Linear Algebra
Test #2
May  5, 2003
Name____________________
R.  Hammack
Score ______


(1) This problem concerns the matrix A = (1   2   2   1)        1   1   1   1       1   0   1   1.
(a) Is (1 )   -1   -1   1 in the null space of A? (1   2   2   1) (1 ) = (-2)   1   1   1   1   -1     0   1   0   1   1   -1     1                   1, so the answer is NO



(b) Is [2  9  7  2]  in the row space of A?
(1   2   2   1) -> (1    2    2    1 ) -> (1   0   0   1) -> (1   0   0   1) -> (1 ...  1   1         0    -2   -1   0          0   0   2   0         0   0   1   0         0   0   1   0

Now you can see that Row(A) = Span( [1 0 0 1], [0 1 0 0], [0 0 1 0] ).
Then   [2  9  7  2] = 2[1 0 0 1] + 9[0 1 0 0] + 7[0 0 1 0] is in Row(A).
So the answer to the question is YES.

(2)    | 0   0   1   1 | =    2   2   2   2    0   4   4   4    0   0   0   5   -| 0   4   4   4 | = | 2   2   2   2 | = 40     2   2   2   2       0   4   4   4     0   0   1   1       0   0   1   1     0   0   0   5       0   0   0   5
(3) Find the inverse of this matrix without doing any row reductions. (Hint: is it orthogonal?) (1           1            1               )  -          --         -------  2           2      ...              1  -          -                     -------  2          2          0          Sqrt[2]
It's orthogonal, so its inverse is its transpose: (1          1          1          1       )   -          -          -          -  2          2 ...    1         1                        --------   -------  0          0           Sqrt[2]   Sqrt[2]

(4) Suppose W = { (x) | 3 x - 2 y + 7 z = 0 }         y         z.   Find W^⊥.

Note that W is the plane orthogonal to  (3 )    -2    7, so  W^⊥ = Span ( (3 ))                      -2                      7.

(5) Consider the matrix A = (-1   1    0 )       0    1    0       0    1    -1
(a) Find the eigenvalues of A.

det(A - λI) = | -1 - λ   1             0           | = (-1 - λ) | 1 - λ    ...            1             -1 - λ                       0             1             -1 - λ

Eigenvlaues are -1 and 1.



(b) Find the eigenspaces of A.


To find the eigenvectors for 1:  Compute Null(A - I)
( -2   1    0  ) -> ( -2   0    2  ) -> ( 1    0    -1 )     0    0    0            0    1    -2           0    1    -2    0    1    -2           0    0    0            0    0    0
x = z
y = 2z
z = free
Eigenspace: (z  ) = z(1) = Span((1))    2 z      2          2    z        1          1



To find the eigenvectors for -1: Compute Null(A + I)
( 0   1   0 ) -> ( 0   1   0 )    0   2   0           0   0   0    0   1   0           0   0   0
x = free
y = 0
z = free

Eigenspace: (x) = x(1) + z(0) = Span((1), (0))   0      0      0          0    0   z      0      1          0    1


(c) Diagonalize the matrix A, that is find an invertible matrix P and a diagonal matrix D with P^(-1) A P = D.

(1   1   0)^(-1) (-1   1    0 ) (1   1   0) = (1    0    0 )   2   0   0        0    1    0    2   0   0     0    -1   0   1   0   1        0    1    -1   1   0   1     0    0    -1

(6) Suppose !, = {u _ 1, u _ 2, u _ 3}is an orthogonal basis for ^3having the property that || u _ 1 || = 1, || u _ 2 || = 2, and || u _ 3 || = 3. Suppose also that v ∈ ^3satisfies v · u _ 1 = 2,  v · u _ 2 = 5, and  v · u _ 3 = 6. Find [v] _ !,.

Notice that:
1 = || u _ 1 || = (u _ 1 · u _ 1)^(1/2), so u _ 1 · u _ 1 = 1
2 = || u _ 2 || = (u _ 2 · u _ 2)^(1/2), so u _ 2 · u _ 2 = 4
3 = || u _ 3 || = (u _ 3 · u _ 3)^(1/2), so u _ 3 · u _ 3 = 9
Now since the basis is orthogonal,  [v] _ !, = (v · u  ) = (2) = ( )                        1      -            ------------ ...                     3            --------------            u  · u             3           3


(7) Suppose !, = {u _ 1, u _ 2, u _ 3}is a basis for ^3,  and  T : ^3 -> ^2 is a linear transformation for which T(u _ 1) = (3)              0,  T(u _ 2) = (1)              1, and T(u _ 3) = (0 )              -1. Moreover, suppose v is a vector in  ^3for which  FormBox[RowBox[{v, =, Cell[TextData[{3, Cell[BoxData[u ]], +2, Cell[BoxData[u ]], +5, Cell[Box ...                                                    1                      2                      3.  Find T(v).

FormBox[RowBox[{T(v),  , =,  , RowBox[{RowBox[{T, (, Cell[TextData[{3, Cell[BoxData[u ]], +2,  ...                                                                          0       1       -1     -3


(8) Suppose A is a matrix which has 5 rows, and the rows are linearly independent. Suppose also that Null(A) is two-dimensional. How many columns does A have?

Since the rows are linearly independent, the 5 rows form a basis for the row space.
Thus rank(A) = dim(Row(A)) = 5.

By Rank Theorem: Rank(A) + Nullity(A) = n = (number of columns)
Thus (number of columns) = 5 + 2 = 7


(9) Suppose A is a 5×5 matrix with a one-dimensional null space.   Find det(A).

Since the null space is one dimensional, it follows that Null(A) = span(x) for some nonzero vector x.
This means that Nullity(A) = 1. Since Nullity(A) is not zero, A is not invertible, by the invertible matrix theorem. Hence det(A) = 0.