Lecture Note on Linear Algebra 4. Matrix Equations

Lecture Note on Linear Algebra 4. Matrix Equations Wei-Shi Zheng, 2012 [email protected]

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What Do You Learn from This Note

The product between a matrix and a vector. We will also discuss something about the existence of solutions of a linear system (‚5•§|) • 35). Basic concept: matrix equation(Ý •§)

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Matrix Equation

Definition 1 (product of matrix and vector(Ý †•þƒm ¦{)).  Let c1  c2    A = (~a1 ~a2 · · · ~an ) be an m × n matrix (~ai ∈ Rm )and ~c =  ..  ∈ Rn .  .  cn Then define the product of A and c as   c1  c2    A~c = (~a1 ~a2 · · · ~an )  ..  = c1~a1 + c2~a2 + · · · + cn~an ,  .  cn which is the linear combination of ~a1 , . . . , ~an with coefficients c1 , . . . , cn . Example:     4         1 2 −1   1 2 −1 3 3 =4 +3 +7 = 0 −5 3 0 −5 3 6 7 1

Remarks: 1. A~c is defined (k½Â ) only if the number of columns of A is equal to the number of entries of ~c.(Ò´Ý A ê† •þ~c ‘ꘗ) a11 a12 · · · a1n  a21 a22 · · · a2n    2. Let A =  .. .. .. . Then . .  . . . .  am1 am2 · · · amn      Pn    a11 a1n c1 a11 + · · · + cn a1n j=1 cj a1j         .. .. A~c = c1  ... +· · ·+cn  ...  =  =  . . Pn am1 amn c1 am1 + · · · + cn amn j=1 cj amj Definition 2 (matrix equation(Ý •§)). Let A be an m × n matrix and ~b ∈ Rm . Also let x denote an undetermined vector in Rn . Then we call A~x = ~b 

 s1   a matrix equation in x. A solution of A~x = ~b is a vector s =  ...  ∈ Rn sn satisfying A~s = ~b. 

 s1   Remark. If we write A = (~a1 · · · ~an ), ~s =  ... , then we have sn   s1   A~s = (~a1 · · · ~an )  ...  = s1~a1 + · · · + sn~an = ~b. sn   s1   In other words, s =  ...  is a solution of A~x = ~b iff (s1 , · · · , sn ) is a sn solution of the vector equation x1~a1 + · · · + xn~an = ~b. So we have the following fact:   a11 x1 + · · · +a1n xn = b1 ··· ⇔ x1~a1 + · · · + xn~an = ~b ⇔ A~x = ~b,  am1 x1 + · · · +amn xn = bm 2

that is, they have the same solution set. Now we can view a system of linear equations in three different but equivalent ways: 1. A matrix equation 2. A vector equation 3. A collection of linear equations We finally present some properties of the product of matrix and vector: Theorem 3. Let A be an m × n matrix and ~u, ~v ∈ Rn , λ ∈ R. Then 1. A(~u + ~v ) = A~u + A~v ; 2. A(λ~u) = λ(A~u).     u1 v1     Proof. Suppose that A = (~a1 · · · ~an ), ~u =  ...  and ~v =  ... . un vn 1. A(~u + ~v ) = = = =

(u1 + v1 )~a1 + · · · + (un + vn )~an u1~a1 + v1~a1 + · · · + un~an + vn~an (u1~a1 + · · · + un~an ) + (v1~a1 + · · · + vn~an ) A~u + A~v

2. A(λ~u) = = = =

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(λu1 )~a1 + · · · + (λun )~an λ(u1~a1 ) + · · · + λ(un~an ) λ(u1~a1 + · · · + un~an ) λ(A~u)

A Theorem on Consistent (ƒ ƒN tion for Matrix Equation

) Solu-

The next theorem tells under what conditions of A, the solution set of A~x = ~b is non–empty (consistent) for any ~b.

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Theorem 4. Let A = (~a1 · · · ~an ) be an m × n matrix. Then the following statements are equivalent. 1. ∀ ~b ∈ Rm , A~x = ~b has a solution. 2. ∀ ~b ∈ Rm , ~b is a linear combination of ~a1 , . . . , ~an . 3. Rm = Span{~a1 , . . . , ~an }. 4. A has a pivot position (Ì ˜) in every row. Proof. Proof. We say all the above statements are equivalent means that any one of the statements can be derived from any other one (XJþ¡ o‡·K d§@oÙ¥?Û˜‡·KÑŒ±dÙ¦3‡·Kƒ˜íÑ). Statements (·K)1.–3. are obviously equivalent ( d5w ´„) since they are logically equivalent. So our focus on proving the equivalent between statement 1 and statement 4. Let U be the matrix in REF (reduced echelon form, {z F/) such that U ∼ A (=U †A d). So there is a series of matrices op1

op2

op

l A = A0 −→ A1 −→ · · · −→ Al = U,

where op1 , . . . , opl are elementary row operations. We can perform row reduction (1z{C†) algorithm on [A ~b] and obtain a reduced augmented ~ matrix [U d]. So, if statement 4 is true, there is no pivot in the augmented column, so A~x = ~b has a solution for any ~b, and statement 1 is true. If statement 1 is true, we finally wish to prove statement 4 is true. We reach this claim by contradiction (‡ {) Assume that there exits a row of A which has no pivot position. Then, the last row of U must be zero. That is there is no solution for the linear system corresponding to augmented matrix ~ Note that the solutions between these two systems (corresponding to [U d]. A and U respectively) are equivalent, so there is also no solution for the linear system A~x = ~b too. This yields contradiction to statement 1, so statement 4 is true. (•Ò´`§ÏLÐ C† §ü‡‚5XÚ )´˜ §¤± d XÚÃ)§ XÚ•Ã). ù‡(؆®• ·K1gñ§¤±3 ù«œ¹eb ·K4Ã)´† ")

Reference David C. Lay. Linear Algebra and Its Applications (3rd edition). Pages 41∼50.

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The High, by Turner

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