Section 6.1 Eigenvalues and Eigenvectors: Problem...

Section 6.1 Eigenvalues and Eigenvectors: Problem 16 Results for this submission Entered Answer Preview Result Message -8 -8 correct [1,0,1,0],[0.707106781186547,0,0.707106781186547,0] [1, 0, 1, 0], [1/sqrt(2), 0, 1/sqrt(2), 0] incorrect You have entered vectors which are not orthogonal. 0 0 correct [-1,0,1,0],[0,-1,0,1] [-1, 0, 1, 0], [0, -1, 0, 1] incorrect You have entered vector(s) which are not of unit length. At least one of the answers above is NOT correct. (1 point) The matrix M = [[-4, 0, -4, 0], [0, -4, 0, -4], [-4, 0, -4, 0], [0, -4, 0, -4]] has two distinct eigenvalues lambda1 < lambda2. Find the eigenvalues and an orthonormal basis for each eigenspace. lambda1 = -8 Orthonormal basis of eigenspace: [1,0,1,0],[1/sqrt(2),0,1/sqrt(2),0] lambda2 = 0 Orthonormal basis of eigenspace: [-1,0,1,0],[0,-1,0,1] To enter a basis into WeBWorK, place the entries of each vector inside of brackets, and enter a list of these vectors, separated by commas.

Transcript

00:01 Hello, we are given this matrix m.

00:08 Minus 4, 0, minus 4, 0, 0, minus 4, 0, minus 4, minus 4, 0, minus 4, 0, and 0, minus 4, 0, and minus 4.

00:30 Let us call this column vector c1 and this vector is c2.

00:41 This vector is again, we can see that it is c1 and this vector is again c2.

00:46 So the columns of the matrix are c1, c2, c1, c2.

00:50 Now our job, we are asked to find the eigenvalues for this matrix and an orthonormal basis for each eigenspace corresponding to the eigenvalues.

01:01 Now the general method is to use the characteristic polynomial and find its root, but because of the spatial structure of this matrix, we can directly find the eigenvectors and eigenvalues.

01:13 So let us first observe that m times 1, 1, 1, 1.

01:24 This vector is equal to 1 times the first column plus 1 times the second column plus 1 times the third column plus 1 times the fourth column.

01:35 This is equal to c1 plus c2 plus c1 plus c2, which is equal to minus 8, minus 8, minus 8, minus 8, which is equal to minus 8 times 1, 1, 1, 1.

01:53 We have this.

01:54 Secondly, observe, we can see that m times 1, minus 1, 1, minus 1, this thing, we can, because of the spatial structure, we can see this.

02:10 E is c1 minus the second column plus the third column minus the fourth column, which is c1 minus c2 plus c1 minus c2, which is 2 times c1 minus c2, which is equal to 2 times c1 minus c2 is the vector minus 4, 4, minus 4, 4, which is equal to minus 8 times 1.

02:54 And if we take minus 4 common from this vector, we get 1, minus 1, 1, minus 1.

03:02 So we see that, so we have this, so we see that minus 8 is a, well, we'll come to it.

03:11 Let's first see some more.

03:13 Now, m times 1, 0, minus 1, 0, which is equal to c1 plus 0 times c2 minus c1 plus 0 times c2.

03:35 Note that the columns of the matrix are c1, c2, c1, c2.

03:40 So we can find this thing is c1 plus 0 times c2 minus c1 plus 0 times c2, which is equal to 0, which is equal to 0 times 1, 0, minus 1, 0.

03:53 And we have that m times 0, 1, 0, minus 1 is equal to c1 times 0 plus c2, c1 times 0 minus c2, which is again 0, which is 0 times 0, 1, 0, minus 1.

04:14 So we see that it has, m has two eigenvalues, 0 and minus 8...

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