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 ÷µ^3having the property that || u _ 1 || = 1, || u _ 2 || = 2, and || u _ 3 || = 3. Suppose also that v ∈ ÷µ^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 ÷µ^3,  and  T : ÷µ^3 -> ÷µ^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  ÷µ^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.