Det(A+B) ? det(A) + det(B)

[phyin]

det(A+B) ?? det(A) + det(B)

I'm guessing greater than but I'm not too sure. I need a proof on this so I can be assured of it and then use the statement to prove something else.

any hint (or link to proof) would be much appreciated.

edit:

x*det(A) ?? det(x*A)
what's the relation there?

 

[micromass]

Neither holds, take

$$A=B=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)$$

for one equality and


$$A=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right),~B=-A$$

for the other.

For the last question, we do have the equality

$$det(xA)=x^n detA$$

where A is the rank of the matrix. From this we can deduce that neither of the inequalities between det(xA) and xdet(A) holds true. Just take x>1 and x<1.

 

[phyin]

thanks. i better find an alternative way to my proof ;s

 

[AlephZero]

Another simple counter-example is
$$A = \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \qquad B = \begin{pmatrix} 0 & 0 \\ 0 & b \end{pmatrix}$$

det(A) = det(B) = 0, det(A+B) = ab

 

[blue_raver22]

The correct relationship is actually det(A+B) = det(A) + det(B) only if A and B commute, which means that AB = BA. This is known as the distributive property of determinants.

To prove this, we can use the definition of determinants as the signed volume of a parallelepiped. Let A and B be two n x n matrices. The determinant of A+B is equal to the signed volume of the parallelepiped formed by the column vectors of A and B.

On the other hand, the determinant of A is equal to the signed volume of the parallelepiped formed by the column vectors of A, and the determinant of B is equal to the signed volume of the parallelepiped formed by the column vectors of B.

Now, when we add A and B, we are essentially combining the column vectors of A and B to form a new parallelepiped. The signed volume of this new parallelepiped will be equal to the sum of the signed volumes of the individual parallelepipeds formed by A and B.

Therefore, we can conclude that det(A+B) = det(A) + det(B) when A and B commute.

As for the relation between x*det(A) and det(x*A), we can use the same logic as above. The determinant of x*A is equal to the signed volume of the parallelepiped formed by the column vectors of x*A. This is equivalent to scaling the column vectors of A by a factor of x, which will result in a new parallelepiped with a signed volume of x*det(A). Therefore, the relationship is det(x*A) = x*det(A).