What is the Identity Property of Determinants and its Relation to Linearity?

[twoflower]

Hi all,

I don't understand one thing about linearity of determinants. In the book I have:

<br /> \det \left( \begin{array}{ccc} . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{<i>} \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{[j]} \end{array} \right) = \det \left( \begin{array}{ccc} . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{<i>} \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{[j+i]} \end{array} \right) <br /> </i></i>

And the explanation is:

<br /> \det \left( \begin{array}{ccc} . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{<i>} \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{[j+i]} \end{array} \right) = \det \left( \begin{array}{ccc} . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{<i>} \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{[j]} \end{array} \right) + \det \left( \begin{array}{ccc} . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{<i>} \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ . &amp; . &amp; . \\ \mbox{<i>} \end{array} \right)<br /> </i></i></i></i>

But I can't see how these two matrixes (I mean now left and right side of the bottom equation) can be identical, because when I sum the two matrixes on the right, I won't get the matrix on the left...

Thank you for the explanation.

 

[Galileo]

You shouldn't add the bottom right matrices, since they are determinants. The last determinant is zero because two rows are equal.
To see why the equality is true, expand the first along the last row.

 

[twoflower]

You shouldn't add the bottom right matrices, since they are determinants. The last determinant is zero because two rows are equal.
To see why the equality is true, expand the first along the last row.

This property of determinant is before expaning along rows/columns, so I think it should be possible to see it even simplier.

I know they are determinants, but I suppose that if

A = B + C
then det(A) = det(B) + det(C)

 

[robphy]

Here's a special case that might help:

\vec A\times (\vec B +\vec A) = \vec A\times\vec B +\vec A\times \vec A= \vec A\times\vec B

I know they are determinants, but I suppose that if

A = B + C
then det(A) = det(B) + det(C)

This is generally false.
Let B=\left(\begin{array}{cc} 1 &amp; 0 \\ 0 &amp;0 \end{array} \right) and C=\left(\begin{array}{cc} 0&amp; 0 \\ 0 &amp;1 \end{array} \right). These have zero determinant... so the sum of the determinants is zero. However, the matrix sum has determinant 1.

 

[twoflower]

Thank you, I think I have it. I just have to write the expression for the determinant of the matrix on the left side and I can split it into two determinants equal to the ones on the right side. Thanks.