Suppose A has eigenvalues 0,3,5 withbindependent eigenvectorsu, v, w. a)Give a b

Question

Suppose A has eigenvalues 0,3,5 withbindependent eigenvectorsu, v, w. a)Give a basis for the nullspace and a basis for the columnspace. b)Find a particular solution to Ax=v+w. Find allsolution c)Show that Ax=u has no solution. ( if it had a solution, then_________ would be in the column space.) Please show all your work Suppose A has eigenvalues 0,3,5 withbindependent eigenvectorsu, v, w. a)Give a basis for the nullspace and a basis for the columnspace. b)Find a particular solution to Ax=v+w. Find allsolution c)Show that Ax=u has no solution. ( if it had a solution, then_________ would be in the column space.) Please show all your work

Explanation / Answer

(a) we know that the eigen vectors corresponding to thedistinct eigen values are linearly independent. so, u,v, w are linearly independent. also, A has 3 eigen values says that A is of order 3x3. but one of the eigen values of A is 0 and so, A issingular. in other words, the determinant of A is zero and one row islinearly dependent. the basis of null space of A has one non zerovector. the basis of the column space has two vectors. (b) since u,v,w are linearly independent, Ax = v + w leads toAx is not the vector u. from this, Ax has two non zero equations which give onelinearly independent solution. please post the third question in the next post and send thelink to me. sorry and thank you.

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