55
$\begingroup$

1) Switching two rows or columns causes the determinant to switch sign

2) Adding a multiple of one row to another causes the determinant to remain the same

3) Multiplying a row as a constant results in the determinant scaling by that constant.

Using the geometric definition of the determinant as the area spanned by the columns of the matrix, could someone give me a geometric interpretation of the above theorems? Thanks.

$\endgroup$
9
  • $\begingroup$ @julien Oops, sorry about that. Fixed. $\endgroup$
    – dfg
    Dec 8, 2013 at 15:27
  • 1
    $\begingroup$ What have you tried? For (2), a sketch of the 2-dimensional case should help (hint: shearing!). For (1), you'll have to deal with the sign issues you glossed over in your geometric interpretation - it's the determinant's absolute value that equals the area - and figure out what the sign tells you (hint: left- vs. right-handedness). For (3), a sketch should help also, plus some intuition about how areas change under scalings... $\endgroup$
    – fgp
    Dec 8, 2013 at 15:54
  • $\begingroup$ @fgp For (3), I couldn't use my intuition about areas, because if you multiply the columns by a factor the way the area changes is simple enough, your simply scaling it! But why does multiplying a row (i.e. just coordinates of the columns) affect the area the same way? $\endgroup$
    – dfg
    Dec 8, 2013 at 15:58
  • $\begingroup$ @fgp For 2, the shearing analogy helps explain why adding one column to another doesn't cause a difference in shearing, but I don't see how shearing helps with understanding adding rows. $\endgroup$
    – dfg
    Dec 8, 2013 at 16:14
  • 1
    $\begingroup$ I have tried to put "row operation" determinant geometrically parallelogram into Google Books. Already the first hit returns a book having some pictures which might help your intuition. You can also try to search web with the same query. $\endgroup$ Jan 22, 2014 at 11:45

1 Answer 1

46
+50
$\begingroup$

Row operations can be thought of as acting on the ENTIRE space by reflection, shearing, or dilation.

For 2, have you heard of Cavalieri's principle? It says that shearing things while holding each cross section steady maintains volume.

For 3, you are just stretching or shrinking along each axis.

For 1, you are just reflecting in the plane $x_i=x_j$.

Edit: Every row operation is the effect of multiplying on the left by an elementary matrix. We can think of this elementary matrix as a map from $R^n$ to $R^n$, which changes every vector in $R^n$ including the column vectors. Thus, each row operation corresponds to a way of changing the whole space.

The row operation in 1 interchanges two rows. This corresponds to interchanging two coordinates in the space. It is not obvious, but it has been shown that interchanging two coordinates is the same thing as reflecting the entire space around the subset where the two coordinates are equal. This does not change volume.

The row operation in 3 corresponds to stretching one coordinate by the multiple given, which multiplies volume by the same amount.

The operation in 2 can be thought of as follows: say that you add a multiple of the second row to the first. Imagine the space sliced into 'pancakes', one for each value of the second coordinate. The map doesn't interchange pancakes, it just slides each pancake 'horizontally'. This doesn't change the volume of anything.

$\endgroup$
15
  • 1
    $\begingroup$ I don't understand what you mean by your first statement ("Row operations can be thought of as..."). Could you expand on it? $\endgroup$
    – dfg
    Jan 12, 2014 at 17:33
  • $\begingroup$ Also for 3), whats $x_i$ and $x_j$? $\endgroup$
    – dfg
    Jan 12, 2014 at 17:39
  • 1
    $\begingroup$ I'm thinking of the $i$th row as the $x_i$ coordinates of the column vectors. Multiplying the row by 3 multiplies the $x_i$ coordinate of each vector by 3; but multiplying the entire *space's$ $x_i$ coordinate bh 3 has the same effect. $\endgroup$ Jan 12, 2014 at 20:04
  • $\begingroup$ For 3), $i$ and $j$ are the two rows being switched. $\endgroup$ Jan 12, 2014 at 20:05
  • 1
    $\begingroup$ Technically it is a hyperplane. In 2-d, it would be the line $y=x$, while in 3-d it is one of the planes $y=x$, $x=z$, or $y=z$. Reflecting (1,2,3) across $y=z$ gives (1,3,2). $\endgroup$ Jan 13, 2014 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.