Properties of Determinants

June 5, 2019

Every square matrix has a determinant associated with it. The determinant will be a test for invertibility, since it is faster(?) to compute it than to use eliminations to check if we have a full rank matrix. If determinant of some matrix is 0 then it is not invertible. And determinant got a lot more than that. We use $\|A\|$ or $\det A$ to represent matrix $A$’s determinant.

There’re three properties of a determinant, and we can infer another seven from the basic ones:

1. $\det I=1$.

2. Exchanging rows reverse the sign of determinant. Therefore $\det P= 1/-1$ if the number of exchanges is even/if odd

3. (a) If we multiply a row of the matrix with factor $t$, then the determinant will also be multiplied by $t$, that is $\begin{vmatrix} ta & tb\\ c & d \end{vmatrix} =t\begin{vmatrix} a & b\\ c & d \end{vmatrix}$

   (b) $\begin{vmatrix} a+a' & b+b'\\ c & d \end{vmatrix} =\begin{vmatrix} a & b\\ c & d \end{vmatrix} +\begin{vmatrix} a' & b'\\ c & d \end{vmatrix}$ note that this is not saying $\det(A+B)=\det(A)+\det(B)$. This is linear in each row, separately.

4. 2 equal rows $\rightarrow \det=0$. We can infer this from property 2. When we have two rows the same in a matrix, we can exchange those rows and get a inverse determinant. But the matrix in fact isn’t changed at all just by switching rows. Adding these two determinant will give us zero.

5. Subtracting a multiple of one row from the other does not change the determinant. This is saying, by property 3b. $\begin{vmatrix} a & b\\ c-ka & d-kb \end{vmatrix} =\begin{vmatrix} a & b\\ c & d \end{vmatrix} +\begin{vmatrix} a & b\\ -ka & -kb \end{vmatrix}$. And from property 3a., $\begin{vmatrix} a & b\\ c & d \end{vmatrix} +\begin{vmatrix} a & b\\ -kb & -kb \end{vmatrix} =\begin{vmatrix} a & b\\ c & d \end{vmatrix} -k\begin{vmatrix} a & b\\ a & b \end{vmatrix}$, we factor out the $-k$, we finally see by property 4 same rows give us zero determinant, so it does not change the determinant of the original matrix.

6. If any rows of a matrix is zero, then the determinant will be 0. We can use $t=0$ from 3a to obtain this.

7. Determinant of an upper triangular matrix $U$ is the product of all its diagonal items $d_1*d_2*…*d_n$. Suppose all $d_i$ are non-zeros. Then we can get a symmetric matrix by subtracting all numbers above the diagonal line. Note this does not affect the determinant by property 5. And we can use 3a to factor out each row’s $d_i$ one by one, and what’s left in the matrix is identity $I$. If there’s any zero $d_i$, we can make the whole row to zero by elimination, and get $\det=0$.

8. $\det A=0$ iff $A$ is singular. For <-, If A is singular, we can get a row of zero. If A is non-singular, we get D and what’s happened in property 7.

9. $\det{AB}=\det A *\det B$This can be proved when later we know eigenvalues. We are just multiplying the eigenvalues of these two matrices. From this we can get $\det I=\det {A^{-1}A}=\det A*\det A^{-1}=1$, then $\det A^{-1}=\frac{1}{\det A}$. Similar things happen for $A^2$ and etc. But note that $\det 2A=2^n\det A$ because we have 2 on every row. By property 3a we can get this, where $n$ is the number of rows.

10. $\det A^\top=\det A$ $\begin{aligned} \begin{vmatrix} A^{\top } \end{vmatrix} & =\begin{vmatrix} A \end{vmatrix}\\ \begin{vmatrix} U^{\top } L^{\top } \end{vmatrix} & =\begin{vmatrix} LU \end{vmatrix}\\ \begin{vmatrix} U^{\top } \end{vmatrix}\begin{vmatrix} L^{\top } \end{vmatrix} & =\begin{vmatrix} L \end{vmatrix}\begin{vmatrix} U \end{vmatrix} \end{aligned}$ and both triangular, tranposed determinants are the same. One implication is that from property 6, any column is zero will lead to determinant as 0.