage.
- If the bottom line of the matrix we obtain only consists of zeros, then the determinant is zero, and thus the determinant of the original matrix was zero.
- Otherwise, the matrix has been transformed into an upper triangular matrix,
all of whose diagonal elements are 1. But now we can transform this matrix
into the identity matrix I
_{n}( by elementary row operations of the type S_{ij}(c). Since we know that det(I_{n}) must be 1, we then ﬁnd a unique value for the determinant of the original matrix A. In particular, in this case det(A) ≠ 0.

**Theorem 34**There is only one determinant function and it is uniquely given by our algorithm. Furthermore, a matrix A ∈ M(n × n, F) is regular if and only if det(A) ≠ 0.

In particular, using these methods it is easy to see that the following theorem is true.

**Theorem 35**Let A, B ∈ M(n×n, F). Then we have det(A· B) = det(A) · det(B).

*Proof*

If either A or B is singular, then A · B is singular. This can be seen by thinking about the linear mappings

**V → V**which A and B represent. At least one of these mappings is singular. Thus the dimension of the image is less than n, so the dimension of the image of the composition of the two mappings must also be less than n. Therefore A · B must be singular. That means, on the one hand, that det(A · B) = 0. And on the other hand, that either det(A) = 0 or else det(B) = 0. Either way, the theorem is true in this case.

If both A and B are regular, then they are both in GL(n, F). Therefore, as we have seen, they can be written as products of elementary matrices. It suffices then to prove that det(S

_{1})det(S

_{2}) = det(S

_{1}S

_{2}), where S

_{1}and S

_{2}are elementary matrices. But our arguments above show that this is, indeed, true.

Remembering that A is regular if and only if A ∈ GL(n, F), we have:

**Corollary**

If A ∈ GL(n, F) then det(A

^{−1}) = (det(A))^{−1}.In particular, if det(A) = 1 then we also have det(A

^{−1}) = 1. The set of all such matrices must then form a group.

Another simple corollary is the following.

**Corollary**

Assume that the matrix A is in block form, so that the linear mapping which it represents splits into a direct sum of invariant subspaces (see theorem 29). Then det(A) is the product of the determinants of the blocks.

*Proof*

If

That is, for the matrix A

_{i}

^{*}, all the blocks except the i-th block are replaced with identity-matrix blocks. Then A = A

_{1}

^{*}· · · A

_{p}

^{*}, and it is easy to see that det(A

_{i}

^{*}) = det(A

_{i}) for each i.

**Deﬁnition**

The special linear group of order n is defined to be the set

SL(n, F) = {A ∈ GL(n, F) : det(A) = 1}.

**Theorem 36**Let A' = C

^{−1}AC. Then det(A ′ ) = det(A).

*Proof*

This follows, since det(C

^{−1}) = (det(C))

^{−1}.

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