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\lecture{5}{Linear algebra : column space of matrix A, solution space of Ax=0, relationship between them}{Ayush Choure}


\textbf{Last Lecture} : Given vectors $v_1$,$v_2$,$v_3$,\ldots,$v_n$ the space spanned by them is $\sum^{n}_{i=1}\alpha_i v_i$. Thus $\sum^{n}_{i=1}\alpha_i v_i$ is a subspace.

Consider the space of vectors $\lbrace x : Ax = \overline 0 \rbrace$.

Looking by the column perspective
\begin{center}
	$ Ax = b \Rightarrow A^{1}x_{1} + A^{2}x_{2} + \ldots + A^{n}x_{n} = b $
\end{center}
 
here,
	$A^{i}$ is the $i^{th}$ column of $A$ and
	$x_{i}$ is the $i^{th}$ component of vector $x$

which means $b$ is in the column space of $A$. Hence, in the given situation 
\begin{center}
$A^{1}x_{1} + A^{2}x_{2} + \ldots + A^{n}x_{n} = \overline 0$
\end{center}

\textbf{Assume} : $A^{1}, A^{2}, \ldots , A^{k}$ are a basis for the space spanned by $A^{1}, A^{2}, \ldots , A^{n}$, $k \leq n$.

which implies,
\begin{center}
$A^{k+1} = \sum^{k}_{j=1}\alpha^{k+1}_{j}A^{j}$\\
$A^{k+2} = \sum^{k}_{j=1}\alpha^{k+2}_{j}A^{j}$\\
\ldots

$A^{n} = \sum^{k}_{j=1}\alpha^{n}_{j}A^{j}$\\
\end{center}

hence, dimension of this space is atleast $n-k$
\begin{center}
$dim \lbrace x: Ax = \overline 0 \rbrace \geq n-k$
\end{center}

now, to prove that it is EXACTLY $n-k$.

\text{Proof}: Take an $x_{0}$ such that $Ax_{0} = \overline 0$. Also let $U_{i}$ be such that,


$U_{k+1}$ is 
\begin{displaymath}
\begin{matrix}
\begin{pmatrix}
\alpha^{k+1}_{1}\\
\alpha^{k+2}_{2}\\
\vdots\\
\alpha^{k+1}_{k}\\
1\\
0\\
\vdots\\
0
\notag
\end{pmatrix}
\end{matrix}
\end{displaymath}


$U_{k+2}$ is 
\begin{displaymath}
\begin{matrix}
\begin{pmatrix}
\alpha^{k+2}_{1}\\
\alpha^{k+2}_{2}\\
\vdots\\
\alpha^{k+2}_{k}\\
1\\
0\\
\vdots\\
0
\notag
\end{pmatrix}
\end{matrix}
\end{displaymath}

Now consider the vectors $x$ and $x\prime$ such that,
\begin{center}
$x = \lbrack x_{1} \ldots x_{n} \rbrack ^{T}$ 

$x\prime = x - \lbrace x_{k+1}U_{k+1} + \ldots + x_{n}U_{n} \rbrace$
\end{center}

but
\begin{center}
$Ax = \overline 0$ \qquad and

$A\lbrace x_{k+1}U_{k+1} + \ldots + x_{n}U_{n} \rbrace = \overline 0$
\end{center}

because given that, $x$ and $U_{i}$ are from the null space of $A$. Therefore

\begin{center}
$Ax\prime = \overline 0$
\end{center}

note that because of way in which $x\prime$ is defined, its last $n-k$ components are zero. Hence above equation means that the combination of first $k$ columns is also zero. Since the first $k$ columns are linearly independent, therefore, the linear combination is \textit{trivial}.Hence,
\begin{center}
$x_{1}\prime = x_{2}\prime = \ldots = x_{n}\prime $
\end{center}

therefore, the dimension of  $ \lbrace x : Ax_{0} = \overline 0 \rbrace$ is EXACTLY $n-k$. 

In the next lecture, we will analyse the row perspective.



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