Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 15

MATH152 April 2013

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Question A 15

Assume that the matrix B was obtained by applying a sequence of elementary row operations to the matrix A. Circle all the correct statements that are true independent of what the given A and B are. (a) $\displaystyle \det A=\det B$ (b) If B is invertible then A is also invertible. (c) If $v={\begin{bmatrix}1\\0\end{bmatrix}}$ is an eigenvector of A, then v is also an eigenvector of B. (d) If $\lambda =0$ is an eigenvalue of A, then $\lambda =0$ is also an eigenvalue of B.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Elementary row operations are row switching row multiplication by a non-zero constant row addition Do any of these change the value of the determinant?

Hint 2
A matrix is invertible if its determinant is non-zero. Can elementary row operations transform a matrix that had a non-zero determinant to a matrix with determinant zero?

Hint 3

Eigenvectors are very specific to a matrix. Multiplying a matrix A with e1 returns the first column of A. That is, if aj is the j-th column of A, $A={\begin{bmatrix}a_{1}&|&a_{2}&|&\dots &|&a_{n}\end{bmatrix}}$ then $Ae_{1}={\begin{bmatrix}a_{1}&|&a_{2}&|&\dots &|&a_{n}\end{bmatrix}}e_{1}=a_{1}$ Since v = e1 is an eigenvector of the matrix A we know that ${\begin{aligned}Av&=\lambda v\\Ae_{1}&=\lambda e_{1}={\begin{bmatrix}\lambda \\0\\\vdots \\0\end{bmatrix}}\end{aligned}}$ Hence a1 = λ e1. In other words, the only non-zero entry of the first column of A is in the first row. Can you perform elementary row operations to move that entry into another row, and hence find a counter example?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. (b) and (d) are the only correct statements independent of $A$ and $B$. (a) Recall that scaling a row is an elementary row operation which scales the determinant by the same factor, so if the determinant of $A$ is non-zero, $\displaystyle \det A=\det B$ can fail to hold. (b) All elementary row operations can be represented by invertible matrices, so since $B$ is a product of these matrices, if $B$ is invertible, then so is $A$. Alternatively, recall that elementary row operations can only scale the determinant of a matrix by a non-zero factor. Thus, if $\displaystyle \det B$ is non-zero (i.e. it is invertible) then $\displaystyle \det A$ is non-zero (so is invertible as well). (c) This is false, elementary row operations do change eigenvectors. For a simple example, let $A={\begin{bmatrix}1&0\\0&1\end{bmatrix}}$ and $B={\begin{bmatrix}1&0\\1&1\end{bmatrix}}$. Subtracting the first row from the second row reduces B to A. However, $\displaystyle Av=v$, so v is an eigenvector of A, while $Bv={\begin{bmatrix}1&0\\1&1\end{bmatrix}}{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}1\\1\end{bmatrix}}$ which is not a multiple of v. Hence v is not an eigenvector of B. (d) If $\displaystyle \lambda =0$ is an eigenvalue of A, then A is not invertible and the determinant of A is 0. As in (b), elementary row operations only scale the determinant by a constant non-zero factor. Therefore, the $\displaystyle \det(B)=c(\det(A))=c(0)=0$ and hence B is not invertible. Hence $\displaystyle \lambda =0$ is also an eigenvalue of B. Alternatively, applying an elementary row operation can be represented by multiplication of the original matrix by an elementary matrix (which is invertible). Suppose $\displaystyle B=EA$. Then if $\displaystyle Av=0$, we have $\displaystyle Bv=E(Av)=E(0)=0$, so 0 is an eigenvalue for $\displaystyle B$ as well (in this case, v is also an eigenvector for this special eigenvalue).

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