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Assignment 4 Solutions
1) Let us calculate the determinant of A − λ Id by expanding along the top row:
2 [ ]
4 3 2
det det 0 0 2 7 0 0 1 2 2 0 0 4 0 0 (10 ) det 0 2 7 ( 9) det 0 2 7 0 0 0 1 2 0 1 2 (10 ) ( 2 ) (2 ) 7( 2 ) 9 4( 2 )(2 ) ( 28) 8 19 24 12.
A Id
⎡ −^ − ⎤
− = ⎢^ ⎥
⎡ −^ − ⎤ ⎡
= − ⎢^ − − − ⎥^ − − ⎢ − − − + −
= − ⎡⎣^ − − − + − − ⎤⎦+ − − − − −
- I will prove this in two ways. First, I will give a proof which should be understandable to all. Second, I will give a more formal proof using induction for those students who have seen induction elsewhere. a) Let A be an n n x matrix: 11 12 1 21 22 2
1 2
n n
n n nn
a a a a a a A a a a
= ⎢^ ⎥
Then, to find the characteristic polynomial, we need to calculate the
determinant of A − λ Id. To do this, let’s see what happens if we expand
the top row:
21 23 2 22 2 31 33 3 11 12 2 1 3 21 2( 1) 1 1 ( 1)
det( ) ( ) det det ... ... ... ... det ...
n n n n nn n n nn n n n n n
a a a a a a a a A Id a a a a a a a a a a a a
−
−
−^ ⎡^ ⎤
− = − ⎢^ ⎥^ − ⎢^ ⎥
+ ± ⎢^ ⎥
continuing on the expansion another step would be messy. However,
looking at the pattern, we see the first matrix has (n-1) terms with λ and
that the others will all have (n-2) such terms. So, if we expand it out, we can see that we’ll end up with
det( A − λ Id ) = ( a 11 − λ)( a 22 − λ)...( ann − λ) +stuff
The first bit is precisely a polynomial of degree n, and the stuff is a whole bunch of polynomials of degree (n-1) and less.
b) Proof by Induction Base Case: I will prove that a 2 × 2 matrix has a characteristic polynomial of degree 2. (Technically, I could start a n=1 like usual but that’s a rather trivial case and I think that starting at n=2 is more convincing. Both prove the same point for matrices which are bigger than 1 × 1 )
Let 11 12. It’s characteristic polynomial is 21 22
a a A a a
=^ ⎡^ ⎤
11 12 21 22 11 22 21 12 (^211 22 11 22 21 )
det( ) det
( )( ) ( ) (
a a A Id a a a a a a a a a a a a
which is clearly a quadratic polynomial.
Inductive Case: Assume that all ( n − 1) × ( n − 1 )matrices have characteristic polynomials of degree ( n − 1)with a leading coefficient of
± 1. Then, as above, we look at the determinant of A^ −^ λ Id.
21 23 2 22 2 31 33 3 11 12 2 1 3 21 2( 1) 1 1 ( 1) 11 1 12 2
det( ) ( ) det det ... ... ... ... det ... ( ) det( ) det ..
n n n n nn n n nn n n n n n
a a a a a a a a A Id a a a a a a a a a a a a a A Id a A
−
−
+ ± ⎢^ ⎥
. + a 1 (^) n det An
Now, by the inductive hypothesis, det ( A 1 − λ Id )is a polynomial of
degree ( n −1) with a leading coefficient of± 1. Furthermore, the matrices
A 12 through A 1 n have only( n − 2 )entries involving a λ. Thus, since the
formula for a determinant never involves multiplying a term by itself, each of these matrices have determinants which give polynomials of degree at
most (. Thus, the final sum is, as desired, a polynomial of degree
with leading coefficient
n − 2 ) n
This time, as expected, there is only one free variable and so one
eigenvector: (^3)
v
= ⎢^ −⎥
JG
At this point, checking our work is straight forward:
1
2
3
v
v
v
⎡^ − −^ −^ ⎤ ⎡ −^ ⎤ ⎡− ⎤
⎡ −^ −^ −^ ⎤ ⎡− ⎤ ⎡− ⎤
⎡^ − −^ −^ ⎤ ⎡ ⎤ ⎡− ⎤
JG
JJG
JG
b) Again, the characteristic polynomial:
3 2
det( ) 2 3 2 4 2 5 2 3 5 2 2 3 4 2 2 4 3 3 2 2 5 2 2 2 6 11 6 1 3 2
A Id
⎡ −^ − ⎤
− = ⎢^ − − ⎥
So this gives the eigenvalues λ = 1, λ= 2, and λ= 3. We move on to calculating eigenvectors:
A Id
⎡^ − ⎤
− = ⎢^ − ⎥
→ ⎢^ ⎥
So the eigenvector associated to 1 is (^1)
v
= ⎢^ ⎥
JG
A Id
⎡^ − ⎤
− = ⎢^ − ⎥
→ ⎢^ ⎥
So the eigenvector associated to 2 is (^2)
v
= ⎢^ ⎥
JJG
A Id
− = ⎢^ − ⎥
→ ⎢^ −⎥
So the eigenvector associated to 3 is (^3)
v
= ⎢^ ⎥
JG
Checking our work, we see
1 1
2 2
3 3
Av v
Av v
Av v
⎡^ − ⎤ ⎡ ⎤ ⎡ ⎤
= ⎢^ − ⎥ ⎢ ⎥^ = ⎢ ⎥=
= ⎢^ − ⎥ ⎢ ⎥^ = ⎢ ⎥=
= ⎢^ − ⎥ ⎢ ⎥^ = ⎢ ⎥=
JG JG
JJG JGJ
JG JG
As expected.
a. First, I claim that if A = PUP −^1 then det A = det U.
3 2 [ ]
3 2
det( ) det 3 4 0 3 1 3 4 3 6 0 3 4 2 0 7 12 6 24 6 7 6 18
A Id
− = ⎢^ − − ⎥
= ⎡⎣^ − + − + ⎤⎦− −
To factor this polynomial, we check all factors of 18. It turns out that is a factor. Next we divide the characteristic polynomial by that factor:
( λ^ −^3
2 3 2 3 2 2 2
So now we know that the characteristic polynomial is
. Using the quadratic equation, we get a further
factoring into ( )
( λ^ −^3 )(^ −λ^2 +^4 λ+^6 )
λ − 3 ⎡ ⎣^ λ− − ( 2 + 10 ) ⎤ ⎡⎦ ⎣λ^ − −( 2 − 10 )⎤⎦. Since there are
three distinct eigenvalues, the associated eigenvectors will be linearly independent (see question 5). Therefore, the matrix is diagonalizable. Getting the eigenvectors for these eigenvalues is messy but the method holds.
A Id
⎡ −^ − ⎤
− = ⎢^ − ⎥
⎡^ − − ⎤
→ ⎢^ ⎥
So 3 (choosing integer coefficients).
v
⎡^ − ⎤
= ⎢^ − ⎥
JG
A Id
→ ⎢^ − − ⎥
⎢ − +^ − ⎥
→ ⎢^ − ⎥
→ ⎢^ ⎥
3 2
det det 3 4 0 3 1 3 1 4 3 0 6 12 3 0 6 4 6 11 6 1 2 3
A Id
− = ⎢^ − − ⎥
Since we have three distinct eigenvalues (1, 2, and 3), the matrix is diagonalizable. To find the eigenvectors:
A Id 0
⎡^ − − ⎤
− = ⎢^ − ⎥
→ ⎢^ −⎥
So 1.
v
JG
A Id
− = ⎢^ − ⎥
→ ⎢^ −⎥
So 2 (using integer coefficients).
v
JJG
A Id 0
⎡ −^ − ⎤
− = ⎢^ − ⎥
→ ⎢^ − ⎥
So 3 (using integer coefficients).
v
JG
Using these eigenvectors, we form the matrices 1 2 5 1 0 0 1 3 2 and 0 2 0 1 3 2 0 0 3
P D
= ⎢^ ⎥^ = ⎢ ⎥
