Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 24

MATH152 April 2016

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Question A 24
A $3\times 3$ matrix A with real entries has been typed into MATLAB. The result of the command = eig(A) is (after some slight formatting changes to make it fit better in the exam): V = 0.8165 + 0.0000i 0.8165 + 0.0000i 0.5774 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.5774 + 0.0000i 0.4082 - 0.4082i 0.4082 + 0.4082i 0.5774 + 0.0000i D = 1.0000 + 2.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 2.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i -1.0000 + 0.0000i Circle all true statements below: (a) A has no real eigenvalues. (b) All eigenvalues of A have negative real parts. (c) The eigenvectors of A are a basis for $\mathbb {R} ^{3}$ . (d) Eigenvectors of A associated to distinct complex eigenvalues are linearly independent. (e) $[1,1,1]^{T}$ is an eigenvector of A.

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Hint
Definition of eigenvectors and eigenvalues. In Matlab, command $[V,D]=eig(A)$ returns diagonal matrix $D$ of eigenvalues and matrix $V$ whose columns are the corresponding right eigenvectors, so that $AV=VD$ Note also that there are no repeated diagonal entries in $D$ . This can be used for the question.

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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. Following the hint, the diagonal matrix $D$ is the eigenvalues and the column of $V$ are the are the corresponding right eigenvectors. Therefore, A has 3 eigenvalues: $1+2i$ , $1-2i$ , $-1$ , and their corresponding eigenvectors are $[0.8165,0,0.4082-0.4082i]^{T}$ , $[0.8165,0,0.4082+0.4082i]^{T}$ , $[0.5774,0.5774,0.5774]^{T}$ . The eigenvectors in $V$ are normalized so that the 2-norm of each one is $1$ . So $(a)$ is wrong. We have $-1$ $(b)$ is wrong. We have $1+2i$ $(c),(d)$ is correct. A has three different eigenvalues and hence the three eigenvectors are independent, and they can form a set of basis. $(e)$ is correct. $[1,1,1]^{T}$ is a scalar multiples of the last column vectors of $A$ . The scalar multiplication of any eigenvector is a eigenvector as well, so $(e)$ is correct Thus, the answer is $\color {blue}(c),(d),(e)$ .