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Given two matrices $A,B \in M_{4 \times 4} (\mathbb{R})$ such that $\det (AB) \ne 0$ what is the rank of the matrix B? How to solve this I have no idea how to approach this problem.

Martin Sleziak
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Guestinc
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2 Answers2

4

Hint: use the fact that $\det(AB)=\det(A)\det(B)$.

Alijah Ahmed
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  • Thus the rank of b will be equal to 4? – Guestinc Mar 09 '15 at 13:28
  • I guess, but I can not really confirm. I think a determinant $\neq 0$ means that your matrix is of full rank but I am not entirely into determinants. Can some one else confirm or deny this? – Pedro Mar 09 '15 at 13:36
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    @Pedro See: http://math.stackexchange.com/questions/355644/what-does-it-mean-to-have-a-determinant-equal-to-zero or http://math.stackexchange.com/questions/507638/intuition-behind-matrix-being-invertible-iff-determinant-is-non-zero (And probably several other posts on this site.) And maybe also ProofWiki. – Martin Sleziak Mar 09 '15 at 14:18
  • @MartinSleziak Thanks! So the answer is 4, since the matrix is invertible and thus of full rank? – Pedro Mar 09 '15 at 14:20
  • But am i right with the rank of B? – Guestinc Mar 09 '15 at 14:21
  • I guess it must be 4. Since you have only independent columns and the only solution in the nullspace of A = {0}, which means all your dimensions are part of your column space, and none of them are part of the null space. The number of independent columns is really what a rank of a matrix means. The more independent columns, the more "necessary dimensions" your matrix carries. Since all columns are needed to express the system, none of those columns can be expressed as a linear combination of te previous ones. – Pedro Mar 09 '15 at 14:25
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The matrix $AB$ is invertible, so has rank$~4$. The rank of a product of matrices can never be more than the rank of either of those matrices, so $B$ cannot have rank less than $4$. Now consider the size of $B$.

Marc van Leeuwen
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