Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 27/Solution 1

Recall that the definition of an inverse of $A$ denoted $A^{-1}$ is

$\displaystyle {}AA^{-1}=A^{-1}A=I$

where $I$ is the identity. Now assume we want to solve $A\mathbf {x} =0$ for some $\mathbf {x}$. If $A$ is invertible then by the above property we can left multiply by $A^{-1}>$ to get

$\mathbf {x} =A^{-1}0=0$

so we conclude that if $A$ is invertible then $A\mathbf {x} =0$ only has $\mathbf {x} =0$, the trivial solution and that option c is correct.

Now assume we wanted to find eigenvalues $A\mathbf {v} =\lambda \mathbf {v}$. If we wanted to find zero eigenvalues then we'd need to solve $A\mathbf {v} =0$ but as we saw above, the only solution to this if $A$ is invertible is $\mathbf {v} =0$ which is not a valid eigenvector. Therefore, if $A$ is invertible then it does not have any zero eigenvalues and option d is correct.

A square matrix $A$ of size $n\times n$ always has $n$ eigenvalues. If the matrix has rank $r$ then there are always $r$ non-zero eigenvalues and $n-r$ zero eigenvalues. Since in this case we have shown that there are no zero eigenvalues then $n-r=0$ or $n=r$, thus the matrix $A$ has rank $n$ and option a is correct.

Recall that rank has to do with the number of pivots in a row reduced matrix. If $A$ has full rank (rank $n$) then there will be $n$ pivots which means two things. We have shown that all invertible matrices have rank $n$. Firstly this means that the row reduced form of $A$ is the identity matrix (since it has $n$ pivots) so option b is correct and secondly it means that there are no rows of zeros which means that $\det(A)\not =0$, so option e is correct.

Therefore, from starting with the definition of an inverse we were able to deduce that all of the options are correct.