[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) $A,~C$