• Nov 28th 2013, 01:16 PM
CrispyPlanet
Question 1: Let A be a k x k matrix and let B be an (n-k) x (n-k) matrix. Let

$E= \begin{array}{cc} I_k & 0 \\ 0 & B\end{array}$

Show that det(E) = det(B)

My attempt: $det(E)=det(BI_k) - det(00)=det(B)det(I_k)=det(B)*1=det(B)$

I'm not sure this is correct, as I don't know the rules for finding the determinant of a matrix populated itself with matrices. Can anyone point me in the right direction?

Thanks,
CP
• Nov 28th 2013, 02:30 PM
romsek
Quote:

Originally Posted by CrispyPlanet
Question 1: Let A be a k x k matrix and let B be an (n-k) x (n-k) matrix. Let

$E= \begin{array}{cc} I_k & 0 \\ 0 & B\end{array}$

Show that det(E) = det(B)

My attempt: $det(E)=det(BI_k) - det(00)=det(B)det(I_k)=det(B)*1=det(B)$

I'm not sure this is correct, as I don't know the rules for finding the determinant of a matrix populated itself with matrices. Can anyone point me in the right direction?

Thanks,
CP

One way you could show this is realizing that B is diagonizable by similarity transforms. Those (n-k)x(n-k) transform matrices can be extended to k x k by filling off diagonal elements with 0 and diagonal elements with 1 and using the new k x k transform matrix on E won't affect the Ik block. Then what you will have is a the product of a unitary matrix transpose with a diagonal matrix, the elements being the eigenvalues of E, and a unitary matrix. The determinant of this is the product of the matrix determinants which is 1 * det(E) * 1.

det(E) is read off as the product of it's diagonal elements which is just 1n * det(B) = det(B)

There are probably easier ways. See this and look for the section on Block matrices
• Nov 28th 2013, 02:42 PM
CrispyPlanet