01:640:250 - INTRODUCTORY LINEAR ALGEBRA - SPRING 2024
IZAR ALONSO
Extra practice for previous topics - brief solutions
1. Multiple possibilities! See review session.
2. A is not invertible if and only if c = −10/3.
3. RREF of A is


1 0 2 −1
R = 0 1 −2 3 
0 0 0
0
(a) Let ⃗v1 and ⃗v2 be the two vectors in W. We first must check that W is LI. We can do
this by showing that that the two vectors are not parallel or finding the REF of [⃗v1 ⃗v2 ]
and seeing that is has two nonzero rows. Then, we check that the dimension of W is
2 (because RREF has 2 nonzero rows). Finally, we check that the vectors in W are in
Null(A) (A⃗v1 = 0 and A⃗v2 = 0).
(b) Let ⃗v1 and ⃗v2 be the two vectors in W. We first must check that W is LI. We can do
this by showing that that the two vectors are not parallel or finding the REF of [⃗v1 ⃗v2 ]
and seeing that is has two nonzero rows. Then, we check that the dimension of W is
2 (because RREF has 2 nonzero rows). Finally, we check that the vectors in W are in
Col(A) (⃗v1 = ⃗c1 + ⃗c2 , ⃗v2 = 2⃗c1 − ⃗c2 , where ⃗c1 , ⃗c2 are the first and second columns of A).
(c) Let ⃗v1 and ⃗v2 be the two vectors in W. We first must check that W is LI. We can do
this by showing that that the two vectors are not parallel or finding the REF of [⃗v1 ⃗v2 ]
and seeing that is has two nonzero rows. Then, we check that the dimension of W is
2 (because RREF has 2 nonzero rows). Finally, we check that the vectors in W are in
Row(A) (⃗v1 = ⃗r1 + ⃗r2 , ⃗v2 = 2⃗r1 + 3⃗r2 , where ⃗r1 , ⃗r2 are the first and second rows of R).
Suggested review problems – final exam – brief solutions
2−t
−1
1. (a) Characteristic polynomial det
= t2 + t − 2 = (t − 1)(t + 2).
4
−3 − t
Eigenvalues/Eigenvectors:
1
1
λ1 = 1, v1 =
and λ2 = −2, v2 =
(or any nonzero multiples of these vectors).
1
4
1 1
λ1 0
1 0
(b) Take P = v1 v2 =
and D =
=
.
1 4
0 λ2
0 −2
 
1
T
T
T

2  = 14. The
2. (a) Av = vv v = v v v = 14v since v v = 1 2 3
3
eigenvalue
is
14
.


1 2 3
(b) A =  2 4 6  = v 2v 3v .
3 6 9
(c) Since rank A = dim Col A = 1, we have dim Null A = 2.
1
2
IZAR ALONSO


(1 − t)
2
3
 = −t2 (t − 14)
2
(4 − t)
6
(d) characteristic polynomial det 
3
6
(9 − t)
Eigenvalues are λ1 = 14 (multiplicity 1 ) and λ2 = 0 (multiplicity 2).
(e) Let u1 , u2 be a basis for Null A. They give a basis for the 0 eigenspace of A. Then
3
{u1 , u2 , v} is an
eigenvector basis for R . Hence A is diagonalizable.
(f) Take P = u1 u2 v . Then the diagonal entries of D are 0, 0, 14.
3. Cauchy-Schwarz inequality: |u · v| ≤ ∥u∥∥v∥ triangle inequality: ∥u + v∥ ≤ ∥u∥ + ∥v∥.
Assume Cauchy-Schwarz. Then
∥u+v∥2 = (u+v)·(u+v) = ∥u∥2 +2u·v+∥v∥2 ≤ ∥u∥2 +2∥u∥∥v∥+∥v∥2 = (∥u∥+∥v∥)2 .


1
v·x
= − 17  2 .
4. (a) Calculate v·x = 1+0−3 = −2 and v·v = 1+4+9 = 14. Then y = v·v
3
 




1
8
1
Hence z =  0  + 17  2  = 17  2 . Check: v · z = (1/7)(1 · 8 + 2 · 2 − 3 · 4) = 0.
3
−4
−1


x1
(b) The vector  x2  is in V ⊥ when x1 +2x2 +3x3 = 0. So V ⊥ = Null(A), where A = vT .
x3


−2
Since x2 and x3 are the free variables, a basis for the null space is u1 =  1  , u2 =
0


−3
 0 .
1
(c) Apply Gram-Schmidt to the vectors u1 , u2 from
 (b):
−3
6
1
2
−6 . Check: v1 · v2 = (1/5)((−2) ·
v
=
u
−
v
=
v1 = u1 and v2 = u2 − vv11 ·u
2
·v1 1
5 1
5
5
(−3) + 1 · (−6) + 0 · 5) = 0
v · v1 = 1 · (−2) + 2 · 1 + 3 · 0 = 0, and v · v2 =
 (1/5)(1
 · (−3) + 2 · (−6) + 3 · 5) = 0. 
−2
−3
For an orthonormal set use q1 = ∥v11 ∥ v1 = √15  1  and q2 = ∥v12 ∥ v2 = √170  −6 .
0
5
−2
−8
√
(d) The general formula is c1 = x · q1 = √
and
c
=
x
·
q
=
since
∥q
∥
=
∥q
∥
2
2
1
2 = 1.

 70

 5 

−2
−3
80
−8 
1 
−2 


1 + 70
−6 = 70
20  = z.
Check: c1 q1 + c2 q2 = 5
0
5
−40
1 1 1 0 0
5. Let R = rref(A) =
, basic variables x1 , x4 and free variables x2 .x3 , x5 .
0 0 0 1 1
(a) dim Row A = dim Col A = 2, dim Null A = 3.
1
0
2
(b) For Col A = R , use any orthonormal basis, such as
and
.
0
1
√
√
For Row A = Row R, use 1 1 1 0 0 / 3 and 0 0 0 1 1 / 2.
For Null A, the equations are x1 + x2 + x3 = 0 and x4 + x5 = 0, so it has a basis
3






−1
−1
0
 1 
 0 
 0 











u1 =  0  , u2 =  1  , u3 = 
 0 . Here u1 ⊥ u3 and u2 ⊥ u3 . Replace u2 by
 0 
 0 
 −1 
0
0
1


−1/2
 −1/2 


1
1

1 
u
=
u
−
v2 = u2 − uu21 ·u
u
=
2
·u1 1
2 1

. To get an orthnomal basis {w1 , w2 , w3 },

0 
0
take






−1/2
−1
0
 1 

 0 
q 
 −1/2 




1
2
1
1
1
1




1 
w1 = ∥u1 ∥ u1 = √2 
 0  , w2 = ∥v2 ∥ v2 = 3 
 , w3 = ∥u3 ∥ u3 = √2  0 .

 0 
 −1 
0 
0
0
1
 


1
1
6. The columns of Q must be an orthnormal basis for R3 . One solution is √13  1  , √12  0 .
1
−1
(There are infinitely many other solutions,)
7. (a) True: There are four distinct eigenvalues, hence four independent eigenvectors.
(b) False: Eigenvalue 0 has algebric multiplicity 2, but only one eigenvector if nullity
A = 1.
(c) True: Since nullity A = 4 − 2 = 2, there is a linearly independent set of eigenvectors.
(d) True: Symmetry guarantees that the algebraic multiplicity = geometric multiplicity
for eigenvalues, so there is a basis of eigenvectors. The diagonal matrix could have entries
2, 0, 0, 0, or 2, 2, 0, 0, or 2, 2, 2, 0, depending on the multiplicities.


(7 − t)
0
0
0 (4 − t)
2  = (7−t) t2 − 7t + 10 =
8. (a) Characteristic polynomial det 
0
1 (3 − t)
−(t − 2)(t − 5)(t − 7)
Eigenvalues: λ1 = 2, λ2 = 5, λ3 = 7. Eigenvectors:

5
A − λ 1 I3 =  0
0

2
A − λ 2 I3 =  0
0

0
A − λ 3 I3 =  0
0


0 0
1 0
2 2 → 0 1
1 1
0 0


0
0
1
−1
2 → 0
1 −2
0


0
0
0
−3
2 → 0
1 −4
0





0
0
0
1  so eigenvector v1 =  −1  . (Check: Av1 =  −2  = 2v1 )
0
1
2

 


0
0
0
0
1 −2  so eigenvector v2 =  2  . (Check: Av2 =  10  = 5v2 )
0
0
1
5

 
 
1 0
1
7
0 1  so eigenvector v3 =  0  . (Check: Av3 =  0  = 7v3 )
0 0
0
0
(b) Take P = [v1 , v2 , v3 ]. Then P is invertible since the eigenvectors are independent

2 0 0
(the eigenvalues are distinct); alternate argument: det P = 1. Take D =  0 5 0 
0 0 7
(diagonal matrix of eigenvalues).
4
IZAR ALONSO




2 0
0
1
0 0
0 . Then P −1 =  −1
1 0  and
9. (a) Take P = [v1 , v2 , v3 ] and D =  0 1
0 0 −1
2 −1 1


2 0
0
0  (lower triangular, with eigenvalues on the diagonal).
A = P DP −1 =  1 1
−5 2 −1
 
3
(b) Let y = P −1 x =  1 . Then x = P y = 3v1 + 1v2 + 7v3 . Apply An to each
7
eigenvector:
An x = 3 · 2n v1 + 1 · 1n v2 + 7 · (−1)n v2 ≈ 3 · 2n v1 when n is large.
10. Since Ax = 2x we have 2x · y = (Ax) · y = (Ax)T y = xT (Ay) = x · Ay since A = AT .
But Ay = 3y, so we get 2x · y = 3x · y. This forces x · y = 0.