# Theorems and definitions in linear algebra

This article collects the main theorems and definitions in linear algebra.

## Vector Spaces

Let $V$ be a set on which two operations (vector addition and scalar multiplication) are defined. If the listed axioms are satisfied for every $\vec u$, $\vec v$, and $\vec w$ in $V$ and every scalar (real number) $c$ and $d$, then $V$ is called a vector space:

1. $\vec u + \vec v\text{ is in }V\text{.}$
2. $\vec u + \vec v = \vec v + \vec u$
3. $\vec u + (\vec v + \vec w) = (\vec u + \vec v) + \vec w$
4. $V\text{ has a }\mathbf{zero}\text{ }\mathbf{vector}\text{ }\vec 0\text{ such that for every }\vec u\text{ in }V\text{, }\vec u + \vec 0 = \vec u$
5. $\text{For every }\vec u\text{ in }V\text{, there is a vector in }V\text{ denoted by }-\vec u\text{ such that }\vec u + (-\vec u) = \vec 0\text{.}$

Scalar Multiplication:

1. $c\vec u\text{ is in }V\text{.}$
2. $c(\vec u + \vec v) = c\vec u + c\vec v$
3. $(c + d)\vec u = c\vec u + d\vec u$
4. $c(d\vec u) = (cd)\vec u$
5. $1(\vec u) = \vec u$

### Subspaces

If $W$ is a nonempty subset of a vector space $V$, then $W$ is a subspace of $V$ if and only if the following closure conditions hold:

1. $\text{If }\vec u\text{ and }\vec v\text{ are in }W\text{, then }\vec u + \vec v\text{ is in }W\text{.}$
2. $\text{If }\vec u\text{ is in }W\text{ and }c\text{ is any scalar, then }c\vec u\text{ is in }W\text{.}$

### Linear combinations

A vector $\vec v$ in a vector space $V$ is called a linear combination of the vectors $\vec u_1$, $\vec u_2$, $\dots$ , $\vec u_k$ in $V$ if $\vec v$ can be written in the form $\vec v = c_1\vec u_1 + c_2 \vec u_2 + \dots + c_k\vec u_k$, where $c_1$, $c_2$, $\dots$ , $c_k$ are scalars.

### Systems of linear equations

#### Cramer's Rule

If a system of $n$ linear equations in $n$ variables has a coefficient matrix with a nonzero determinant $\left\vert A\right\vert$, then the solution of the system is given by

$x_1 = \frac{\det(A_1)}{\det(A)}, \qquad x_2 = \frac{\det(A_2)}{\det(A)} , \qquad \dots , \qquad x_n = \frac{\det(A_n)}{\det(A)}$,

────────────────────────────────────────────────────────────────────────────────────────────────────where the $i$th column of $A_i$ is the column of constants in the system of equations.

### Bases

A set of vectors $S = \{\vec v_1$, $\vec v_2$, $\dots$ , $\vec v_n\}$ in a vector space $V$ is called a basis if the following conditions are true:

1. $S$ spans $V$.
2. $S$ is linearly independent.

## Linear transformations and matrices

P.S. coefficient of the differential equation, differentiability of complex function,vector space of functionsdifferential operator, auxiliary polynomial, to the power of a complex number, exponential function.

### Definition of a Linear Transformation

Let $V$ and $W$ be vector spaces. The function $T:V\to W$ is called a linear transformation of $V$ into $W$ if the following two properties are true for all $\vec u$ and $\vec v$ in $V$ and for any scalar $c$.

1. $T(\vec u + \vec v) = T(\vec u) + T(\vec v)$
2. $T(c\vec u) = cT(\vec u)$

### ${\color{Blue}~2.1}$ N(T) and R(T) are subspaces

Let V and W be vector spaces and I: VW be linear. Then N(T) and R(T) are subspaces of V and W, respectively.

### ${\color{Blue}~2.2}$ R(T)= span of T(basis in V)

Let V and W be vector spaces, and let T: V→W be linear. If $\beta={v_1,v_2,\ldots,v_n}$ is a basis for V, then

$\mathrm{R(T)}=\mathrm{span}(T(\beta\mathrm{))}=\mathrm{span}({T(v_1),T(v_2),\ldots,T(v_n)})$.

### ${\color{Blue}~2.3}$ Dimension theorem

Let V and W be vector spaces, and let T: V → W be linear. If V is finite-dimensional, then

$\mathrm{nullity}(T)+\mathrm{rank}(T)=\dim(V).$

### ${\color{Blue}~2.4}$ one-to-one ⇔ N(T) = {0}

Let $T:V\to W$ be a linear transformation. Then $T$ is one-to-one if and only if $\operatorname{ker}(T) = \{\vec 0\}$.

### ${\color{Blue}~2.5}$ one-to-one ⇔ onto ⇔ rank(T) = dim(V)

Let V and W be vector spaces of equal (finite) dimension, and let T:VW be linear. Then the following are equivalent.

(a) T is one-to-one.
(b) T is onto.
(c) rank(T) = dim(V).

### ${\color{Blue}~2.6}$ ∀ ${w_1,w_2,\ldots,w_n}=$ exactly one T (basis),

Let V and W be vector space over F, and suppose that ${v_1, v_2,\ldots,v_n}$ is a basis for V. For $w_1, w_2,\ldots,w_n$ in W, there exists exactly one linear transformation T: V→W such that $\mathrm{T}(v_i)=w_i$ for $i=1,2,\ldots,n.$
Corollary. Let V and W be vector spaces, and suppose that V has a finite basis ${v_1,v_2,\ldots,v_n}$. If U, T: V→W are linear and $U(v_i)=T(v_i)$ for $i=1,2,\ldots,n,$ then U=T.

### ${\color{Blue}~2.7}$ T is vector space

Let V and W be vector spaces over a field F, and let T, U: V→W be linear.

(a) For all $a$F, $a\mathrm{T}+\mathrm{U}$ is linear.
(b) Using the operations of addition and scalar multiplication in the preceding definition, the collection of all linear transformations form V to W is a vector space over F.

### ${\color{Blue}~2.8}$ linearity of matrix representation of linear transformation

Let V and W be finite-dimensional vector spaces with ordered bases β and γ, respectively, and let T, U: V→W be linear transformations. Then

(a)$[T+U]_\beta^\gamma=[T]_\beta^\gamma+[U]_\beta^\gamma$ and
(b)$[aT]_\beta^\gamma=a[T]_\beta^\gamma$ for all scalars $a$.

### ${\color{Blue}~2.9}$ composition law of linear operators

Let V,W, and Z be vector spaces over the same field f, and let T:V→W and U:W→Z be linear. then UT:V→Z is linear.

### ${\color{Blue}~2.10}$ law of linear operator

Let v be a vector space. Let T, U1, U2$\mathcal{L}$(V). Then
(a) T(U1+U2)=TU1+TU2 and (U1+U2)T=U1T+U2T
(b) T(U1U2)=(TU1)U2
(c) TI=IT=T
(d) $a$(U1U2)=($a$U1)U2=U1($a$U2) for all scalars $a$.

### ${\color{Blue}~2.11}$ [UT]αγ=[U]βγ[T]αβ

Let V, W and Z be finite-dimensional vector spaces with ordered bases α β γ, respectively. Let T: V⇐W and U: W→Z be linear transformations. Then

$[UT]_\alpha^\gamma=[U]_\beta^\gamma[T]_\alpha^\beta$.

Corollary. Let V be a finite-dimensional vector space with an ordered basis β. Let T,U∈$\mathcal{L}$(V). Then [UT]β=[U]β[T]β.

### ${\color{Blue}~2.12}$ law of matrix

Let A be an m×n matrix, B and C be n×p matrices, and D and E be q×m matrices. Then

(a) A(B+C)=AB+AC and (D+E)A=DA+EA.
(b) $a$(AB)=($a$A)B=A($a$B) for any scalar $a$.
(c) ImA=AIm.
(d) If V is an n-dimensional vector space with an ordered basis β, then [Iv]β=In.

Corollary. Let A be an m×n matrix, B1,B2,...,Bk be n×p matrices, C1,C1,...,C1 be q×m matrices, and $a_1,a_2,\ldots,a_k$ be scalars. Then

$A\Bigg(\sum_{i=1}^k a_iB_i\Bigg)=\sum_{i=1}^k a_iAB_i$

and

$\Bigg(\sum_{i=1}^k a_iC_i\Bigg)A=\sum_{i=1}^k a_iC_iA$.

### ${\color{Blue}~2.13}$ law of column multiplication

Let A be an m×n matrix and B be an n×p matrix. For each $j (1\le j\le p)$ let $u_j$ and $v_j$ denote the jth columns of AB and B, respectively. Then
(a) $u_j=Av_j$
(b) $v_j=Be_j$, where $e_j$ is the jth standard vector of Fp.

### ${\color{Blue}~2.14}$ [T(u)]γ=[T]βγ[u]β

Let V and W be finite-dimensional vector spaces having ordered bases β and γ, respectively, and let T: V→W be linear. Then, for each u ∈ V, we have

$[T(u)]_\gamma=[T]_\beta^\gamma[u]_\beta$.

### ${\color{Blue}~2.15}$ laws of LA

Let A be an m×n matrix with entries from F. Then the left-multiplication transformation LA: Fn→Fm is linear. Furthermore, if B is any other m×n matrix (with entries from F) and β and γ are the standard ordered bases for Fn and Fm, respectively, then we have the following properties.
(a) $[L_A]_\beta^\gamma=A$.
(b) LA=LB if and only if A=B.
(c) LA+B=LA+LB and L$a$A=$a$LA for all $a$∈F.
(d) If T:Fn→Fm is linear, then there exists a unique m×n matrix C such that T=LC. In fact, $\mathrm{C}=[L_A]_\beta^\gamma$.
(e) If W is an n×p matrix, then LAE=LALE.
(f ) If m=n, then $L_{I_n}=I_{F^n}$.

### ${\color{Blue}~2.16}$ A(BC)=(AB)C

Let A,B, and C be matrices such that A(BC) is defined. Then A(BC)=(AB)C; that is, matrix multiplication is associative.

### ${\color{Blue}~2.17}$ T−1is linear

Let V and W be vector spaces, and let T:V→W be linear and invertible. Then T−1: W →V is linear.

### ${\color{Blue}~2.18}$ [T−1]γβ=([T]βγ)−1

Let V and W be finite-dimensional vector spaces with ordered bases β and γ, respectively. Let T:V→W be linear. Then T is invertible if and only if $[T]_\beta^\gamma$ is invertible. Furthermore, $[T^{-1}]_\gamma^\beta=([T]_\beta^\gamma)^{-1}$

Lemma. Let T be an invertible linear transformation from V to W. Then V is finite-dimensional if and only if W is finite-dimensional. In this case, dim(V)=dim(W).

Corollary 1. Let V be a finite-dimensional vector space with an ordered basis β, and let T:V→V be linear. Then T is invertible if and only if [T]β is invertible. Furthermore, [T−1]β=([T]β)−1.

Corollary 2. Let A be an n×n matrix. Then A is invertible if and only if LA is invertible. Furthermore, (LA)−1=LA−1.

### ${\color{Blue}~2.19}$ V is isomorphic to W ⇔ dim(V)=dim(W)

Let W and W be finite-dimensional vector spaces (over the same field). Then V is isomorphic to W if and only if dim(V)=dim(W).

Corollary. Let V be a vector space over F. Then V is isomorphic to Fn if and only if dim(V)=n.

### ${\color{Blue}~2.20}$ ??

Let V and W be finite-dimensional vector spaces over F of dimensions n and m, respectively, and let β and γ be ordered bases for V and W, respectively. Then the function $~\Phi$: $\mathcal{L}$(V,W)→Mm×n(F), defined by $~\Phi(T)=[T]_\beta^\gamma$ for T∈$\mathcal{L}$(V,W), is an isomorphism.

Corollary. Let V and W be finite-dimensional vector spaces of dimension n and m, respectively. Then $\mathcal{L}$(V,W) is finite-dimensional of dimension mn.

### ${\color{Blue}~2.21}$ Φβ is an isomorphism

For any finite-dimensional vector space V with ordered basis β, Φβ is an isomorphism.

### ${\color{Blue}~2.22}$ ??

Let β and β' be two ordered bases for a finite-dimensional vector space V, and let $Q=[I_V]_{\beta'}^\beta$. Then
(a) $Q$ is invertible.
(b) For any $v\in$ V, $~[v]_\beta=Q[v]_{\beta'}$.

### ${\color{Blue}~2.23}$ [T]β'=Q−1[T]βQ

Let T be a linear operator on a finite-dimensional vector space V,and let β and β' be two ordered bases for V. Suppose that Q is the change of coordinate matrix that changes β'-coordinates into β-coordinates. Then

$~[T]_{\beta'}=Q^{-1}[T]_\beta Q$.

Corollary. Let A∈Mn×n(F), and le t γ be an ordered basis for Fn. Then [LA]γ=Q−1AQ, where Q is the n×n matrix whose jth column is the jth vector of γ.

### Principal Axes Theorem

For a conic whose equation is $ax^2 + bxy + cy^2 + dx + ey + f = 0$, the rotation given by $X=PX'$ eliminates the $xy$-term if $P$ is an orthogonal matrix, with $\left\vert P\right\vert = 1$, that diagonalizes $A$. That is,

$P^TAP = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}$,

────────────────────────────────────────────────────────────────────────────────────────────────────where $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$. The equation of the rotated conic is given by

$\lambda_1(x')^2 + \lambda_2(y')^2 + \begin{bmatrix} d & e\\ \end{bmatrix}PX' + f = 0$.

### ${\color{Blue}~2.27}$ p(D)(x)=0 (p(D)∈C∞)⇒ x(k)exists (k∈N)

Any solution to a homogeneous linear differential equation with constant coefficients has derivatives of all orders; that is, if $x$ is a solution to such an equation, then $x^{(k)}$ exists for every positive integer k.

### ${\color{Blue}~2.28}$ {solutions}= N(p(D))

The set of all solutions to a homogeneous linear differential equation with constant coefficients coincides with the null space of p(D), where p(t) is the auxiliary polynomial with the equation.

Corollary. The set of all solutions to s homogeneous linear differential equation with constant coefficients is a subspace of $\mathrm{C}^\infty$.

### ${\color{Blue}~2.29}$ derivative of exponential function

For any exponential function $f(t)=e^{ct}, f'(t)=ce^{ct}$.

### ${\color{Blue}~2.30}$ {e−at} is a basis of N(p(D+aI))

The solution space for the differential equation,

$y'+a_0y=0$

is of dimension 1 and has $\{e^{-a_0t}\}$as a basis.

Corollary. For any complex number c, the null space of the differential operator D-cI has {$e^{ct}$} as a basis.

### ${\color{Blue}~2.31}$ $e^{ct}$ is a solution

Let p(t) be the auxiliary polynomial for a homogeneous linear differential equation with constant coefficients. For any complex number c, if c is a zero of p(t), then to the differential equation.

### ${\color{Blue}~2.32}$ dim(N(p(D)))=n

For any differential operator p(D) of order n, the null space of p(D) is an n_dimensional subspace of C.

Lemma 1. The differential operator D-cI: C to C is onto for any complex number c.

Lemma 2 Let V be a vector space, and suppose that T and U are linear operators on V such that U is onto and the null spaces of T and U are finite-dimensional, Then the null space of TU is finite-dimensional, and

dim(N(TU))=dim(N(U))+dim(N(U)).

Corollary. The solution space of any nth-order homogeneous linear differential equation with constant coefficients is an n-dimensional subspace of C.

### ${\color{Blue}~2.33}$ ecit is linearly independent with each other (ci are distinct)

Given n distinct complex numbers $c_1, c_2,\ldots,c_n$, the set of exponential functions $\{e^{c_1t},e^{c_2t},\ldots,e^{c_nt}\}$ is linearly independent.

Corollary. For any nth-order homogeneous linear differential equation with constant coefficients, if the auxiliary polynomial has n distinct zeros $c_1, c_2, \ldots, c_n$, then $\{e^{c_1t},e^{c_2t},\ldots,e^{c_nt}\}$ is a basis for the solution space of the differential equation.

Lemma. For a given complex number c and positive integer n, suppose that (t-c)^n is athe auxiliary polynomial of a homogeneous linear differential equation with constant coefficients. Then the set

$\beta=\{e^{c_1t},e^{c_2t},\ldots,e^{c_nt}\}$

is a basis for the solution space of the equation.

### ${\color{Blue}~2.34}$ general solution of homogeneous linear differential equation

Given a homogeneous linear differential equation with constant coefficients and auxiliary polynomial

$(t-c_1)^n_1(t-c_2)^n_2\cdots(t-c_k)^n_k,$

where $n_1, n_2,\ldots,n_k$ are positive integers and $c_1, c_2, \ldots, c_n$ are distinct complex numbers, the following set is a basis for the solution space of the equation:

$\{e^{c_1t}, te^{c_1t},\ldots,t^{n_1-1}e^{c_1t},\ldots,e{c_kt},te^{c_kt},\ldots,t^{n_k-1}e^{c_kt}\}$.

### Definition of an Orthogonal Matrix

A square matrix $P$ is called orthogonal if it is invertible and if

$P^{-1} = P^T$.

### Real Spectral Theorem

If $A$ is an $n\times n$ symmetric matrix, then the following properties are true:

1. $A$ is diagonalizable.
2. All eigenvalues of $A$ are real.
3. If $\lambda$ is an eigenvalue of $A$ with multiplicity $k$, then $\lambda$ has $k$ linearly independent eigenvectors. That is, the eigenspace of $\lambda$ has dimension $k$.

Also, the set of eigenvalues of $A$ is called the spectrum of $A$.

## Elementary matrix operations and systems of linear equations

### Elementary matrix operations

The three elementary row operations are the following:

1. Interchange two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of a row to another row.

### Elementary matrix

An $n \times n$ matrix is called an elementary matrix if it can be obtained from the identity matrix $I_n$ by a single elementary row operation.

### Rank of a matrix

The rank of a matrix A is the number of pivot columns after the reduced row echelon form of A.

### Invertible Matrices

$\text{If }A\text{ is }n\text{ × }n\text{, then the following statements are equivalent:}$

1. $A \text{ is invertible.}$
2. $A\vec x = \vec b\text{ has a unique solution for every }n \times 1\text{ column matrix }\vec b\text{.}$
3. $A\vec x = \vec 0\text{ has only the trivial solution.}$
4. $A\text{ is row-equivalent to }I_{n}\text{.}$
5. $A\text{ can be written as the product of elementary matrices.}$
6. $\det(A) \ne 0$
7. $\operatorname{rk}(A) = n\text{ number of columns.}$
8. $\operatorname{nul}(A) = 0$
9. $\text{All of the }n\text{-row vectors of }A\text{ are linearly independent.}$
10. $\text{All of the }n\text{-column vectors of }A\text{ are linearly independent.}$

## Determinants

If

$A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$

is a 2×2 matrix with entries form a field F, then we define the determinant of A, denoted det(A) or |A|, to be the scalar $ad-bc$.

＊Theorem 1: linear function for a single row.
＊Theorem 2: nonzero determinant ⇔ invertible matrix

Theorem 1: The function det: M2×2(F) → F is a linear function of each row of a 2×2 matrix when the other row is held fixed. That is, if $u,v,$ and $w$ are in F² and $k$ is a scalar, then

$\det\begin{pmatrix} u + kv\\ w\\ \end{pmatrix} =\det\begin{pmatrix} u\\ w\\ \end{pmatrix} + k\det\begin{pmatrix} v\\ w\\ \end{pmatrix}$

and

$\det\begin{pmatrix} w\\ u + kv\\ \end{pmatrix} =\det\begin{pmatrix} w\\ u\\ \end{pmatrix} + k\det\begin{pmatrix} w\\ v\\ \end{pmatrix}$

Theorem 2: Let A $\in$ M2×2(F). Then thee deter minant of A is nonzero if and only if A is invertible. Moreover, if A is invertible, then

$A^{-1}=\frac{1}{\det(A)}\begin{pmatrix} A_{22}&-A_{12}\\ -A_{21}&A_{11}\\ \end{pmatrix}$

## Diagonalization

Characteristic polynomial of a linear operator/matrix

### ${\color{Blue}~5.1}$ diagonalizable⇔basis of eigenvector

A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis β for V consisting of eigenvectors of T. Furthermore, if T is diagonalizable, $\beta= {v_1,v_2,\ldots,v_n}$ is an ordered basis of eigenvectors of T, and D = [T]β then D is a diagonal matrix and $D_{jj}$ is the eigenvalue corresponding to $v_j$ for $1\le j \le n$.

### ${\color{Blue}~5.2}$ eigenvalue⇔det(A-λIn)=0

Let A∈Mn×n(F). Then a scalar λ is an eigenvalue of A if and only if det(AIn)=0

### ${\color{Blue}~5.3}$ characteristic polynomial

Let A∈Mn×n(F).
(a) The characteristic polynomial of A is a polynomial of degree n with leading coefficient(-1)n.
(b) A has at most n distinct eigenvalues.

### ${\color{Blue}~5.4}$ υ to λ⇔υ∈N(T-λI)

Let T be a linear operator on a vector space V, and let λ be an eigenvalue of T.
A vector υ∈V is an eigenvector of T corresponding to λ if and only if υ≠0 and υ∈N(T-λI).

### ${\color{Blue}~5.5}$ vi to λi⇔vi is linearly independent

Let T be a linear operator on a vector space V, and let $\lambda_1,\lambda_2,\ldots,\lambda_k,$ be distinct eigenvalues of T. If $v_1,v_2,\ldots,v_k$ are eigenvectors of t such that $\lambda_i$ corresponds to $v_i$ ($1\le i\le k$), then {$v_1,v_2,\ldots,v_k$} is linearly independent.

### ${\color{Blue}~5.6}$ characteristic polynomial splits

The characteristic polynomial of any diagonalizable linear operator splits.

### ${\color{Blue}~5.7}$ 1 ≤ dim(Eλ) ≤ m

Let T be alinear operator on a finite-dimensional vectorspace V, and let λ be an eigenvalue of T having multiplicity $m$. Then $1 \le\dim(E_{\lambda})\le m$.

### ${\color{Blue}~5.8}$ S = S1 ∪ S2 ∪ ...∪ Sk is linearly independent

Let T be a linear operator on a vector space V, and let $\lambda_1,\lambda_2,\ldots,\lambda_k,$ be distinct eigenvalues of T. For each $i=1,2,\ldots,k,$ let $S_i$ be a finite linearly independent subset of the eigenspace $E_{\lambda_i}$. Then $S=S_1\cup S_2 \cup\cdots\cup S_k$ is a linearly independent subset of V.

### ${\color{Blue}~5.9}$ ⇔T is diagonalizable

Let T be a linear operator on a finite-dimensional vector space V that the characteristic polynomial of T splits. Let $\lambda_1,\lambda_2,\ldots,\lambda_k$ be the distinct eigenvalues of T. Then
(a) T is diagonalizable if and only if the multiplicity of $\lambda_i$ is equal to $\dim(E_{\lambda_i})$ for all $i$.
(b) If T is diagonalizable and $\beta_i$ is an ordered basis for $E_{\lambda_i}$ for each $i$, then $\beta=\beta_1\cup \beta_2\cup \cup\beta_k$ is an ordered $basis^2$ for V consisting of eigenvectors of T.

Test for diagonlization

## Inner product spaces

### ${\color{Blue}~6.1}$ properties of linear product

Let V be an inner product space. Then for x,y,z\in V and c \in f, the following staements are true.
(a) $\langle x,y+z\rangle=\langle x,y\rangle+\langle x,z\rangle.$
(b) $\langle x,cy\rangle=\bar{c}\langle x,y\rangle.$
(c) $\langle x,\mathit{0}\rangle=\langle\mathit{0},x\rangle=0.$
(d) $\langle x,x\rangle=0$ if and only if $x=\mathit{0}.$
(e) If$\langle x,y\rangle=\langle x,z\rangle$ for all $x\in$ V, then $y=z$.

### ${\color{Blue}~6.2}$ law of norm

Let V be an inner product space over F. Then for all x,y\in V and c\in F, the following statements are true.
(a) $\|cx\|=|c|\cdot\|x\|$.
(b) $\|x\|=0$ if and only if $x=0$. In any case, $\|x\|\ge0$.
(c)(Cauchy-Schwarz In equality)$|\langle x,y\rangle|\le\|x\|\cdot\|y\|$.
(d)(Triangle Inequality)$\|x+y\|\le\|x\|+\|y\|$.

### ${\color{Blue}~6.3}$ span of orthogonal subset

Let V be an inner product space and $S=\{v_1,v_2,\ldots,v_k\}$ be an orthogonal subset of V consisting of nonzero vectors. If $y$∈span(S), then

$y=\sum_{i=1}^n{\langle y,v_i \rangle \over \|v_i\|^2}v_i$

### ${\color{Blue}~6.4}$ Gram-Schmidt process

Let V be an inner product space and S=$\{w_1,w_2,\ldots,w_n\}$ be a linearly independent subset of V. DefineS'=$\{v_1,v_2,\ldots,v_n\}$, where $v_1=w_1$ and

$v_k=w_k-\sum_{j=1}^{k-1}{\langle w_k, v_j\rangle\over\|v_j\|^2}v_j$

Then S' is an orhtogonal set of nonzero vectors such that span(S')=span(S).

### ${\color{Blue}~6.5}$ orthonormal basis

Let V be a nonzero finite-dimensional inner product space. Then V has an orthonormal basis β. Furthermore, if β =$\{v_1,v_2,\ldots,v_n\}$ and x∈V, then

$x=\sum_{i=1}^n\langle x,v_i\rangle v_i$.

Corollary. Let V be a finite-dimensional inner product space with an orthonormal basis β =$\{v_1,v_2,\ldots,v_n\}$. Let T be a linear operator on V, and let A=[T]β. Then for any $i$ and $j$, $A_{ij}=\langle T(v_j), v_i\rangle$.

### ${\color{Blue}~6.6}$ W⊥ by orthonormal basis

Let W be a finite-dimensional subspace of an inner product space V, and let $y$∈V. Then there exist unique vectors $u$∈W and $z$∈W such that $y=u+z$. Furthermore, if $\{v_1,v_2,\ldots,v_k\}$ is an orthornormal basis for W, then

$u=\sum_{i=1}^k\langle y,v_i\rangle v_i$.

S=\{v_1,v_2,\ldots,v_k\} Corollary. In the notation of Theorem 6.6, the vector $u$ is the unique vector in W that is "closest" to $y$; thet is, for any $x$∈W, $\|y-x\|\ge\|y-u\|$, and this inequality is an equality if and onlly if $x=u$.

### ${\color{Blue}~6.7}$ properties of orthonormal set

Suppose that $S=\{v_1,v_2,\ldots,v_k\}$ is an orthonormal set in an $n$-dimensional inner product space V. Than
(a) S can be extended to an orthonormal basis $\{v_1, v_2, \ldots,v_k,v_{k+1},\ldots,v_n\}$ for V.
(b) If W=span(S), then $S_1=\{v_{k+1},v_{k+2},\ldots,v_n\}$ is an orhtonormal basis for W(using the preceding notation).
(c) If W is any subspace of V, then dim(V)=dim(W)+dim(W).

### ${\color{Blue}~6.8}$ linear functional representation inner product

Let V be a finite-dimensional inner product space over F, and let $g$:V→F be a linear transformation. Then there exists a unique vector $y$∈ V such that $\rm{g}(x)=\langle x, y\rangle$ for all $x$∈ V.

### ${\color{Blue}~6.9}$ definition of T*

Let V be a finite-dimensional inner product space, and let T be a linear operator on V. Then there exists a unique function T*:V→V such that $\langle\rm{T}(x),y\rangle=\langle x, \rm{T}^*(y)\rangle$ for all $x,y$ ∈ V. Furthermore, T* is linear

### ${\color{Blue}~6.10}$ [T*]β=[T]*β

Let V be a finite-dimensional inner product space, and let β be an orthonormal basis for V. If T is a linear operator on V, then

$[T^*]_\beta=[T]^*_\beta$.

### ${\color{Blue}~6.11}$ properties of T*

Let V be an inner product space, and let T and U be linear operators onV. Then
(a) (T+U)*=T*+U*;
(b) ($c$T)*=$\bar c$ T* for any c∈ F;
(c) (TU)*=U*T*;
(d) T**=T;
(e) I*=I.

Corollary. Let A and B be n×nmatrices. Then
(a) (A+B)*=A*+B*;
(b) ($c$A)*=$\bar c$ A* for any $c$∈ F;
(c) (AB)*=B*A*;
(d) A**=A;
(e) I*=I.

### ${\color{Blue}~6.12}$ Least squares approximation

Let A ∈ Mm×n(F) and $y$∈Fm. Then there exists $x_0$ ∈ Fn such that $(A*A)x_0=A*y$ and $\|Ax_0-Y\|\le\|Ax-y\|$ for all x∈ Fn

Lemma 1. let A ∈ Mm×n(F), $x$∈Fn, and $y$∈Fm. Then

$\langle Ax, y\rangle _m =\langle x, A*y\rangle _n$

Lemma 2. Let A ∈ Mm×n(F). Then rank(A*A)=rank(A).

Corollary.(of lemma 2) If A is an m×n matrix such that rank(A)=n, then A*A is invertible.

### ${\color{Blue}~6.13}$ Minimal solutions to systems of linear equations

Let A ∈ Mm×n(F) and b∈ Fm. Suppose that $Ax=b$ is consistent. Then the following statements are true.
(a) There existes exactly one minimal solution $s$ of $Ax=b$, and $s$∈R(LA*).
(b) The vector $s$ is the only solution to $Ax=b$ that lies in R(LA*); that is, if $u$ satisfies $(AA*)u=b$, then $s=A*u$.