MATH152 April 2010
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Question B 02 (a)

The matrix
 $A=\left[{\begin{array}{ccc}2&1&1\\1&2&1\\1&1&2\end{array}}\right]$
is known to have eigenvalues 1 and 4.
(a) Find the eigenvector of A that corresponds to eigenvalue 4.

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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 the eigenvectors with eigenvalue $\lambda$ satisfy
 $(A\lambda {}I)\mathbf {x} =\mathbf {0}$

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Solution

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We want to find the eigenvector of
 ${\begin{aligned}A={\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}}\end{aligned}}$
with eigenvalue $\lambda =4$. We therefore want to find the vector $\mathbf {x} =[a,b,c]^{T}$ such that
 ${\begin{aligned}(A4I)\mathbf {x} ={\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}=\mathbf {0} \end{aligned}}$
We therefore have three equations with three unknowns
 ${\begin{aligned}2a+b+c&=0\\a2b+c&=0\\a+b2c&=0\end{aligned}}$
We can solve the first equation for c to get
 ${\begin{aligned}c=2ab\end{aligned}}$
and can sub this into the second equation to get
 ${\begin{aligned}a2b+2ab&=0\\3a3b&=0\\b&=a.\end{aligned}}$
Using c=2ab and b=a we conclude that c=a. From the last equation we get,
 ${\begin{aligned}a+a2a=0\end{aligned}}$
which is satisfied for all a. Therefore, let a=1 without loss of generality. This implies that b=c=1 as well. Therefore the eigenvector for the eigenvalue 4 is
 ${\begin{aligned}\mathbf {x} ={\begin{bmatrix}1\\1\\1\end{bmatrix}}.\end{aligned}}$
Of course, any multiple of this is also an eigenvector (expect multiplying by 0).
