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

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

(b) Is [2  9  7  2]  in the row space of A?

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)
(3) Find the inverse of this matrix without doing any row reductions. (Hint: is it orthogonal?)
It's orthogonal, so its inverse is its transpose:

(4) Suppose .   Find .

Note that W is the plane orthogonal to , so  .

(5) Consider the matrix
(a) Find the eigenvalues of A.

Eigenvlaues are -1 and 1.

(b) Find the eigenspaces of A.

To find the eigenvectors for 1:  Compute Null

x = z
y = 2z
z = free
Eigenspace:

To find the eigenvectors for -1: Compute Null(A + I)

x = free
y = 0
z = free

Eigenspace:

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

(6) Suppose is an orthogonal basis for having the property that , , and . Suppose also that satisfies ,  , and  . Find .

Notice that:
, so
, so
, so
Now since the basis is orthogonal,

(7) Suppose is a basis for ,  and   is a linear transformation for which ,  , and . Moreover, suppose v is a vector in  for which  .  Find T(v).

(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.