Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (c)/Solution 1"

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In fact, Eigenvectors corresponding to distinct eigenvalues are linearly independent. Let’s prove it.  
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In fact, eigenvectors corresponding to distinct eigenvalues are linearly independent. Let’s prove it.  
  
First, we prove that any two of these three eigenvectors are linearly independent, without lost of generality, let’s choose <math display="inline">v_1,v_2</math>, we need to show that <math display="inline">av_1+bv_2=0</math> holds only when <math display="inline">a=b=0</math> (definition of linearly independence). Indeed, (1) multiply both sides by <math display="inline">\lambda_1</math> we have <math display="inline">a\lambda_1v_1+b\lambda_1v_2=0</math>; (2) let <math display="inline">A</math> act on both sides we get <math display="inline">aAv_1+bAv_2=a\lambda_1v_1+b\lambda_2v_2=0</math>; (3) Subtracting above equations gives <math display="inline">b(\lambda_1-\lambda_2)v_2=0</math>, now since <math display="inline">\lambda_1\neq\lambda_2</math>, the solutions is <math display="inline">b=0</math>, thus we also get <math display="inline">a=0</math>.
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Suppose that <math>\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n</math> are eigenvectors corresponding to distinct eigenvalues <math>\lambda_1, \lambda_2, \dots, \lambda_n</math>, respectively. Let <math>m</math> be the largest integer for which <math>\mathbf{v}_1, \dots, \mathbf{v}_m</math> are linearly independent. If we had <math>m < n</math>, we could write <math>\mathbf{v}_{m+1} = \sum_{i=1}^{m} c_i \mathbf{v}_i</math> for some <math>c_i</math> not all zero. Now <math>A \mathbf{v}_{m+1} = \lambda_{m+1} \mathbf{v}_{m+1} \implies A\sum_{i=1}^{m} c_i \mathbf{v}_i = \sum_{i=1}^{m} c_i \mathbf \lambda_i \mathbf{v}_i = \lambda_{m+1} \sum_{i=1}^{m} c_i \mathbf{v}_i</math>, whence <math>\sum_{i=1}^{m} c_i \mathbf (\lambda_i - \lambda_{m+1}) \mathbf{v}_i = \mathbf{0}</math>. As the <math>\lambda_i</math> are distinct, this implies (by linear independence) that <math>c_1 = \cdots = c_n = 0</math>, so <math>\mathbf{v}_{m+1} = \mathbf{0}</math>, contradicting the assumption that <math>\mathbf{v}_{m+1}</math> was an eigenvector. We conclude that <math>m = n</math>, so all the eigenvectors are linearly independent.
  
Now let’s prove <math display="inline">v_1,v_2,v_3</math> are linearly independent (in fact it is just induction), i.e., to show <math display="inline">av_1+bv_2+cv_3=0</math> holds only when <math display="inline">a=b=c=0</math>. In the same manner, (1) multiply both sides by <math display="inline">\lambda_1</math> we have <math display="inline">a\lambda_1v_1+b\lambda_1v_2+c\lambda_1v_3=0</math>; (2) let <math display="inline">A</math> act on both sides we get <math display="inline">aAv_1+bAv_2+cAv_3=a\lambda_1v_1+b\lambda_2v_2+c\lambda_3v_3=0</math>; (3) Subtracting above equations gives <math display="inline">b(\lambda_1-\lambda_2)v_2+c(\lambda_1-\lambda_3)v_3=0</math>, now since <math display="inline">\lambda_3\neq\lambda_1\neq\lambda_2</math> and <math display="inline">v_2,v_3</math> are linearly independent according to above argument, the solutions is <math display="inline">b=c=0</math>, thus we also get <math display="inline">a=0</math>.
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Each of the 3 eigenvalues found in part (b) has a corresponding eigenvector, and by the above, they are linearly independent. (Clearly, there cannot be more linearly independent eigenvectors, since the dimension of the matrix is 3.) Thus, the answer is <math display="inline">\color{blue} 3.</math>
 
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All eigenvectors are linearly independent. The solution is <math display="inline">\color{blue} 3</math>.
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Latest revision as of 21:47, 8 March 2018

In fact, eigenvectors corresponding to distinct eigenvalues are linearly independent. Let’s prove it.

Suppose that are eigenvectors corresponding to distinct eigenvalues , respectively. Let be the largest integer for which are linearly independent. If we had , we could write for some not all zero. Now , whence . As the are distinct, this implies (by linear independence) that , so , contradicting the assumption that was an eigenvector. We conclude that , so all the eigenvectors are linearly independent.

Each of the 3 eigenvalues found in part (b) has a corresponding eigenvector, and by the above, they are linearly independent. (Clearly, there cannot be more linearly independent eigenvectors, since the dimension of the matrix is 3.) Thus, the answer is