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Math 601 Solutions to Practice Problems
- Compute the following determinants:
(a) (b)
ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ
Answer:
(a) Expanding along the indicated rows and columns:
ââ ââ â â â â ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ â â â â
a b (^) º º a bº º
a ba b a ba b
œ # $
œ # ' "#^ !%^ $ & "#^ '#
œ # ' % $ & "! œ "!#
(b) Using row reduction:
ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ ââ â â â â â â â â â â â â â â â â
1 œ^ œ^ œ !!! %
œ )
- For what values of +is the following matrix invertible? â â ââ ââ ââ ââ ââ ââ â â
Answer:
The matrix is invertible as long as the determinant is nonzero. Using cofactor expansion:
â â ââ ââ ââ ââ ââ ââ ââ ââ â â â â ââ ââ â â
ˆ (^) a b ‰ ˆ ‰ a ba b
œ + œ + + "
œ + + + ' ) œ + + # '+ ) œ + + # + %
Therefore, the matrix is invertible as long as + Á !ß #ß %.
- Let PÀ ‘ #^ Ä‘#be a linear transformation such that
P (^) ” #"^ • œ (^) ” '$^ • and P (^) ” ""^ • œ” %$•
(a) Find the # ‚ # matrix for P with respect to the basis ß.
œ ” (^) " • ” (^) "•
(b) Find P.
Answers:
(a) We have:
P #"^ œ $ #"^ ! ""
P œ " #
so the matrix is (^) ” •.
(b) We need to express (^) ” %^ • as a linear combination of (^) ” #^ • and (^) ” "•: " " "
" " " μ^ # " % μ^! " # μ! " #
We conclude that (^) ” %^ • ” #^ • ” "•: " " "
œ $ #
(c) (^) º º a ba b, so the eigenvalues are and.
& $ œ^ " œ^ 3^ 3^3
- -^ -^ -
We find the corresponding eigenvectors:
- œ 3 : The eigenspace is the nullspace of (^) ” $ 3& $ 3#^ • , so one eigenvector is ” (^) $ 3# •.
Reasoning: This matrix has zero determinant, so we know that the second row is a multiple of the first. Since a #ß $ 3bmultiplies with the first row to give zero, it must be an eigenvector. We can check to make sure:
” •” • ” • ” • (yes, it works)
& $ $ 3 œ^ " $3 œ 3 $ 3
- œ 3 This should be the complex conjugate of the eigenvector for : (^3) ” (^) $ 3# •
Note: Other answers are correct, though they may look quite different. For example, (^) ” $ 3& •
is also an eigenvector for. 3
- Consider the vector space T (^) $consisting of all polynomials with degree less than $. Let Pbe
the following linear transformation from T (^) $ to T$:
P : Bˆ^ a b ‰ œ a" B : b wa bB
Find the eigenvalues for P. For each eigenvalue, find the corresponding eigenvector.
Answer:
We use the basis e "ß Bß B #f. To start, we find the matrix for P:
P " œ! P B œ " B P B œ #B #B
a b a b ˆ #‰^ #
so Pis represented by the matrix. This matrix is upper triangular, so the
Ô ×
Õ Ø
eigenvalues are the numbers along the diagonal, namely! ", , and #.
- œ! Clearly œ the constant function " is an eigenvector for !.
Ô ×
Õ Ø
!! #!! " B
B œ B
B œ!
Ô × Ô ×
Õ Ø Õ Ø
" #
$
So is an an eigenvector for.
Ô ×
Õ Ø
œ " B "
is free
! " #! " #! " # B œ #B !!!!!!!!!
B œ B
B
Ô × Ô × Ô ×
Õ Ø Õ Ø Õ Ø
" $
$
$
So is an eigenvector for.
Ô ×
Õ Ø
œ " #B B # #
- Let E be the matrix. Find a matrix F such that F œ E.
$
( Hint: Use diagonalization.)
Answer:
We find the eigenvalues of E:
º º a^ ba^ b^ , so the eigenvalues are^ and^.
& œ^ *^ ) œ^ "^ )^ "^ )
- -^ -^ -^ -
so is an eigenvector for is free
- œ " $#^ '%^ μ "!^ #!^ Ê #" "
B œ #B ” • ” • (^) B ” •
" #
so is an eigenvector for. is free
B œ B B
$
"^
$
Therefore, if we use the basis (^) ” • ” • , the matrix becomes (^) ” •. In this basis, it's
" ß^ # E! )
easy to find F:
F œ œ
È
È ”^ •
$ $
Finally, we must switch Fback to the standard basis:
œ
to standard coords.^ change back^ eigenbasismatrix in^ change to new coords. (inverse of (^) ” #"^ $#•)
# % œ # ""
( ”^ •
- Find an orthonormal basis for the following subspace of ‘%:
W œ ß ß
Span
ÚÝ Þá Ý á Û ß ÝÝ áá Ü à
Ô × Ô × Ô ×
ÖÖ Ù ÖÙ Ö Ù ÖÙ Ö ÙÙ
Õ Ø Õ Ø Õ Ø
Answer:
We use the Gram-Schmidt process:
Make the 1st vector a unit vector: l a "ß "ß "ß "b lœ # , so "# a "ß "ß "ß "b is a unit vector.
Make the 2nd vector orthogonal to the 1st: a ""ß "ß "b † "#a "ß "ß "ß "b œ! , so the second vector is already orthogonal to the first. Make the 3rd vector orthogonal to the 1st: a #ß #ß #ß 'b † "#^ a "ß "ß "ß "b œ "#a # # # 'b œ % , and a #ß #ß #ß 'b a b a% "# "ß "ß "ß "b œ a %ß !ß !ß 4 b Ãnew 3rd vector Make the 2nd vector a unit vector: l a "ß "ß "ß "b l œ # , so "#a "ß "ß "ß "b is a unit vector
Make the 3rd vector orthogonal to the 2nd: a %ß !ß !ß %b † "#^ a "ß "ß "ß "b œ "#a % %b œ % , and a %ß !ß !ß %b a b a% "# "ß "ß "ß "b œ a #ß #ß #ß #b à new 3rd vector Make the 3rd vector a unit vector: l a #ß #ß #ß #b l œ % , so "% a #ß #ß #ß #b œ "# a "ß "ß "ß "b is a unit vector.
- Find the Fourier series for the following function:
0 B œ
! B
" B
! B
a b
ÚÝ
Û
Ý
Ü
if if if
1 1 1 1
Answer:
The functions (^) È^ "# 1 ß (^) È" 1 sin 8B, (^) È" 1 cos8B 8 œ "ß #ß $ß á( ) are orthonormal.
¢ 0 B ßa b £ œ (^) ( 0 Ba b (^) È .B œ (^) È ( .B œ (^) È œ Ê
"
(^) Î#
Î# È (^1) 1 1
1 1
1 1 1
¢ 0 B ßa b^ 8B£^ œ^ ( 0 Ba b^ 8B .B œ^ È ( 8B .B œ!
Î#
Î# È 1 È 1 1 1
1 1 sin sin sin 1 (since sin 8Bis an odd function)
¢ 0 B ßa b 8B£ œ (^) ( 0 Ba b 8B .B œ (^) È ( 8B .B
Î#
Î# È 1 È 1 1 1
1 1 cos cos cos 1
œ œ.
" 8B # 8
È ”^8 •^8 È
sin 1 sin Î#
Î#
1
1
Let's try a few values of 8 to find the pattern for :
sinŠ ‹
sinˆ^8 #^1 ‰ "! "! "! "! â
We conclude that:
0 B œ B $B &B â
œ
a b (^) Œ Ê Œ (^) È Œ (^) È Œ (^) È
" " " " È# 1 È 1 cos^ È 1 cos^ È 1 cos
B $B &B (B â 1 1 1 1
cos cos cos cos
- Compute the following integrals:
(a) (^) ( a ba b 1
1 $ sin B # sin #B sin B # sin #B $ sin$B .B
(b) (^) ( a b
1
1 " $ sin B # sin#B .B
Answers:
The functions (^) È^ "# 1 ß sinÈ^ 8B 1 , cosÈ8B 1 ( 8 œ "ß #ß $ß á) are orthonormal.
(a) ¢ $ sin B # sin #Bß sin B # sin #B $ sin$B£
œ ¢$ È^1 sin È^ 8B 1 # È^1 sinÈ^ #B 1 ß È^1 sinÈ^1 B # È^1 sinÈ^ #B 1 $È 1 sinÈ$B 1 £ œ ˆ$^ È^1 ‰ˆ È^1 ‰ ˆ#^ È^1 ‰ˆ #^ È^1 ‰^ a b! ˆ$^ È 1 ‰ œ 1
