Dimension, Rank, Basis, Four Fundamental Subspace

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A. Dimension, Rank, Basis, Four Fundamental Subspace
B. Orthogonality, Projection, Component, Eigenvectors, and Eigenvalues
1. Find the rank of the matrix
a) 𝐴𝐴 = �
1 2 3
2 3 4
3 5 7

b) B = �
0 1 2 1
1 2 3 2
3 1 1 3

2. Let 𝑉𝑉 be a subset of R4 consisting of vectors that are perpendicular to vectors 𝑎𝑎, 𝑏𝑏, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐
where 𝑎𝑎 = < 1, 0, 1, 0 >, 𝑏𝑏 = < 1, 1, 0, 0 >, 𝑐𝑐 =< 0, 1, −1, 0 >,
Namely, 𝑉𝑉 = {𝑥𝑥 ∈ 𝑅𝑅4|𝑎𝑎𝑇𝑇𝑥𝑥 = 0, 𝑏𝑏𝑇𝑇𝑥𝑥 = 0, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶𝑇𝑇𝑥𝑥 = 0}
a. Prove that V is a subspace of 𝑅𝑅4
b. Find a basis for V
c. Determine the Dimension of V
Solution Hint:
a) Observe that the conditions 𝑎𝑎𝑇𝑇𝑥𝑥 = 0, 𝑏𝑏𝑇𝑇𝑥𝑥 = 0, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑇𝑇𝑥𝑥 = 0, combining 𝐴𝐴𝐴𝐴 = 0
Where, A = �
1 0 1 0
1 1 0 0
0 1 −1 0
�. Note that the rows of the matrix A are 𝑎𝑎𝑇𝑇, 𝑏𝑏𝑇𝑇, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑇𝑇. It follows
that the subset V is in the null space 𝑁𝑁(𝐴𝐴) of the matrix 𝐴𝐴. Being the null space 𝑉𝑉 = 𝑁𝑁(𝐴𝐴), is
a subspace of 𝑅𝑅4.
b) To find a basis, we determine the solutions of 𝐴𝐴𝐴𝐴 = 0
Applying elementary row operations to the augmented matrix, we see that,

1 0 1 0 0
1 1 0 0 0
0 1 −1 0 0
� ⟶ �
1 0 1 0 0
0 1 −1 0 0
0 1 −1 0 0
� (𝑅𝑅2 − 𝑅𝑅3) ⟶ �
1 0 1 0 0
0 1 −1 0 0
0 0 0 0 0

Then, Determine the general solution and determine the basis and you will have it.
3. Determine which of the following is a subspace of 𝑅𝑅3.
a) 𝑥𝑥 + 2𝑦𝑦 − 3𝑧𝑧 = 4
b) 𝑥𝑥−1
2 = 𝑦𝑦+2
3 = 𝑧𝑧
4
c) 𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧 = 0 and 𝑥𝑥 − 𝑦𝑦 + 𝑧𝑧 = 1
d) 𝑥𝑥 = −𝑧𝑧 and 𝑥𝑥 = 𝑧𝑧
e) 𝑥𝑥2 + 𝑦𝑦2 = 𝑧𝑧
f) 𝑥𝑥
2 = 𝑦𝑦−3
5
4. Suppose 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟(𝑅𝑅0) = 𝐴𝐴 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑅𝑅0 = �
1 3 5 0 7
2 6 10 1 16
3 9 15 1 23
�. Show that –
a) The row space has dimension 2, matching the rank
b) The column space of 𝑅𝑅0 has also dimension 𝑟𝑟 = 2
c) The null space of 𝑅𝑅0 has dimension 3
d) The null space of 𝑅𝑅0
𝑇𝑇, which can also be called the left null space of 𝑅𝑅0; has dimension 1.
5. Find a basis for each of the four fundamental subspaces associated with the matrix.
A= �
1 2 0 1
0 1 1 0
1 2 0 1

6. Let A be a real 7×3 matrix such that the null space is spanned by the vectors

1
2
0
�,�
2
1
0
� 𝑎𝑎𝑎𝑎𝑎𝑎 �
1
−1
0
�. Find the rank of the matrix A.
7. Let V be a subset of the vector space 𝑅𝑅𝑛𝑛 consisting only of the zero vector of 𝑅𝑅𝑛𝑛, Namely 𝑉𝑉 =
{0}. Then prove that V is a subspace of 𝑅𝑅𝑛𝑛.
8. Let 𝐴𝐴 = �
4 1
3 2
�and consider the following subset V of the 2-dimensional vector space 𝑅𝑅2,
Namely 𝑉𝑉 = {𝑥𝑥 ∈ 𝑅𝑅2|𝐴𝐴𝐴𝐴 = 5𝑥𝑥}
a) Prove that the subset V is a subspace of 𝑅𝑅2
b) Find a basis for V and determine the dimension of 𝑉𝑉
9. The smallest subspace of 𝑅𝑅3 containing the vectors (2, -3, -3) and (0, 3, 2) is the plane whose
equation is 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 + 6𝑧𝑧 = 0. Determine the value of 𝑎𝑎, 𝑏𝑏.
10. Determine The matrix representation of the orthogonal projection operator taking 𝑅𝑅3 onto
the plane 𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧 = 0.
11. Let 𝑢𝑢 = (8, √3, √7, −1, 1) and 𝑢𝑢 = (1, −1, 0, 2, √3). If the orthogonal projection of u onto v
is 𝑎𝑎
𝑏𝑏
𝑣𝑣, then determine a and b.
12. Find the point 𝑞𝑞 in 𝑅𝑅3 on the ray connecting the origin to the point (2, 4, 8) which is closest
to the point (1, 1, 1).
13. Find the eigenvalues and eigenvectors of the following matrix A.
𝐴𝐴 = �
1 −1 0
−1 2 −1
0 −1 1

Show that these eigenvectors are perpendicular. [Hint: It will always be perpendicular when
A is symmetric]
14. Suppose you want a vector to rotate about 90 Degree anti-clockwise. Determine the
transformation matrix that should operate on that vector to produce such result? Determine
for 180, and 270 degrees too.
15. Find the rank and the four eigenvalues of A, where 𝐴𝐴 = �
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1

16. [Page 201, Worked Example 4.1A, Introduction to Linear Algebra (4th Edition), Gilbert
Strang]
Suppose S is a six-dimensional subspace of nine-dimensional space 𝑅𝑅9.
a. What are the possible dimensions of subspaces orthogonal to 𝑆𝑆?
b. What are the possible dimensions of the orthogonal complement 𝑆𝑆^ of S?
c. What is the smallest possible size of a matrix A that has row space S?
d. What is the shape of its null space matrix N?
17. (Bonus Problem)
Find all eigenvalues and eigenvectors of the matrix 𝐴𝐴,
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐴𝐴 =

10001 3 5 7 9 11
1 10003 5 7 9 11
1 3 10005 7 9 11
1 3 5 10007 9 11
1 3 5 7 10009 11
1 3 5 7 9 10011⎦

Solution Hint: Let 𝐵𝐵 = 𝐴𝐴 − 10000𝐼𝐼, where I is the 6 ∗ 6 identity matrix. That is, we have,
𝐵𝐵 =

1 3 5 7 9 11
1 3 5 7 9 11
1 3 5 7 9 11
1 3 5 7 9 11
1 3 5 7 9 11
1 3 5 7 9 11⎦

, since all row are same, 𝐵𝐵 is singular and hence 𝜆𝜆 = 0 is an
eigenvalue of B. With elementary row operation, we find 𝐵𝐵 ⟶

1 3 5 7 9 11
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0 ⎦

By inspection, we see that 𝐵𝐵𝐵𝐵 = 36𝑣𝑣, where 𝑣𝑣 =< 1, 1, 1, 1, 1, 1 >. Thus it yields that 𝜆𝜆 = 36 is
the eigenvalue of 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑣𝑣 is the corresponding eigenvector.

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