Raising the bar

Properties IV-VI

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

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

Property IV. Additivity. Suppose the $latex i$th row of $latex A$ is a sum of two vectors:

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

Denote

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

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

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

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

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

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

$latex \det A=(k+l)\det A^0,$ $latex \det A^\prime=k\det A^0,$ $latex \det A^{\prime\prime}=l\det A^0,$

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

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

Proof. Let

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

(all other rows of these matrices are the same). Consider the next sequence of transformations:

$latex \left(\begin{array}{c}... \\ A_i \\... \\ A_j \\... \end{array}\right)\rightarrow\left(\begin{array}{c}... \\A_i+A_j \\... \\A_j \\... \end{array}\right)\rightarrow\left(\begin{array}{c}... \\A_i+A_j \\... \\A_j-(A_i+A_j) \\... \end{array}\right)$

$latex =\left(\begin{array}{c}... \\A_i+A_j \\... \\-A_i \\... \end{array}\right)\rightarrow\left(\begin{array}{c}... \\A_i+A_j+(-A_i) \\... \\-A_i \\... \end{array}\right)=\left(\begin{array}{c}... \\A_j \\... \\-A_i \\... \end{array}\right).$

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

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