Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 27

MATH152 April 2017

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[hide] Question A 27
The matrix $A$ represents a projection in $\mathbb {R} ^{2}$. It is known that $\lambda _{1}=0$ is one eigenvalue with corresponding eigenvector $v_{1}={\begin{bmatrix}-3&1\end{bmatrix}}^{T}$. Find the matrix $A$.

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[show] Hint
Think what are the possible eigenvalues of a projection matrix, and what is the relation between the first and 2nd eigenvector.

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[show] Solution

If we write $A={\begin{bmatrix}a&b\\c&d\end{bmatrix}}$, we are told that it satisfies: ${\begin{bmatrix}a&b\\c&d\end{bmatrix}}{\begin{bmatrix}-3\\1\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$ which gives the equations: $-3a+b=0,\,\,-3c+d=0$ from which we get that the matrix $A$ has the form $A={\begin{bmatrix}a&3a\\c&3c\end{bmatrix}}$. Using that the eigenvalues of a projection are either zero or one, we can pick the 2nd eigenvector to be one which is orthogonal to the first. Choose $v_{2}={\begin{bmatrix}1\\3\end{bmatrix}}$ with eigenvalue one, so that it satisfies the eigenvalue equation: ${\begin{bmatrix}a&3a\\c&3c\end{bmatrix}}{\begin{bmatrix}1\\3\end{bmatrix}}={\begin{bmatrix}1\\3\end{bmatrix}}$. This gives the equations: $10a=1,\,\,10c=3$. Answer: $\color {blue}A={\begin{bmatrix}{\frac {1}{10}}&{\frac {3}{10}}\\{\frac {3}{10}}&{\frac {9}{10}}\end{bmatrix}}$