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

MATH152 April 2011

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Question A 27
Which of the following statements are true for all invertible $n\times n$ invertible matrices $A$: (a) $A$ has rank $n$. (b) The reduced row echelon form of $A$ is the identity matrix. (c) The homogeneous problem $A\mathbf {x} =0$ has only the trivial solution $\mathbf {x} =0$. (d) $A$ does not have $\lambda =0$ as an eigenvalue. (e) $\det(A)\not =0$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
By isolating one property of an inverse can you use it to deduce other properties that support or eliminate options?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.