10
Feb 19

## Properties IV-VI

Exercise 1. Let $S$ be a linearly independent system of $n-1$ vectors in $R^n$. Then it can be completed with a vector $B$ to form a basis in $R^n.$

Proof. One way to obtain $B$ is this. Let $P$ be a projector onto the span of $S$ and let $Q=I-P.$ Take as $B$ any nonzero vector from the image of $Q.$ It is orthogonal to any element of the image of $P$ and, in particular, to elements of $S.$ Therefore $S$ completed with $B$ gives a linearly independent system $\tilde{S}$. $\tilde{S}$ is a basis because $x=Px+Qx$ for any $x\in R^n.$

Property IV. Additivity. Suppose the $i$th row of $A$ is a sum of two vectors: $A=\left(\begin{array}{c}A_1\\...\\A_i^\prime+A_i^{\prime\prime }\\...\\A_n\end{array} \right).$

Denote $A^\prime=\left(\begin{array}{c}A_1\\...\\A_i^\prime\\...\\A_n\end{array}\right)$, $A^{\prime\prime}=\left(\begin{array}{c}A_1\\...\\A_i^{\prime\prime}\\...\\A_n\end{array}\right)$

(except for the $i$th row, all others are the same for all three matrices). Then $\det A=\det A^\prime+\det A^{\prime\prime}.$

Proof. Denote $S$ the system of $n-1$ vectors $A_1,...,A_{i-1},A_{i+1},...,A_n.$

Case 1. If $S$ is linearly dependent, then the system of all rows of $A$ is also linearly dependent. By Property III the determinants of all three matrices $A,\ A^\prime,\ A^{\prime\prime}$ are zero and the statement is true.

Case 2. Let $S$ be linearly independent. Then by Exercise 1 it can be completed with a vector $B$ to form a basis in $R^n.$ $A_i^\prime,\ A_i^{\prime\prime }$ can be represented as linear combinations of elements of $\tilde{S}.$ We are interested only in the coefficients of $B$ in those representations. So let $A_i^\prime=C+kB,$ $A_i^{\prime\prime}=D+lB,$ where $C$ and $D$ are linear combinations of elements of $S.$ Hence, $A_i^\prime+A_i^{\prime\prime}=C+D+(k+l)B.$

We can use Property II to eliminate $C,$ $D$ and $C+D$ from the $i$th rows of $A^\prime,$ $A^{\prime\prime}$ and $A,$ respectively, without changing the determinants of those matrices. Let $A^0$ denote the matrix obtained by replacing the $i$th row of $A$ with $B.$ Then by Property II and Axiom 1 $\det A=(k+l)\det A^0,$ $\det A^\prime=k\det A^0,$ $\det A^{\prime\prime}=l\det A^0,$

which proves the statement.

Combining homogeneity and additivity, we get the following important property that some people use as a definition:

Property V. Multilinearity. The determinant of $A$ is a multilinear function of its rows, that is, for each $i,$ it is linear in row $A_i,$ when the other rows are fixed.

Property VI. Antisymmetry. If the matrix $A^0$ is obtained from $A$ by changing places of two rows, then $\det A^0=-\det A.$

Proof. Let $A=\left(\begin{array}{c}...\\A_i\\...\\A_j\\...\end{array}\right),$ $A^0=\left(\begin{array}{c}...\\A_j\\...\\A_i\\...\end{array}\right)$

(all other rows of these matrices are the same). Consider the next sequence of transformations: $\left(\begin{array}{c}...\\A_i\\...\\A_j\\...\end{array}\right)\rightarrow\left(\begin{array}{c}...\\A_i+A_j\\...\\A_j\\...\end{array}\right)\rightarrow\left(\begin{array}{c}...\\A_i+A_j\\...\\A_j-(A_i+A_j)\\...\end{array}\right)=\left(\begin{array}{c}...\\A_i+A_j\\...\\-A_i\\...\end{array}\right)\rightarrow\left(\begin{array}{c}...\\A_i+A_j+(-A_i)\\...\\-A_i\\...\end{array}\right)=\left(\begin{array}{c}...\\A_j\\...\\-A_i\\...\end{array}\right).$

By Property II, each of these transformations preserves $\det A.$ Recalling homogeneity, we finish the proof.