Which of the following statements about determinants are correct?

Question

Which of the following statements about determinants are correct?

1. $\det(A^2)>0$, for all invertible matrices $A$

2. $\det(A+A^{-1})=\det(A)+\dfrac{1}{\det(A)}$, for all invertible matrices $A$

3. $\det(vv^T)>0$, for all column vectors $v ≠ 0$

4. $\det(AB^T)=\det((A^T)B)$, for all square matrices $A$ and $B$

Which of the statements are correct? I do not feel secure about which of them that is true.

My answer: My calculations have given me that $2$ and $4$ are true. Am I correct?

Answer 1

• $\det (A^2)=\det A \cdot \det A=(\det A)^2>0$
• $\det (I+I)=\det (2I)=2^n .\det I=2^n$ whereas $\det I+\frac{1}{\det I}=2$
• $vv^T$ has rank $1$, so its eigenvalues are zero [$(n-1)$ times] and its trace and hence $\det vv^T=0$
• $\det (AB^T)=\det A \cdot \det B^T=\det A \cdot \det B=\det A^T.\det B=\det(A^TB)$

Answer 2

For 1, use that $\det(AB)= \det(A)\det(B)$ to find that $\det(A^2) = \det(A)^2 > 0$.

For 2, take $A = I $.

For 3, note that the rows of every matrix $vv^T$ are scalar multiples of $v$. Thus, the determinant will be zero for all vectors with more thant $1$ entry.

For 4, use that $\det(A^T) = \det(A)$, so $\det(AB^T) = \det((AB^T)^T) = \det(BA^T) = \det(A^TB)$.