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View Full Version : det(A+B) ?? det(A) + det(B)


phyin
Sep14-11, 10:06 AM
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
Sep14-11, 10:17 AM
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
Sep14-11, 10:27 AM
thanks. i better find an alternative way to my proof ;s

AlephZero
Sep14-11, 04:17 PM
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