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Linear Algebra Practice Final Exam
Final Exam on Sunday, June 2
Bradley 102, 1:00 – 3:
1 Let
A =
( − 2 12 − 1 5
) .
Find a diagonal matrix D and an invertible matrix P such that A = P DP −^1. Briefly explain yourself.
Solution
First we need to find the eigenvalues. For this, we look at
det(A − λI) = det
( − 2 − λ 12 − 1 5 − λ
) = (λ − 2)(λ − 1),
which means that the eigenvalues are λ = 1 and λ = 2. Since they are different, we know that there are two linearly independent eigenvectors for A; these will form a basis for R^2 , so A is diagonalizable. It remains to find the eigenvectors. Let’s do λ = 1 first. We need to solve (A − I)x = 0 , so we row reduce
A − I =
( − 3 12 − 1 4
) ∼
( 1 − 4 0 0
) .
Now we repeat the process for λ = 2. Row reduce
A − I =
( − 4 12 − 1 3
) ∼
( 1 − 3 0 0
) .
We conclude that we have eigenvectors
b 1 =
( 4 1
) for eigenvalue 1 and b 1 =
( 3 1
) for eigenvalue 2.
The upshot of all of this is that
A =
( − 2 12 − 1 5
)
( 4 3 1 1
) ( 1 0 0 2
) ( 4 3 1 1
)− 1 .
(a) Write down a matrix that is not diagonalizable. Explain.
Solution
The matrix A =
( 0 1 0 0
)
is not diagonalizable. This is because the only eigenvalue of A is 0, but
A
( 0 1
)
( 0 1 0 0
) ( 0 1
)
( 1 0
) 6 = 0,
so the R^2 is not spanned by vectors killed by A.
(b) Explain why the null space of a linear transformation is a subspace.
Solution
Recall that Null(T ) = {x | T (x) = 0 }. So if If x and y are both in Null(T ) and c ∈ R, then
T (x + y) = T (x) + T (y) = 0 + 0 = 0 and T (cx) = cT (x) = c 0 = 0 ,
so x + y and cx are both in Null(T ). This shows that Null(T ) is a subspace.
(c) True or False: If an n × n matrix A is diagonalizable, then every vector x in Rn^ is an eigenvector for A. Explain.
Solution
FALSE! For example, if
A =
( 1 0 0 2
) and x =
( 1 1
)
then Ax =
( 1 0 0 2
) ( 1 1
)
( 1 2
) ,
which is not a multiple of x, so x is not an eigenvector. But A is obviously a diagonal(izable) matrix!
As we saw in class, the pivot columns of A form a basis for Col(A). This means that the vectors
,
,
are a basis for Col(A).
4 Let T : Rn^ → Rn^ be a linear transformation with the property that T (T (x)) = T (x) for every vector x ∈ Rn^ (such a linear transformation is called idempotent).
(a) Write V for the image (or range) of T. In other words,
V = {T (x) | x ∈ Rn}.
If v ∈ V , then what is T (v)?
Solution
If v ∈ V , then v = T (x) for some x. This means that
T (v) = T (T (x)) = T (x) = v.
In other words, v is an eigenvector for T with eigenvalue 1.
(b) If x ∈ Rn, then what is T (x − T (x))?
Solution
Again, we just need to calculate
T (x − T (x)) = T (x) − T (T (x)) = T (x) − T (x) = 0.
In other words, v is an eigenvector for T with eigenvalue 0.
(c) Let C = {c 1 , c 2 ,... , ck} be a basis for V. Then we can add some more vectors, b 1 , b 2 ,... , bl to get a basis B for all of Rn. Show that if you replace b 1 with a 1 = b 1 − T (b 1 ) then you still have a basis.
Solution
Since the new collection has the same number of vectors as the given basis, we just need to show that the new collection spans Rn. Since T (b 1 ) ∈ V , we can write T (b 1 ) as a linear combination of the c’s, which shows that b 1 ∈ span{c 1 , c 2 ,... , ck, b 1 − T (b 1 )}. So whatever the span of the new set of vectors is, it contains all the c’s and all the b’s. This means that the span of the new set of vectors is all of Rn.
(d) In the same way, we can replace each bi with ai = bi − T (bi). What is the matrix of T with respect to the basis {c 1 , c 2 ,... , ck, a 1 , a 2 ,... , al}? (This is an easy question – not much work is needed!)
Solution
As we saw in part (a), each ci is an eigenvector with eigenvalue 1; as we saw in part (b), each ai is an eigenvector with eigenvalue 0. This means that T is diagonalizable, and the matrix is
D =
dkk = 1 0 0 .. .
with 1’s down the diagonal until the kth^ place and 0’s afterwards.
