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Hello everyone.
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In this question, we have a matrix a, which is a square matrix of order 3 by 3, and its entries are along the rows 0 to 0, 0 minus 2, and 4 minus 2, 4.
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This 3 by 3 square matrix a has eigenvalues minus 2, 4.
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We need to find its corresponding eigenvectors.
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Now recall that a square matrix a of order n by n, if it has an eigenvalue lambda with the corresponding eigenvector new, then a satisfies the eigenvalue equation, which is given by a multiplied by new equals to lambda multiplied by new.
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Let us denote it as equation 1.
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Now if we rearrange equation 1 we shall obtain a minus lambda i this whole term multiplied by the eigenvector new equals to 0 where i is the n by n by an identity matrix and 0 is actually the 0 matrix now according to the rules of matrix multiplication this eigen factor new is actually a column vector of dimension n by 1 so let us use this equation denoted as equation 2 to find the eigenvectors of the square matrix which is given and having eigenvalues minus 2, 0 and 4.
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So let us denote the eigenvalues of our matrix a.
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So our matrix a can be written as 0 to 0 0, 0 minus 2.
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4 minus 2 so this is a 3 by 3 square matrix and the corresponding eigen values let us denote it as lambda 1 equals to 0 lambda 2 equals to minus 2 and lambda 3 equals to 4 they are the eigenvalues and in order to use equation 2 we also need the corresponding 3 by 3 identity matrix which we denote it as i so that is nothing but 1 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -1 so this is a 3 -by -3 identity matrix so for a 3 -by -3 identity matrix so for each of this eigenvalue 0 minus 2 and 4 let us find the corresponding eigen vectors using equation 2.
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So for lambda 1 equals to 0 if we use equation 2 we shall have a minus lambda 1 multiplied by the identity matrix this whole term multiplied by the corresponding eigenvector new 1 equals to 0.
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Since lambda 1 equals to 0, we can write a multiplied by new 1 equal to 0.
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So let us write this expression in terms of the matrix a and the eigenvector new 1.
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So we shall obtain 0 to 0, 0, 0, 0, my.
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2, 0, 4 minus 2, 4 multiplied by the eigenvector which is a 3 by 1 column vector and let us denote it as x, y and c.
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So this will be equal to 0, 0.
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So it is another column vector 3 by 1 with entry 0, all 0.
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So, if we actually do the matrix multiplication of the left -hand side by expanding along the rows of the matrix a, we shall obtain the following linear equations.
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The first equation is 0x plus 2y plus 0 times z equals to 0.
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So let us denote it as equation 3.
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The second equation will be 0 times x minus 2y plus 0 times z equals to 0.
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Let us denote it as equation 4 and the third equation will be 4x minus 2y plus 4 z equals to 0.
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So, let us denote it as equation 5.
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Now, if we look at equation 4 or equation 3, we will see that y has to be 0.
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Since in equation 3 and 4, all the coefficients of x or z is 0, so either 2y equals to 0 from equation 3 or minus 2y equals to 0...