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.

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1 Comment »

  1. […] Now, applying the linearity we discussed last time, […]

    Pingback by The Laplace/Cofactor Expansion « Mathematics Prelims — June 17, 2009 @ 3:22 pm | Reply


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