Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 02 (b)

MATH152 April 2010

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Question B 02 (b)
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. (b) Find two linearly independent eigenvectors that correspond to eigenvalue 1.

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Hint
Recall that eigenvectors with eigenvalue $\lambda$ satisfy $(A-\lambda {}I)\mathbf {x} =\mathbf {0}$

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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. We want to find two eigenvectors of ${\begin{aligned}A={\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}}\end{aligned}}$ with eigenvalue $\lambda =1$. Just like in B2(a) we want to find the vectors $\mathbf {x} =[a,b,c]^{T}$ such that ${\begin{aligned}(A-1I)\mathbf {x} ={\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}=\mathbf {0} \end{aligned}}$ We therefore have three equations with three unknowns ${\begin{aligned}a+b+c&=0\\a+b+c&=0\\a+b+c&=0\end{aligned}}$ Notice that these are all the same equation, therefore we effectively only have one equation. Notice that if we row reduced the matrix $(A-1I)\mathbf {x}$ we'd get two rows of all zeros, corresponding exactly to the situation of having repeated linear equations. This tells us that two of the variables will be free parameters. Contrast this to B2(a) where we had only one free parameter in the end. The emergence of two free parameters tells us that there will be two eigenvectors. We will take the two free parameters to be a and b, though you can pick any two of a,b, and c to be the free parameters. First Eigenvector: a=1, b=0 To find the first eigenvector we will use set the first free parameter to 1 and the second to 0 (a=1,b=0). Note that you can pick any values you want (except for a=0,b=0 because that will lead to the eigenvector being zero). With a=1 and b=0 we have ${\begin{aligned}a+b+c=1+0+c=0\end{aligned}}$ and therefore, c=-1. Therefore the first eigenvector for the eigenvalue 1 is ${\begin{aligned}\mathbf {x} _{1}={\begin{bmatrix}1\\0\\-1\end{bmatrix}}.\end{aligned}}$ Second Eigenvector: a=0, b=1 To find the second eigenvector we will use set the first free parameter to 0 and the second to 1 (a=0,b=1). Note that you can pick any values you want (except for a=0,b=0 because that will lead to the eigenvector being zero) as long as they are different from the first values you chose, otherwise you will get the same eigenvector. With a=0 and b=1 we have ${\begin{aligned}a+b+c=0+1+c=0\end{aligned}}$ and once again, c=-1. Therefore the second eigenvector for the eigenvalue 1 is ${\begin{aligned}\mathbf {x} _{2}={\begin{bmatrix}0\\1\\-1\end{bmatrix}}.\end{aligned}}$ Any non-trivial linear combination of these two vectors is also an eigenvector, which you can easily verify.

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