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	Comments on: Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix	</title>
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				By: Find an Orthonormal Basis of the Range of a Linear Transformation &#8211; Problems in Mathematics				</title>
				<link>https://yutsumura.com/row-equivalent-matrix-bases-for-the-null-space-range-and-row-space-of-a-matrix/#comment-1636</link>
		<dc:creator><![CDATA[Find an Orthonormal Basis of the Range of a Linear Transformation &#8211; Problems in Mathematics]]></dc:creator>
		<pubDate>Mon, 26 Jun 2017 01:42:05 +0000</pubDate>
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					<description><![CDATA[[&#8230;] Since the both columns contain the leading $1$&#8217;s, we conclude that [left{, begin{bmatrix} 1 \ 0 \ 1 end{bmatrix}, begin{bmatrix} -1 \ 1 \ 1 end{bmatrix} ,right}] is a basis of the range of $A$ by the leading $1$ method. [&#8230;]]]></description>
		<content:encoded><![CDATA[<p>[&#8230;] Since the both columns contain the leading $1$&#8217;s, we conclude that [left{, begin{bmatrix} 1 \ 0 \ 1 end{bmatrix}, begin{bmatrix} -1 \ 1 \ 1 end{bmatrix} ,right}] is a basis of the range of $A$ by the leading $1$ method. [&#8230;]</p>
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				By: Range, null space, rank, and nullity of a linear transformation from $R^2$ to $R^3$ &#8211; Problems in Mathematics				</title>
				<link>https://yutsumura.com/row-equivalent-matrix-bases-for-the-null-space-range-and-row-space-of-a-matrix/#comment-879</link>
		<dc:creator><![CDATA[Range, null space, rank, and nullity of a linear transformation from $R^2$ to $R^3$ &#8211; Problems in Mathematics]]></dc:creator>
		<pubDate>Thu, 16 Mar 2017 15:49:15 +0000</pubDate>
		<guid isPermaLink="false">https://yutsumura.com/?p=1970#comment-879</guid>
					<description><![CDATA[[&#8230;] that these vectors are linearly independent, thus a basis for the range. (Basically, this is the leading 1 method.) Hence we have [calR(T)=calR(A)=Span left{begin{bmatrix} 1 \ 1 \ 0 end{bmatrix}, [&#8230;]]]></description>
		<content:encoded><![CDATA[<p>[&#8230;] that these vectors are linearly independent, thus a basis for the range. (Basically, this is the leading 1 method.) Hence we have [calR(T)=calR(A)=Span left{begin{bmatrix} 1 \ 1 \ 0 end{bmatrix}, [&#8230;]</p>
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