3. Let S, T ? ?(V). (a) Prove that T is invertible if and...

3. Let S, T ? ?(V). (a) Prove that T is invertible if and only if 0 is not an eigenvalue for T. (b) Prove that if ST is invertible, then so are S and T. (c) Prove that ST and TS have the same eigenvalues. (d) Illustrate with an example that ST and TS need not have the same eigenvectors.

Transcript

00:01 In this problem, let st belongs to lv.

00:07 And in first part, that is let tv linear operator, linear operator on finite dimensional space, dimensional space v and t is the invertible if there exists t inverse such that t into t inverse is equal to t inverse t is equal to lv let t be invertible and let zero be an eigenvalue and v lambda not equal to 0 then t v lambda is 0 into v lambda is equal to 0 and t t inverse is equal to lv is equal to this implies t t lambda v is equal to lv which is equal to to v for all v belongs to capital v and this implies the t inverse t v lambda is equal to v lambda now we have now we have t inverse t v lambda is equal to t inverse t of v lambda that is t inverse zero is equal to zero and zero is not an eigen value now let zero v n eigen value for t and let v n eigen value for t and let v n and not equal to then t v node is equal to 0 into v node is equal to 0 and t can't be invertible because if it were there would be would have to exist a linear operator a linear operator a linear operator t inverse such that for every v belongs to v, t inverse t, v is equal to capital v.

03:34 But since t inverse is a linear operator and they always map 0 to 0.

03:54 Then t inverse t v node is equal to zero not equal to v and t is not an invertible operator in next part that is b is if st is invertible invertible then both s and t are invertible.

04:40 Since st is invertible and there exists r belongs to lv that satisfy rst is equal to i and str is equal to i where i is the identity map.

05:10 I is the identity map on v suppose we have small v belongs to null t then meaning that it is belongs to capital v which satisfies t v is equal to zero then we have v is equal to lv and r s t v then it is r s t v that is r s 0 that is r as 0 is equal to 0 therefore we have v belongs to 0 and so null t belongs to 0 so we conclude that set equal to 0 so we conclude that set equal null t is equal to zero thus if s t is invertible so are s and t now in c part suppose each matrix is n y n and we consider our matrices to all be acting on some n dimensional.

07:24 Then st has range equal to full space and since st is equal to i, thus the range of t must also have dimension...

Question

3. Let S, T ? ?(V). (a) Prove that T is invertible if and only if 0 is not an eigenvalue for T. (b) Prove that if ST is invertible, then so are S and T. (c) Prove that ST and TS have the same eigenvalues. (d) Illustrate with an example that ST and TS need not have the same eigenvectors.

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