Further properties of determinants¶
Proposition 5.18 (Related exercises: Exercise 5.2, Exercise 6.11)
(Product formula) For two \(n \times n\)-matrices \(A, B\), we have the following formula:
I.e., the determinant of a product (of square matrices of the same size) is the product of the two individual determinants.
In particular, this shows
even though \(AB \ne BA\)!
Proof. We don’t include a full proof, but only observe that one checks this by direct computation when \(A\) is an elementary matrix. This implies the formula if \(A\) is invertible, since then \(A\) is a product of elementary matrices. If \(A\) is not invertible, then one shows that \(AB\) is also not invertible (for any \(B\)), and therefore both \(\det A = 0\) and \(\det (AB) = 0\), so the formula holds in this case too. See, e.g., (Nicholson 1995, Theorem 3.2.1) for a proof. ◻
Remark 5.19
The determinant is therefore multiplicative, but it is not additive: one has
e.g.
Proposition 5.20 (Related exercises: Exercise 5.5, Exercise 5.7)
Let \(A\) be an upper triangular matrix or a lower triangular matrix, i.e., of the form
resp.
Here * stands for an arbitrary entry. Then
Proof. If one of the entries on the main diagonal, i.e., \(a_{11}, \dots, a_{nn}\) is zero, then the columns of \(A\) are linearly dependent, so that \(A\) is not invertible and \(\det A = 0\). If instead all \(a_{ii} \ne 0\), we can divide the \(i\)-th row by \(a_{ii}\), and assume the entries on the main diagonal are all 1. Then, adding appropriate multiples of the rows to the rows above (resp. below in the case of a lower triangular matrix), which does not affect the determinant, gives \(A \leadsto {\mathrm {id}}\), so that \(\det A = 1\), so the claim holds in this case. ◻
Proposition 5.21
For \(A \in {\mathrm {Mat}}_{n \times n}\), we have
i.e., the determinant does not change when passing from \(A\) to its transpose (Definition 4.89).
Proof. For small matrices (of size at most \(3 \times 3\)), this can be proved directly from the formulae in §Subsection 5.2.1.
In general, one may argue like this: if \(A\) is not invertible, then \(A^T\) is not invertible either (by Lemma 4.91). In this case, both sides of the equation are zero. If \(A\) is invertible, it is a product of elementary matrices: \(A = U_1 \dots U_n\). We then have \(A^T = U_n^T \dots U_1^T\). By the product formula (Proposition 5.18), we may therefore assume \(A\) is an elementary matrix. In this case, one checks the claim by inspection:
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for \(A = \left ( \begin{array}{ccccccc} 1 & & & & & & \\ & \ddots & & & & & \\ & & 0 & & 1 & & \\ & & & \ddots & & & \\ & & 1 & & 0 & & \\ & & & & & \ddots & \\ & & & & & & 1 \end{array} \right )\), we have \(A^T = A\), so the claim clearly holds.
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Likewise, for \(A = \left ( \begin{array}{ccccccc} 1 & & & & & & \\ & \ddots & & & & & \\ & & 1 & & & & \\ & & & r & & & \\ & & & & 1 & & \\ & & & & & \ddots & \\ & & & & & & 1 \end{array} \right )\), we have \(A = A^T\), so again the claim holds obviously.
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The matrix \({\left ( \begin{array}{ccccccc} 1 & & & & & & \\ & \ddots & & & & & \\ & & 1 & & & & \\ & & & \ddots & & & \\ & & r & & 1 & & \\ & & & & & \ddots & \\ & & & & & & 1 \end{array} \right )}\) is a lower triangular matrix, and its transpose an upper triangular matrix. Both have determinant 1 according to Proposition 5.20.
◻
Remark 5.22
We introduced the determinant using row operations. The preceding result implies that one can replace the word “row” in all of the above by the word “column”. Applying that, say, to Remark 5.9 we obtain that \(\det A = 0\) if \(A\) has two identical columns.
Proposition 5.23 (Related exercises: Exercise 5.4, Exercise 6.16, Exercise 6.7, Exercise 6.1, Exercise 6.9)
Let \(A = (a_{ij}) \in {\mathrm {Mat}}_{n \times n}\). Then, for any \(i\), one can compute the determinant using “cofactor expansion” along the \(i\)-th row. That is, the following identity holds, where \(c_{ij}\) are the cofactors of \(A\) (Definition 5.14):
Similarly, one can compute it using cofactor expansion along the \(j\)-th column, for any \(j\):
For a proof of this, see, e.g. (Nicholson 1995, sec. 3.6).
Example 5.24
For example, we expand the determinant along the second row:
The choice of the second row (as opposed to the others) is arbitrary, and the result is the same if we choose another row. However, the presence of the \(a_{22} = 0\) simplifies the computation.