# Mathematics Prelims

## June 16, 2009

### Determinants Are Linear in Rows and Columns

Filed under: Algebra,Linear Algebra — cjohnson @ 8:55 pm

One easy consequence of our definition of determinant from last time is that any singular matrix must have determinant zero. Suppose $A$ is a singular $n \times n$ matrix and that $P$ is the matrix which puts $A$ into row reduced form. Then we have

$\displaystyle \det(A)$

$\displaystyle = \det(P \, P^{-1} \, A)$

$\displaystyle = \det(P) \, \det(P^{-1}) \, \det(A)$

$\displaystyle = \det(P) \, \det(A) \, \det(P^{-1})$

$\displaystyle = \det(P \, A) \det(P^{-1})$

If $A$ is singular, once we put it in row reduced form it must have a row of zeros. We can now break $PA$ up into a product of elementary matrices, one of which will have be of the form $I_{(0R_i)}$. We know that this matrix will have determinant zero, so the product will be zero, and thus $\det(A) = 0$. Likewise, if $\det(A) = 0$, we can write $A$ as a product of elementary row matrices and one will be $I_{(0R_i)}$, so $A$ is singular. Now we know that a matrix is singular if and only if its determinant is zero.

Suppose now that the first row of $A$ can be written as $\alpha + \beta$ for some vectors $\alpha$ and $\beta$. We wish to show that

$\displaystyle \det(A) = \det \left[ \begin{array}{c} \alpha + \beta \\ R_2 \\ \vdots \\ R_n \end{array} \right] = \det \left[ \begin{array}{c} \alpha \\ R_2 \\ \vdots \\ R_n \end{array} \right] + \det \left[ \begin{array}{c} \beta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

First suppose that the rows $R_2$ through $R_n$ form a linearly dependent set. Then our matrix is singular so has determinant zero. The determinants on the right in the above equation are zero too, so our we have our result.

Suppose now that $R_2$ through $R_n$ are linearly independent. We can then extend these to a basis for $\mathcal{F}_{1 \times n}$ by adding a vector, call it $\zeta$. Then there exist scalars $a_i, b_i$ for $i = 1, ..., n$ such that

$\alpha = a_1 \zeta + a_2 R_2 + ... + a_n R_n$

$\beta = b_1 \zeta + b_2 R_2 + ... + b_n R_n$

Some simple manipulations from last time give us the following.

$\displaystyle \det \left[ \begin{array}{c} \alpha \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

$\displaystyle = \det \left[ \begin{array}{c} a_1 \zeta + a_2 R_2 + ... + a_n R_n \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

$\displaystyle = \det \left[ \begin{array}{c} a_1 \zeta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

$\displaystyle = a_1 \det \left[ \begin{array}{c} \zeta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

And likewise,

$\displaystyle \det \left[ \begin{array}{c} \beta \\ R_2 \\ \vdots \\ R_n \end{array} \right] = b_1 \det \left[ \begin{array}{c}\zeta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

Now we combine these results,

$\displaystyle \det \left[ \begin{array}{c} \alpha + \beta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

$\displaystyle = \det \left[ \begin{array}{c} (a_1 + b_1) \zeta + (a_2 + b_2) R_2 + ... + (a_n + b_n) R_n \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

$\displaystyle = (a_1 + b_1) \det \left[ \begin{array}{c} \zeta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

$\displaystyle = a_1 \det \left[ \begin{array}{c} \zeta \\ R_2 \\ \vdots \\ R_n \end{array} \right] + b_1 \det \left[ \begin{array}{c} \zeta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

$\displaystyle = \det \left[ \begin{array}{c} \alpha \\ R_2 \\ \vdots \\ R_n \end{array} \right] + \det \left[ \begin{array}{c} \beta \\ R_2 \\ \vdots \\ R_n \end{array} \right]$

Since we can swap rows without altering the determinant, this result holds for any rows.  A similar argument shows the result also holds for columns.