[선형대수학] 행렬 연산 문제풀이

 

1. Prove Theorem 2.2(i) : $ \left ( AB \right ) C=A \left ( BC \right ) $.

pf) Let $ A=[a _ {ij} ] $, $ B=[b _ {jk} ] $, $ C=[c _ {kl} ] $, and let $ AB=S=[s _ {ik} ] $, $ BC=T=[t _ {jl} ] $. Then

$ s _ {ik} = \sum\limits _ {j=1} ^ {m} a _ {ij} b _ {jk} $ and $ t _ {jl} = \sum\limits _ {k=1} ^ {n} b _ {jk} c _ {kl} $

Multiplying $ S=AB $ by $ C $, the $ il- $entry of $ \left ( AB \right ) C $ is

$ s _ {i1} c _ {1l} +s _ {i2} c _ {2l} + \cdots +s _ {in } c _ {nl} = \sum\limits _ {k=1} ^ {n} s _ {ik} c _ {kl} = \sum\limits _ {k=1} ^ {n} \sum\limits _ {j=1} ^ {m} \left ( a _ {ij} b _ {jk} \right ) c _ {kl} $

On the other hand, multiplying $ A $ by $ T=BC $, the $ il- $entry of $ A \left ( BC \right ) $ is

$ a _ {i1} t _ {1l} +a _ {i2} t _ {2l} + \cdots +a _ {\in } t _ {nl} = \sum\limits _ {j=1} ^ {m} a _ {ij} t _ {jl} = \sum\limits _ {j=1} ^ {m} \sum\limits _ {k=1} ^ {n} a _ {ij} \left ( b _ {jk} c _ {kl} \right ) $

The above sums are equal ; that is, corresponding elements in $ \left ( AB \right ) C=A \left ( BC \right ) $ are equal. Thus, $ \left ( AB \right ) C=A \left ( BC \right ) $.

 

2. Prove Theorem 2.2(ii): $ A ( B+C)=AB+AC $.

Let $ A=[a _ {ij} ] $, $ B=[b _ {jk} ] $, $ C=[c _ {jk} ] $, and let $ D=B+C=[d _ {jk} ],~E=AB=[e _ {ik} ] $, $ F=AC=[f _ {ik} ] $. Then

$ d _ {jk} =b _ {jk} +c _ {jk} $, $ e _ {ik} = \sum\limits _ {j=1} ^ {m} a _ {ij} b _ {jk} $, $ f _ {ik} = \sum\limits _ {j=1} ^ {m} a _ {ij} c _ {jk} $

Thus, the $ ik- $entry of the matrix $ AB+AC $ is

$ e _ {ik} +f _ {ik} = \sum\limits _ {j=1} ^ {m} a _ {ij} b _ {jk} + \sum\limits _ {j=1} ^ {m} a _ {ij} c _ {jk} = \sum\limits _ {j=1} ^ {m} a _ {ij} ( b _ {jk} +c _ {jk} ) $

On the other hand, the $ ik- $entry of $ AD=A \left ( B+C \right ) $ is

$ a _ {i1} d _ {1k} +a _ {i2} d _ {2k} + \cdots +a _ {im} t _ {mk} = \sum\limits _ {j=1} ^ {m} a _ {ij} d _ {jk} = \sum\limits _ {j=1} ^ {m} a _ {ij} \left ( b _ {jk} +c _ {jk} \right ) $

Thus, $ A ( B+C)=AB+AC $, because the corresponding elements are equal.

 

3. Prove Theorem 2.3(iv): $ \left ( AB \right ) ^ {T} =B ^ {T} A ^ {T} $.

pf) Let $ A=[a _ {ik} ] $ and $ B=[b _ {kl} ] $. Then the $ ij- $entry of $ AB $ is

$ a _ {i1} b _ {1j} +a _ {i2} b _ {2j} + \cdots +a _ {im} b _ {mj} $

This is the $ ji- $entry (reverse order) of $ \left ( AB \right ) ^ {T} $. Now column $ j $ of $ B $ becomes row $ j $ of $ B ^ {T} $, and row $ i $ of $ A $ becomes column $ i $ of $ A ^ {T} $. Thus, the $ ij- $entry of $ B ^ {T} A ^ {T} $ is

$ [b _ {1j} ,~b _ {2j} ,~ \cdots ,~b _ {mj} ][a _ {i1} ,~a _ {i2} ,~ \cdots ,~a _ {im} ] ^ {T} =b _ {1j} a _ {i1} +b _ {2j} a _ {i2} + \cdots +b _ {mj} a _ {im} $

Thus, $ \left ( AB \right ) ^ {T} =B ^ {T} A ^ {T} $ on because the corresponding entries are equal.

 

4. Let $ A= \left[ \matrix {1 & 3\\4 & -3} \right] $

(a) Find a nonzero column vector $ u= \left[ \matrix {x\\y} \right] $ such that $ Au=3u $.

(b) Describe all such vectors.

sol)

(a) First set up the matrix equation $ Au=3u $, and then write each side as a single matrix (column vector) as follows:

$ \left[ \matrix {1 & 3\\4 & -3} \right] \left[ \matrix {x\\y} \right] =3 \left[ \matrix {x\\y} \right] $ and then $ \left[ \matrix {x+3y\\4x-3y} \right] = \left[ \matrix {3x\\3y} \right] $

Set the corresponding elements equal to each other to obtain a system of equations:

$$ x+3y=3x\\ 4x-3y=3y $$

or

$$ 2x-3y=0\\ 4x-6y=0 $$

or

$$ 2x-3y=0 $$

The system reduces to one nondegenerate linear equation in two unknowns, and so has an infinite number of solutions. To obtain a nonzero solution, let, say, $ y=2 $; then $ x=3 $. Thus, $ u= ( 3,~2) ^ {T} $ is a desired nonzero vector.

(b) To find the general solution, set $ y=t $, where $ t $ is a parameter. Substitute $ y=t $ into $ 2x-3y=0 $ to obtain $ x= \frac {3} {2} t $. Thus, $ u= \left ( \frac {3} {2} t,~t \right ) ^ {T} $ represents all such solutions.

 

5. Let $ A $ and $ B $ be invertible matrices (with the same size). Show that $ AB $ is also invertible and $ \left ( AB \right ) ^ {-1} =B ^ {-1} A ^ {-1} $. [Thus, by induction, $ \left ( A _ {1} A _ {2} \cdots A _ {m} \right ) ^ {-1} =A _ {m} ^ {-1} \cdots A _ {2} ^ {-1} A _ {1} ^ {-1} $. ]

sol) Using the associativity of matrix multiplication, we get

$ \left ( AB \right ) \left ( B ^ {-1} A ^ {-1} \right ) =A \left ( BB ^ {-1} \right ) A ^ {-1} =AIA ^ {-1} =AA ^ {-1} =I $

$ \left ( B ^ {-1} A ^ {-1} \right ) \left ( AB \right ) =B ^ {-1} \left ( AA ^ {-1} \right ) B=B ^ {-1} IB=B ^ {-1} B=I $

Thus, $ \left ( AB \right ) ^ {-1} =B ^ {-1} A ^ {-1} $

 

Diagonal and Triangular Matrices

A square matrix $ D=[d _ {ij} ] $ is diagonal if its nondiagonal entries are all zero. Such a matrix is sometimes denoted by $ D= \rm diag \it \left ( d _ {11} ,d _ {22} , \cdots ,d _ {nn} \right ) $

6. Write out the diagonal matrices $ A= \rm diag \left ( 4,~-3,~7 \right ) $, $ B= \rm diag \left ( 2,~-6 \right ) $.

sol)

$ A= \left[ \matrix {4 & 0 & 0\\0 & -3 & 0\\0 & 0 & 7} \right] $, $ B= \left[ \matrix {2 & 0\\0 & -6} \right] $

 

7. Find a $ 2 \times 2 $ matrix $ A $ such that $ A ^ {2} $ is diagonal but not A.

sol) Let $ A= \left[ \matrix {1 & 2\\3 & -1} \right] $. Then $ A ^ {2} = \left[ \matrix {7 & 0\\0 & 7} \right] $, which is diagonal.

 

8. Let $ A=[a _ {ij} ] $ and $ B=[b _ {ij} ] $ upper triangular matrices. Prove that $ AB $ is upper triangular with diagonal $ a _ {11} b _ {11} ,~a _ {22} b _ {22} ,~ \cdots ,~a _ {nn} b _ {nn} $.

sol) Let $ AB=[c _ {ij} ] $. Then $ c _ {ij} = \sum\limits _ {k=1} ^ {n} a _ {ik} b _ {kj} $ and $ c _ {ii} = \sum\limits _ {k=1} ^ {n} a _ {ik} b _ {ki} $. Suppose $ i>j $. Then, for any $ k $, either $ i>k $ or $ k>j $, so that either $ a _ {ik} =0 $ or $ b _ {kj} =0 $. Thus, $ c _ {ij} =0 $, and $ AB $ is upper triangular. Suppose $ i=j $. Then, for $ k<i $, we have $ a _ {ik} =0 $; and, for $ k>i $, we have $ b _ {ki} =0 $. Hence, $ c _ {ii} =a _ {ii} b _ {ii} $, as claimed. [This proves one part of Theorem 2.5(i); the statements for $ A+B $ and $ kA $ are left as exercises.]

 

9. Determine whether or not each of the following matrices is symmetric—that is, $ A ^ {T} =A $. or skew-symmetric—.that is, $ A ^ {T} =-A $:

(a) $ A= \left[ \matrix {5 & -7 & 1\\-7 & 8 & 2\\1 & 2 & -4} \right] $

(b) $ B= \left[ \matrix {0 & 4 & -3\\-4 & 0 & 5\\3 & -5 & 0} \right] $

(c) $ C= \left[ \matrix {0 & 0 & 0\\0 & 0 & 0} \right] $

(sol)

(a) By inspection, the symmetric elements (mirror images in the diagonal) are $ -7 $ and 7, 1 and 1, 2 and 2. Thus, A is symmetric, because symmetric elements are equal.

(b) By inspection, the diagonal elements are all 0, and the symmetric elements, 4 and $ -4 $, $ -3 $ and 3, and 5 and $ -5 $, are negatives of each other. Hence, $ B $ is skew-symmetric.

(c) Because $ C $ is not square, $ C $ is neither symmetric nor skew-symmetric.

 

Orthogonal Matrices

(정의) A real matrix $ A $ is orthogonal if $ A ^ {T} =A ^ {-1} $. that is, if $ AA ^ {T} =A ^ {T} A=I $. Thus, $ A $ must necessarily be square and invertible.

Generally speaking, vectors $ u _ {1} ,~u _ {2} ,~ \cdots ,~u _ {m} $ in $ R ^ {n} $ are said to form an orthonormal set of vectors if the vectors are unit vectors and are orthogonal to each other; that is,

$ u _ {i} \cdot u _ {j} = { \begin {cases} 0~ & ( i \neq j)\\1 & ( i=j)\end {cases} } $

In other words, $ u _ {i} \cdot u _ {j} = \delta _ {ij} $ where $ \delta _ {ij} $ is the Kronecker delta function:

 

THEOREM : Let $ A $ be a real matrix. Then the following are equivalent:

(a) $ A $ is orthogonal.

(b) The rows of $ A $ form an orthonormal set.

(c) The columns of $ A $ form an orthonormal set.

 

10. Let A be an arbitrary $ 2 \times 2 $ (real) orthogonal matrix.

(a) Prove: If $ ( a,~b) $ is the first row of $ A $, then $ a ^ {2} +b ^ {2} =1 $ and

$ A= \left[ \matrix {a & b\\-b & a} \right] $ or $ A= \left[ \matrix {a & b\\b & -a} \right] $.

 

(b) Prove Theorem 2.7: For some real number $ \theta $,

$ A= \left[ \matrix {\cos \theta & \sin \theta \\-\sin \theta & \cos \theta } \right] $ or $ A= \left[ \matrix {\cos \theta & \sin \theta \\\sin \theta & -\cos \theta } \right] $

 

(sol)

(a) Suppose $ \left ( x,~y \right ) $ is the second row of $ A $. Because the rows of $ A $ form an orthonormal set, we get

$ a ^ {2} +b ^ {2} =1 $, $ x ^ {2} +y ^ {2} =1 $, $ ax+by=0 $

Similarly, the columns form an orthogonal set, so

$ a ^ {2} +x ^ {2} =1 $, $ b ^ {2} +y ^ {2} =1 $, $ ab+xy=0 $

Therefore, $ x ^ {2} =1-a ^ {2} =b ^ {2} $, whence $ x=\pm b $.

Case (i): $ x=b $. Then $ b \left ( a+y \right ) =0 $, so $ y=-a $.

Case (ii): $ x=-b $. Then $ b \left ( y-a \right ) =0 $, so $ y=a $.

This means, as claimed,

$ A= \left[ \matrix {a & b\\-b & a} \right] $ or $ A= \left[ \matrix {a & b\\b & -a} \right] $.

 

(b) Because $ a ^ {2} +b ^ {2} =1 $, we have $ -1 \leq a \leq 1 $. Let $ a=\cos \theta $. Then $ b ^ {2} =1-\cos ^ {2} \theta $, so $ b=\sin \theta $. This proves the theorem.

 

 

11. Find a $ 2 \times 2 $ orthogonal matrix $ A $ whose first row is a (positive) multiple of $ \left ( 3,~4 \right ) $.

sol) Normalize $ ( 3,~4) $ to get $ \left ( \frac {3} {5} ,~ \frac {4} {5} \right ) $. Then, by Problem beyond,

$ A= \left[ \matrix { \frac {3} {5} & \frac {4} {5} \\- \frac {4} {5} & \frac {3} {5} } \right] $ or $ A= \left[ \matrix { \frac {3} {5} & \frac {4} {5} \\ \frac {4} {5} & - \frac {3} {5} } \right] $.

 

12. Find a $ 3 \times 3 $ orthogonal matrix $ P $ whose first two rows are multiples of $ u _ {1} = \left ( 1,~1,~1 \right ) $ and $ u _ {2} = \left ( 0,~-1,~1 \right ) $, respectively. (Note that, as required, $ u _ {1} $ and $ u _ {2} $ are orthogonal.)

sol)

First find a nonzero vector $ u _ {3} $ orthogonal to $ u _ {1} $ and $ u _ {2} $ say (cross product) $ u _ {3} =u _ {1} \times u _ {2} = \left ( 2,~-1,~-1 \right ) $. Let $ A $ be the matrix whose rows are $ u _ {1} $, $ u _ {2} $, $ u _ {3} $ ; and let $ P $ be the matrix obtained from $ A $ by normalizing the rows of $ A $. Thus,

$ A= \left[ \matrix {1 & 1 & 1\\0 & -1 & 1\\2 & -1 & -1} \right] $ and $ P= \left[ \matrix { \frac {1} {\sqrt {3} } & \frac {1} {\sqrt {3} } & \frac {1} {\sqrt {3} } \\0 & - \frac {1} {\sqrt {2} } & \frac {1} {\sqrt {2} } \\ \frac {2} {\sqrt {6} } & - \frac {1} {\sqrt {6} } & - \frac {1} {\sqrt {6} } } \right] $

 

13. Suppose $ A $ is invertible. Show that if $ AB=AC $, then $ B=C $. Give an example of a nonzero matrix $ A $ such that $ AB=AC $ but $ B \neq C $.

(sol)

$ A= \left[ \matrix {1 & 2\\1 & 2} \right] $, $ B= \left[ \matrix {0 & 0\\1 & 1} \right] $, $ C= \left[ \matrix {2 & 2\\0 & 0} \right] $

 

14. Find $ 2 \times 2 $ invertible matrices $ A $ and $ B $ such that $ A+B \neq O $ and $ A+B $ is not invertible.

sol)

$ A= \left[ \matrix {1 & 2\\0 & 3} \right] $, $ B= \left[ \matrix {4 & 3\\3 & 0} \right] $

 

15. Show (a) $ A $ is invertible if and only if $ A ^ {T} $ is invertible.

(b) The operations of inversion and transpose commute; that is, $ \left ( A ^ {T} \right ) ^ {-1} = \left ( A ^ {-1} \right ) ^ {T} $ .

(c) If $ A $ has a zero row or zero column, then $ A $ is not invertible.

 

16. Suppose $ A $ is a square matrix. Show that

(a) $ A+A ^ {T} $ is symmetric,

(b) $ A-A ^ {T} $ is skew-symmetric,

(c) $ A=B+C $, where $ B $ is symmetric and $ C $ is skew-symmetric.

 

17. Write $ A= \left[ \matrix {4 & 5\\1 & 3} \right] $ as the sum of a symmetric matrix $ B $ and a skew-symmetric matrix $ C $.

 

18. Suppose $ A $ and $ B $ are symmetric. Show that the following are also symmetric:

(a) $ A+B $

(b) $ kA $, for any scalar $ k $

(c) $ A ^ {2} $

(d) $ A ^ {n} $, for $ n>0 $

(e) $ f ( A) $, for any polynomial $ f ( x) $.

 

19. Find a $ 2 \times 2 $ orthogonal matrix $ P $ whose first row is a multiple of

(a) $ ( 1,~2,~3) $ and $ ( 0,~-2,~3) $

(b) $ ( 1,~3,~1) $ and $ ( 1,~0,~-1) $.

sol)

(a) $ \left[ \matrix {1/ \sqrt {14} & 2/ \sqrt {14} & 3/ \sqrt {14} \\0 & -2/ \sqrt {13} & 3/ \sqrt {13} \\12/ \sqrt {157} & -3/ \sqrt {157} & -2/ \sqrt {157} } \right] $

(b) $ \left[ \matrix {1/ \sqrt {11} & 3/ \sqrt {11} & 1/ \sqrt {11} \\1/ \sqrt {2} & 0 & -1/ \sqrt {2} \\3/ \sqrt {22} & 3/ \sqrt {22} & 3/ \sqrt {22} } \right] $

 

20. Suppose $ A $ and $ B $ are orthogonal matrices. Show that $ A ^ {T} $, $ A ^ {-1} $, $ AB $ are also orthogonal.

 

21. Which of the following matrices are normal? $ A= \left[ \matrix {3 & -4\\4 & 3} \right] $, $ B= \left[ \matrix {1 & -2\\2 & 3} \right] $, $ C= \left[ \matrix {1 & 1 & 1\\0 & 1 & 1\\0 & 0 & 1} \right] $.

sol) $ A,~C $