Without calculation, find one eigenvalue and two linearly...

Without calculation, find one eigenvalue and two linearly independent eigenvectors of A = egin{bmatrix} 2 & 2 & 2 \ 2 & 2 & 2 \ 2 & 2 & 2 end{bmatrix}. Justify your answer. One eigenvalue of A is ? = [ ] because [ ? ] Two linearly independent eigenvectors of A are [ ] because [ ? ] (Use a comma to separate answers as needed.) each vector is in the column space of A. the entries of each vector are equal. the entries of each vector are positive. the entries of each vector sum to 0.

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00:01 As in this question, given matrix is here, given matrix a is equals to here 3 cross 3 matrix which is 2 2 2, all 3 rows are 2 2 and 2 here.

00:13 So first of all we have to find here eigenvalue.

00:17 So for the first part as the second column, second column, second column is multiple, multiple of the first column of the first column, multiple of the first column or same as the first column.

00:43 So determinant of a equals to 0 implies a is not invertible, a is not invertible and if a is not invertible this one is implying here 0 is an eigenvalue, 0 is an eigenvalue of a.

01:07 Hence this one is answer for the first part.

01:11 Now come to the second part here.

01:13 The eigenvector corresponding to eigenvalue 0 is x here.

01:19 So augmented matrix let x be eigenvector corresponding to eigenvalue, eigenvalue 0.

01:46 So ax, so ax equals to 0.

01:52 So augmented matrix can be written as here 2 2 2 0, 2 2 2 0, 2 2 and 2 0 here.

02:03 So as here row first, row first can be written as here, when we divide row first with 2.

02:13 We can write it as here 1 1 1 0 rest two rows will be same here 2 2 0 0 here.

02:24 Now replace second row with second row minus 2 into first row and third row with third row minus 2 into first row.

02:35 We can write it as here 1 0 0 1 0 0 1 0 0 0 0 0 here...

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