MATH 323 Theorems

MATH 323 Theorems

Chapter 1

Theorem 1.3.2

For all $\alpha ,\beta \in \mathbb{ℝ}$ and for all $A,B,C$, the indicated operations are defined:

1. $A+B=B+A$ (commutativity of addition)
2. $(A+B)+C=A+(B+C)$ (associativity of addition)
3. $(AB)C=A(BC)$ (associativity of multiplication)
4. $A(B+C)=AB+AC$ (right distributivity)
5. $(A+B)C=AC+BC$ (left distributivity)
6. $(\alpha \beta )A=\alpha (\beta A)$
7. $\alpha (AB)=(\alpha A)B=A(\alpha B)$
8. $(\alpha +\beta )A=\alpha A+\beta A$
9. $\alpha (A+B)=\alpha A+\alpha B$

Theorem 1.3.3

If $A$ and $B$ are nonsingular $n\times n$ matrices, then $AB$ is also nonsingular and $(AB{)}^{-1}=B^{-1}{A}^{-1}$

(note the opposite order on the right-hand side)

Corollary

For nonsingular matrices $A_1,\dots ,A_k$, $A_1{A}_2\dots A_k$ is also nonsingular and $(A_1{A}_2\dots A_k{)}^{-1}=A_k^{-1}\dots A_2^{-1}{A}_1^{-1}$

Theorem 1.4.2

Equivlent Conditions for Nonsingularity

Let $A$ be a $n\times n$ matrix. Then the following are equivalent:

1. $A$ is nonsingular
2. $A\overrightarrow{x}=\overrightarrow{0}$ has only trivial solution $\overrightarrow{0}$
3. $A$ is row equivalent to identity matrix of size $n$ ($A\sim I_n$)

Proof

(1) → (2): Let $\hat{x}$ be solution of $A\overrightarrow{x}=\overrightarrow{0}$. So $A\hat{x}=\overrightarrow{0}$ can become $A^{-1}A\hat{x}=A^{-1}\overrightarrow{0}$. $I\hat{x}=\overrightarrow{0}$, so $\hat{x}=\overrightarrow{0}$

(2) → (3): Rewrite $A\overrightarrow{x}=\overrightarrow{0}$ in row echelon form as $U\overrightarrow{x}=\overrightarrow{0}$. If one diagonal entry of $U$ is 0, then there is at least one free variable and infinitely many solutions, one of which is nonzero. This leads to a contradiction, so all diagonal entries of $U$ must be 1, so rref will be identity matrix.

(3) → (1): If

$A$

is row equivalent to

$I$

, then there exists a sequence of elementary matrices

$E_1,\dots ,E_k$

such that

$\prod _{i=k}^1{E}_i=A$

. Therefore,

$A^{-1}=\prod _{k=1}^k{E}_i^{-1}$

(note reverse order), so

$A$

is nonsingular.

Q.E.D.

Corollary

The $n\times n$ system of equations $A\overrightarrow{x}=\overrightarrow{b}$ has a unique solution iff $A$ is nonsingular.

Proof

• $A$ is nonsingular, $A\overrightarrow{x}=\overrightarrow{b}$ can be rewritten as $\hat{x}=A^{-1}\overrightarrow{b}$
• Assume unique solution $\hat{x}$ exists. If $A$ were singular then homogeneous system $A\overrightarrow{x}=\overrightarrow{0}$ would have a nonzero solution $\overrightarrow{y}\ne \overrightarrow{0}$ (by Theorem 1.4.2). If this were the case, ... there would be another solution, which is a contradiction.
  Q.E.D.

Chapter 2

Theorem 2.1.1

The determinant can be expressed as a cofactor expansion using any row or column of $A$:

$\left|A\right|=\sum _{k=1}^n{a}_{ik}{A}_{ik}=\sum _{k=1}^n{a}_{kj}{A}_{kj}$
$1\le i,j\le n$

Theorem 2.1.2

If $A$ is a $n\times n$ matrix, then $\left|A^T\right|=\left|A\right|$

Theorem 2.1.3

If $A$ is a triangular matrx, then $\left|A\right|$ is equal to the product of the diagonal entries of $A$.

Proof

Proof by Induction.

Basis step. determinant of $2\times 2$ matrix is product of diagonals.

Inductive step. Assuming determinant of

$n-1\times n-1$

matrix is product of diagonals,

$n\times n$

can be formed by adding a nonzero row to the top, and a zero column with

$a_{1}1$

nonzero to the

$n-1\times n-1matrix$

. The determinant would just be

$a_{1}1A_{11}$

.

Q.E.D.

Theorem 2.1.4

For a $n\times n$ matrix $A$

1. If $A$ has a row or column consisting entirely of 0s, then $\left|A\right|=0$
2. If $A$ has two identical rows or two identical columns, then $\left|A\right|=0$

Theorem x.x.x

A $n\times n$ matrix $A$ is singular iff $|A|=0$.

Proof

$A$ can be reduced to row echelon form by a finite number of row operations. This means that $U=\left(\prod _{i=k}^1{E}_i\right)A$ is an upper triangular matrix whose determinant is $(\prod _{i=k}^1|E_i|)|A|=\prod _{i=1}^n{u}_{ii}$. The determinants of elementary matrices will never be zero, so if the determinant of $A$ is zero, then at least one row of $U$ must be all 0.

Such a matrix cannot be row-equivalent to

$I$

and therefore cannot have an inverse.

Q.E.D.

Theorem x.x.x

$\left|AB\right|=\left|A\right|\left|B\right|$

Proof

If $B$ is singular, it follows from #Theorem 1.5.2 that $AB$ is also singular (see exercise 14 of Section 1.5). Therefore $\left|AB\right|=0$ if $A$ or $B$ is singular.

Let

$B$

be nonsingular, so

$B=\prod _{i=k}^1{E}_i$

Therefore,

$\left|AB\right|=\left|A\prod _{i=k}^1{E}_i\right|$

. By separating the elementary matrices from the determinant, we arrive that

$\left|A\right|\prod _{i=k}^1\left|E_i\right|=\left|A\right|\left|B\right|$

.

Q.E.D.