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This is from my textbook, let's say we have a matirx A and its RREF matrix R $$A=\begin{pmatrix}1 & 0 &3 \\2 & 1 & 5\\1 & 0 & 3 \end{pmatrix}$$ $$R=\begin{pmatrix}1 & 0 &3 \\0 & 1 & -1\\0 & 0 & 0 \end{pmatrix}$$

I can understand the column vector $V_3=3V_1-V_2$ as first and second columns are pivot columns, so the basis would be $\begin{pmatrix}1\\2\\1\end{pmatrix} and \begin{pmatrix}0\\1\\0\end{pmatrix} $ that span column space $C(A)$

But the textbook also says any two column vectors of A could be a basis for $C(A)$, how come?I'm really confused about it

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Note that the rank is 2. Also $V_1=\frac{1}{3}V_2+\frac{1}{3}V_3$, and $V_2=3V_1-V_3$.

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Quite simply any two column vectors of $A$ are linearly independent but the three aren't. This can be seen by inspection of your matrix $R$. As rank(R)=2 any two columns form a basis of your column space.

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  • $\begingroup$ My example looks like a special case in 3 vectors, what I really meant was let's say we have one more vector $V_4$, so $3v_1-v_2-v_3+v_4=0$, now we can't say any two vectors span C(A), can we? $\endgroup$ – whoisit Apr 10 '16 at 12:55
  • $\begingroup$ I believe Emilio answered your question? Sorry I was away from my computer $\endgroup$ – K.Power Apr 10 '16 at 15:55
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You have found that $3v_1-v_2-v_3=0$ so the three columns $\{v_1,v_2,v_3\}$ of the matrix are linearly dependent vector, but $\{v_1,v_2\}$, $\{v_1,v_3\}$ and $\{v_2,v_3\}$ are linearly independent, so any of these couple of vectors span the same space as $\{v_1,v_2,v_3\}$

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  • $\begingroup$ My example looks like a special case in 3 vectors, what I meant was let's say $3v_1-v_2-v_3+v_4=0$ , now we can't say any two vectors span C(A), can we? $\endgroup$ – whoisit Apr 10 '16 at 12:51
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    $\begingroup$ Not in general, you need the maximum number of vectors that are linearly independent. $\endgroup$ – Emilio Novati Apr 10 '16 at 12:53

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