Solutions to Math 20F final exam, March 20, 2009

1) L equals

 1   0   0   0
 0   1   0   0
-3  -5   1   0
 2  -1   0   1

Move P_24 two places to the right, so it can be next to the  A   (cf. PA=LU), 
E_31 doesn't change, but E_21 changes to E_41 as P_24 moves past it.
L is thus the inverse of the product E_42E_32E_31E_41, which is easily
computed without the necessity of any matrix multiplication.

2A)  charpoly(A) is  det(A - xI), 
and the determinant can be computed in many ways.

2B)  Let  P be the matrix

-1   0   1
 0   1   0
 1   0   1

The last two columns are a basis for the 1-eigenspace of  A.
A basis for the NS(A-I) is computed in the usual way, 
noting that the two free variables for
the matrix A-I are the coefficients for the two basis vectors 
that span the null space.

2C)  The first column of the matrix P in part (2B) is a 
basis for the (-1)-eigenspace of  A.
Thus matrices  P and D that work for part (2C) are 
the matrix P in part (2B) and the diagonal matrix
D defined by

-1   0   0
 0   1   0
 0   0   1

That's because the diagonal elements of  D  are the eigenvalues 
of the corresponding eigenvectors
comprising the columns of  P.  
We knew in advance that the matrix A was diagonalizable because
A is symmetric.


3A)  Let x1, x2, x3  be the first three columns of  A.  
Then an orthogonal basis for CS(A) 
consists of the column vectors
v1 = x1 = (1,1,2,0),  v2 = (-1,-1,1,1), and  v3 = (1/3, 1/3, -1/3, 1).
These can be found via Gram-Schmidt.

3B) (6,6,0,0) =(Proj of u onto v1) + (Proj of u onto v2) + (Proj of u onto v3)
since v1, v2, v3 is an orthogonal basis.
The projection of  u  onto  CS(A) yields the closest point in CS(A)  to  u.
For example, the closest point on the real axis to the 
point (17,36) in R^2 is the projection (17,0).

4)  Any three linearly independent vectors in  W  will work.  
Note that W is the set of vectors
(a,b,c, -a-b-c) = a(1,0,0,-1) + b(0,1,0,-1) + c(0,0,1,-1), 
and the three vectors on the right
form a basis for W, since they span W and are obviously independent.
Another method is to note that W is the null space of the 4 by 4 
matrix whose 16 entries are all equal to 1.

5)  Let v be the 9 by 1 vector with 1's in the 
third, fifth, seventh, and eighth entries, and zeros
in the other five entries.  Assuming Jan's statement 
is true for U, we have Uv = 0.
Since PA=LU, we have  PAv = LUv = 0 .
Multiply both sides by the inverse of P to see 
that Av = 0, thus proving Jan's statement for A.