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Quiz 9 - November 12, 2015.
- Let S be the back shift operator in R∞. That is, S(a 1 , a 2 , a 3 ,... ) = (a 2 , a 3 , a 4 ,... ). Let T = S^2 + S − 12 I. Hint: null T is spanned by eigenvectors of S.
(a) Compute dim null T.
(b) Compute a basis for null T.
Solution: T = S^2 + S − 12 I = p(S), where p(x) = x^2 + x − 12. Since p(x) = (x + 4)(x − 3), T = (S + 4I)(S − 3 I). Recall from the Midterm test that every scalar λ is an eigenvalue of S, that null(S − λI) = span((1, λ, λ^2 , λ^3 ,... )) and that dim null(S−λI) = 1. Define vλ = (1, λ, λ^2 , λ^3 ,... ). Then it is clear by intuition that dim null T = 2 and that null T is spanned by (v− 4 , v 3 ). Why is the above statement “clear by intuition”? First of all, I told you that null T is spanned by eigenvectors of S. If vλ ∈ null T , then
T vλ = (S + 4I)(S − 3 I)vλ = (S − 3 I)(S + 4I)vλ = 0.
Clearly λ− 4 and λ 3 ∈ null T , so dim null T is at least 2. However, suppose that λ 6 = 3. (S − 3 I)vλ = λSvλ − 3 vλ = (λ − 3)vλ 6 = 0 since λ 6 = 3. Then (λ−3)vλ null(S +4I). That is, (λ−3)vλ is an eigenvector of S with eigenvalue −4, so λ = −4. Similarly, if λ 6 = −4, then it must equal 3. You were meant to realize that the operator T is a(n easily factored) polynomial of the shift operator, so that you would quickly write T = (S + 4 I)(S − 3 I). Then you look at the expression and think: “Eigenvectors of S with eigenvalues 3 and −4 are in the null space of T , so the dimension of the null space is 2 and it is spanned by (1, 3 , 9 , 27 ,... ) and (1, 4 , 16 , 64 ,... ). Now I’ll start over again. T = S^2 + S − 12 I = (S + 4I)(S − 3 I). Suppose that v ∈ null T and that v 6 = 0. Then (S + 4I)(S − 3 I)v = 0. If w = (S − 3 I)v 6 = 0, then w must be an eigenvector of S with eigenvalue −4, so (S − 3 I)v = av− 4 for some scalar a. If u = (S + 4I)v 6 = 0, then u must be an eigenvector of S with eigenvalue 3, so u = bv 3 for some scalar b. Then
w = (S − 3 I)v = Sv − 3 v = av− 4 and u = (S + 4I)v = Sv + 4v = bv 3 , so u − w = Sv + 4v − Sv + 3v = 7v and finally
v =
(bv 3 − av− 4 ).
That is, v is a linear combination of the two eigenvectors v 3 and v− 4 of S. Thus null T = span(v 3 , v− 4 ) and dim null T = 2.
- Define T ∈ L(R^3 , R^4 ) by T (x, y, z) = (x+y+z, x−y+2z, x+y+3z, x−y+4z). Find w~ = (x 1 , x 2 , x 3 , x 4 ) ∈ R^4 such that ‖ w~ − (1, 0 , 2 , 0)‖ is minimized and compute ‖ w~ − (1, 0 , 2 , 0)‖. I suggest the following steps.
(a) Compute a basis for range T.
(b) Compute an orthonormal basis for range T.
(c) Compute w~ = Prange T (1, 0 , 2 , 0).
(d) Compute ‖ w~ − (1, 0 , 2 , 0)‖.
Solution: I’ll solve this problem using the suggested steps.
(a) range T is spanned by T e 1 , T e 2 and T e 3. That is, range T = span(v 1 , v 2 , v 3 ) where v 1 = (1, 1 , 1 , 1), v 2 = (1, − 1 , 1 , −1) and v 3 = (1, 2 , 3 , 4). Then (v 1 , v 2 , v 3 ) is a basis of range T. It is not immediately obvious that v 3 ∈/ span(v 1 , v 2 ) but I won’t waste time proving it since the Gram- Schmidt method will eliminate v 3 automatically if it is the span of the previous two.
(b) Apply Gram-Schmidt to (v 1 , v 2 , v 3 ) to compute an orthonormal basis of range T. I will use the notation ei for these vectors. Don’t confuse them with my use of ei ∈ R^3 above.
- Clearly e 1 = 12 (1, 1 , 1 , 1).
x + y + z = 5/ 4 x − y + 2z = − 1 / 4 x + y + 3z = 7/ 4 and x − y + 4z = 1/ 4.
Adding lines 1 & 2 and 3 & 4 gives
2 x + 3z = 1 2 x + 7z = 2
Subtracting line 1 from line 2 above gives z = 1/4. Then 2x = 1/4, so x = 1/8. Finally, y = 5/ 4 − 1 / 4 − 1 /8 = 7/8. Thus T (1/ 8 , 7 / 8 , 1 /4) = (5/ 4 , − 1 / 4 , 7 / 4 , 1 /4).
