Famous Theorems of Mathematics/Algebra/Matrix Theory

An m&times;n matrix M is a function $$M:A\rightarrow F$$ where A = {1,2...m} &times; {1,2...n} and F is the field under consideration.

An m&times;n matrix (read as m by n matrix), is usually written as:


 * $$A=\left(\begin{matrix}

a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \end{matrix}\right)$$

For other related definitions please see this link.

Basic Proofs
1. The set of all m×n matrices forms an abelian group under matrix addition.


 * Proof: Clearly the sum of two m×n matrices is another m×n matrix. If A and B are two matrices of equal order then working with their (i,j)th entries we have $$(A+B)_{i,j}=(A_{i,j})+(B_{i,j})=(B_{i,j})+(A_{i,j})=(B+A)_{i,j}$$ which proves A+B = B+A i.e. commutativity. For associativity we proceed similarly so that A + (B + C) = (A + B) + C. Also the m×n matrix with all entries zero is the additive identity. For every matrix A, the matrix -A whose (i,j)th entry is $$-A_{i,j}$$ is the inverse. So matrices of same order form an abelian group under addition.

2. Scalar Multiplication has the following properties:
 * 1. Left distributivity: (α+β)A = αA+βA.
 * 2. Right distributivity: α(A+B) = αA+αB.
 * 3. Associativity: (αβ)A=α(βA)).
 * 4. 1A = A.
 * 5. 0A= 0.
 * 6. (-1)A = -A.


 * (0,1,-1,α & β are scalars; A & B are matrices of equal order, 0 is the zero matrix.)


 * Proof: Start with the left hand side of (1). We will work with the (i,j)th entries. Clearly $$((\alpha+\beta)A)_{i,j}=(\alpha+\beta)\cdot A_{i,j}=(\alpha A)_{i,j}+(\beta A)_{i,j}$$ and so (1) is proved. Similarly (2) can be proved. Associativity follows as $$((\alpha \beta)A)_{i,j}=(\alpha \beta)\cdot A_{i,j}=\alpha(\beta A_{i,j})$$. (4), (5) and (6) follow directly from the definition.

3. Matrix multiplication has the following properties:
 * 1. Associativity: A(BC) = (AB)C.
 * 2. Left distributivity: A(B+C) = AB+AC.
 * 3. Right distributivity: (A+B)C = AC+BC.
 * 4. IA = A = AI.
 * 5. α(BC) = (αB)C = B(αC).


 * (α is a scalar; A, B & C are matrices, I is the identity matrix. A,B,C & I are of orders m&times;n, n&times;p, p&times;r & m&times;m respectively.)


 * Proof: We work with the (i,j)th entries and prove (1) only. The proofs for the rest are similar. Now $$(A(BC))_{i,j}=\sum_{k=1}^n A_{i,k}(BC)_{k,j}=\sum_{k=1}^n A_{i,k}\Big( \sum_{l=1}^p B_{k,l}C_{l,j}\Big)=\sum_{k=1}^n \sum_{l=1}^p A_{i,k}B_{k,l}C_{l,j}$$ and also $$((AB)C)_{i,j}=\sum_{l=1}^p (AB)_{i,l}C_{l,j}=\sum_{l=1}^p \Big( \sum_{k=1}^n A_{i,k}B_{k,l}\Big) C_{l,j}=\sum_{k=1}^n \sum_{l=1}^p A_{i,k}B_{k,l}C_{l,j}$$ so that (i,j)th entries on the two sides are equal.

4. ''Let A and B be m&times;n matrices. Then:''
 * (i) $$(kA)^T$$ = $$kA^T$$
 * (ii) $$(A+B)^T = A^T + B^T$$
 * (iii) $$(AB)^T = B^TA^T$$


 * Sketch of Proof: Work with the (i,j) entries as in the previous proofs.

5. Any system of linear equations has either no solution, exactly one solution or infinitely many solutions.


 * Proof: Suppose a linear system Ax = b has two different solutions given by X and Y. Then let Z = X - Y. Clearly Z is non zero and A(X + kZ) = AX + kAZ = b + k(AX - AY) = b + k(b - b) = b so that X + kZ is a solution to the system for every possible value of k. Since k can assume infinitely many values so clearly we have an infinite number of solutions.

6. Any triangular matrix A satisfying $$AA^T = A^TA$$ is a diagonal matrix.


 * Proof: Suppose A is lower triangular. Now the (i,i)th entry of $$AA^T$$ is given by $$\sum_{k=1}^n (A_{i,k})(A^T_{k,i})= \sum_{k=1}^n (A_{i,k})(A_{i,k}) = \sum_{k=1}^n (A_{i,k}^2) = \sum_{k=1}^i (A_{i,k}^2)$$. Also the (i,i)th entry of $$A^TA$$ is given by $$\sum_{k=1}^n (A^T_{i,k})(A_{k,i}) = \sum_{k=1}^n (A_{k,i})(A_{k,i}) = \sum_{k=1}^n (A_{k,i}^2) = \sum_{k=i}^n (A_{k,i}^2)$$. Now as $$AA^T = A^TA$$ so $$\sum_{k=1}^i (A_{i,k}^2) = \sum_{k=i}^n (A_{k,i}^2)$$ and as $$A_{i,i}^2$$ can be subtracted from the two sides we are left with $$\sum_{k=1}^{i-1} (A_{i,k}^2) = \sum_{k=i+1}^n (A_{k,i}^2)$$.

Now if i = 1 then we have $$0 = A_{2,1}^2 + A_{3,1}^2 \cdots A_{n,1}^2 $$ which gives us $$A_{2,1}=A_{3,1}\cdots A_{n,1}=0$$. Similarly for i =2 we have $$ 0 = A_{2,1}^2 = A_{3,2}^2 + A_{4,2}^2 \cdots A_{n,2}^2 $$ so that $$A_{3,2}=A_{4,2}\cdots A_{n,2}=0$$. It is now clear that in this fashion all non diagonal entries of A can be shown to be zero. The proof for an upper triangular matrix is similar.