Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 03 (a)

MATH152 April 2011

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Question B 03 (a)
Let A be a $2\times {2}$ matrix which transforms ${\begin{bmatrix}1\\1\end{bmatrix}}\quad {\text{to}}\quad {\begin{bmatrix}2\\2\end{bmatrix}}$ and transforms ${\begin{bmatrix}1\\-1\end{bmatrix}}\quad {\text{to}}\quad {\begin{bmatrix}0\\0\end{bmatrix}}.$ What are the eigenvalues and eigenvectors of A?

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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
Recall that an eigenvector is a vector such that, left-multiplying it by A produces a multiple of itself, i.e., $A\mathbf {x} =\lambda \mathbf {x} .$ Do you see this property in the provided vectors?

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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. From the hint we recall that an eigenvalue, eigenvector pair satisfies, $A\mathbf {x} =\lambda \mathbf {x} .$ We have in our problem that [1,1] transforms to [2,2], which in matrix form is, $A{\begin{bmatrix}1\\1\end{bmatrix}}={\begin{bmatrix}2\\2\end{bmatrix}}=2{\begin{bmatrix}1\\1\end{bmatrix}}.$ Therefore we see that [1,1] is an eigenvector of A with eigenvalue 2. For the other case we have that [1,-1] gets transformed to [0,0]. Therefore, $A{\begin{bmatrix}1\\-1\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}=0{\begin{bmatrix}1\\-1\end{bmatrix}}$ and so we see that [1,-1] is also an eigenvector with eigenvalue 0. Recall that an $n\times {n}$ matrix has at most n eigenvalues. Therefore since in this case, n=2, we have found all the eigenvalues of the system. We conclude that A has eigenvalues and eigenvectors, ${\begin{aligned}\lambda _{1}&=2,\quad {}\mathbf {x} _{1}={\begin{bmatrix}1\\1\end{bmatrix}}\\\lambda _{2}&=0,\quad {}\mathbf {x} _{2}={\begin{bmatrix}1\\-1\end{bmatrix}}\end{aligned}}.$