Let A be an m x n matrix. The goal of this exercise is to show that the matrix equation AT A = ATT has a solution for all € Rm. This solution is often called the least squares solution to the system A = 6. (a) Show that im(ATA) ≤ im(AT), and conclude from this that dim(im(ATA)) ≤ dim(im(AT)). (b) Show that null(ATA) = null(A). (Hint: Show this by proving containment both ways. The "C" part is the tricky one. Start by assuming E null(AT A), so that AT A = 0. Then multiply this equation on both sides by T, the row vector version of , and use some properties of the transpose to conclude that Ax = 0, so x € null(A).)
Let A be an m x n matrix. The goal of this exercise is to show that the matrix equation AT A = ATT has a solution for all € Rm. This solution is often called the least squares solution to the system A = 6. (a) Show that im(ATA) ≤ im(AT), and conclude from this that dim(im(ATA)) ≤ dim(im(AT)). (b) Show that null(ATA) = null(A). (Hint: Show this by proving containment both ways. The "C" part is the tricky one. Start by assuming E null(AT A), so that AT A = 0. Then multiply this equation on both sides by T, the row vector version of , and use some properties of the transpose to conclude that Ax = 0, so x € null(A).)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Let A be an m × n matrix. The goal of this exercise is to show that the matrix equation A¹ Aỡ = ATT has a
solution for all 7 € Rm. This solution is often called the least squares solution to the system Aỡ = 6.
(a) Show that im(AT A) C im(AT), and conclude from this that dim(im(ATA)) ≤ dim(im(AT)).
(b) Show that null(ATA) = null(A).
(Hint: Show this by proving containment both ways. The "C" part is the tricky one. Start by assuming
☛ E null(ATA), so that Aª Aỡ = Ổ. Then multiply this equation on both sides by xª, the row vector
version of , and use some properties of the transpose to conclude that Aï = Ổ, so x € null(A).)
(c) Using the Rank-Nullity Theorem (that's the result of Q3(e) on this problem set, which you may use here
without proof), conclude from the previous part that in fact dim(im(ATA)) = dim(im(AT)).
(d) Use parts (a) and (c) to conclude that the equation A¹ Añ = ATT has a solution for all be Rm.
(e) This idea is useful in many applications. First, convince yourself that if Aỡ = 6, then is also a solution
to A¹ Az = A¹b. Nothing new there. The interesting thing is when Ax = 6 is inconsistent. In that case,
the solution to AT Aỡ = ATT is the ☞ such that Aỡ is closest to b.
By the result of the important problem from WA1, you know that Ax = 6 is inconsistent if and only if
bim(A) = col(A). So, just as one example, if im(A) is a plane through the origin in R³ and 6 is not
on that plane, then the solution to A¹ Ax = ATT lets you find Ar, the closest point on the plane to 6,
a problem you already know how to solve.
So let's see that it works. Let II be the plane through the origin spanned by the two vectors [1 1 0]T
and [0 1 2], and let B = (5, −1, 6). Using the methods you learned earlier in the course, find closest
point Q on II to B.
Next, find a 3 × 2 matrix A such that II is the image/column space of A, and check that if ☞ is solution
to AT A = ATT, where 6= [5 -1 6], then Ar is has the same coordinates as Q.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc84b9504-04d8-4492-b13a-660d2a574b53%2F9e609b78-fe7c-4f18-b79f-fc74ea0ddf80%2F9e1fji8_processed.png&w=3840&q=75)
Transcribed Image Text:Let A be an m × n matrix. The goal of this exercise is to show that the matrix equation A¹ Aỡ = ATT has a
solution for all 7 € Rm. This solution is often called the least squares solution to the system Aỡ = 6.
(a) Show that im(AT A) C im(AT), and conclude from this that dim(im(ATA)) ≤ dim(im(AT)).
(b) Show that null(ATA) = null(A).
(Hint: Show this by proving containment both ways. The "C" part is the tricky one. Start by assuming
☛ E null(ATA), so that Aª Aỡ = Ổ. Then multiply this equation on both sides by xª, the row vector
version of , and use some properties of the transpose to conclude that Aï = Ổ, so x € null(A).)
(c) Using the Rank-Nullity Theorem (that's the result of Q3(e) on this problem set, which you may use here
without proof), conclude from the previous part that in fact dim(im(ATA)) = dim(im(AT)).
(d) Use parts (a) and (c) to conclude that the equation A¹ Añ = ATT has a solution for all be Rm.
(e) This idea is useful in many applications. First, convince yourself that if Aỡ = 6, then is also a solution
to A¹ Az = A¹b. Nothing new there. The interesting thing is when Ax = 6 is inconsistent. In that case,
the solution to AT Aỡ = ATT is the ☞ such that Aỡ is closest to b.
By the result of the important problem from WA1, you know that Ax = 6 is inconsistent if and only if
bim(A) = col(A). So, just as one example, if im(A) is a plane through the origin in R³ and 6 is not
on that plane, then the solution to A¹ Ax = ATT lets you find Ar, the closest point on the plane to 6,
a problem you already know how to solve.
So let's see that it works. Let II be the plane through the origin spanned by the two vectors [1 1 0]T
and [0 1 2], and let B = (5, −1, 6). Using the methods you learned earlier in the course, find closest
point Q on II to B.
Next, find a 3 × 2 matrix A such that II is the image/column space of A, and check that if ☞ is solution
to AT A = ATT, where 6= [5 -1 6], then Ar is has the same coordinates as Q.
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