Linear Algebra Lecture 9

Linear Algebra Lecture 9

1. Linear independence

2. Spanning a space

3. Basis and Dimension

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Independence

Vectors $X_1,X_2,...X_n$ are independent if no combination gives zero vector, except the zero combination.
$c_1x_1+c_2x_2+...+c_nx_n \ne 0$, except all $c_i=0$.

Suppose $A$ is $m$ by $n$ with $m \lt n$. More unknown $X$s than equations.
The conclusion is there are some non-zero solutions to $Ax=0$, there are some special solutions.
Because starts with a system and does elimination, gets the thing into an echelon form with some pivot columns and some free columns, the reason is there will be free variables, at least $n-m$.

When $v_1,...v_n$ are columns of matrix $A$. They are independent if the null space of $A$ is only the zero vector. They are dependent if $Ac=0$ for some non-zero $c$.

Vectors are independent when rank = n and no free variables, the null space is only zero vector, dependent when rank < n and has free variables.

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Spanning a space

Vectors $v_1,...,v_n$ span a space means the space consists of all combinations of those vectors.
The columns of a matrix span the column space.

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Basis

A basis for a vector space is a sequence of vectors $v_1,v_2,...v_n$ with two properties:
1. They are independent.
2. They span the space.

Example:
Space is $R^3$, one basis is $\begin{bmatrix}1\\0\\0 \end{bmatrix},\begin{bmatrix}0\\1\\0 \end{bmatrix},\begin{bmatrix}0\\0\\1 \end{bmatrix}$

How to test vectors to be a basis?
You would put them in the columns of a matrix, and do elimination, row reduction, and you would see do you get any free variables or are all the columns pivot columns.

In $R^n$, n vectors give basis if the $n \times n$ matrix with those columns is invertible.

There are many bases, but they all have the same number of vectors.If we’re talking about the space $R^n$, then the number of vectors is $n$.

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Dimension

Given a space, then every basis for the space has the same number of vectors. This number is the dimension of the space.

Rank(A) = number of pivot columns = the dimension of the column space C(A)

If you know the dimension of the space, then we have a couple of vectors that are independent, they’ll automatically be a basis.

The dimension of the null space is the number of free variables.

$dimC(A)=r$
$dimN(A) =n-r$