Step 5 For 1 = -1, we found the eigenspace consists of solutions to the following. [20210 0 0 0 0 0 0 0 0 Jxz. Let x2 =

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Step 5 For 1 1 We Found The Eigenspace Consists Of Solutions To The Following 20210 0 0 0 0 0 0 0 0 Jxz Let X2 1 (24.51 KiB) Viewed 36 times

Step 5 For 1 = -1, we found the eigenspace consists of solutions to the following. [20210 0 0 0 0 0 0 0 0 Jxz. Let x2 = s be the free variable in the equation. Also, note that x, = t is a free variable as it does not appear in the equations. Therefore, the 0 -1 Writing this as a system of equations gives 2x2 + 2x2 = 0, or X1 )s set of solutions is given by the vector Thus, the eigenvectors are of the form s 5 Find the spanning set for the eigenspace for 1 = -1. [2] -1 0 1 1 0 1 span It X Submit Skip (you cannot come back)