Image/Column space of a Matrix

**AshleyT** · August 14th 2011, 02:10 PM

Hey there!
I'm really struggling with this question

. The solutions on it don't seem to match up with everything i've read/googled!

Now, I understand how to get the kernel. But I have no idea what they are doing to get the image basis!

I've looked around on wikipedia etc and read that you take the corresponding columns to the pivots - which to me would mean the basis would be: (1 0 3), (-2, 2, -4)...Not (1; 2; 1); (2; 0; 6)

Please see the pdf attachment for the solutions.

Any help would be appreciated!

Thankyou!

**FernandoRevilla** · August 14th 2011, 11:24 PM

Originally Posted by AshleyT

Now, I understand how to get the kernel. But I have no idea what they are doing to get the image basis!

In general, consider $A:\mathbb{R}^m\to\mathbb{R}^n$ and suppose $\{u_1,u_2,\ldots,u_r\}$ is a basis of $\ker A$ and $B=\{u_1,u_2,\ldots,u_r,u_{r+1},\ldots,u_m\}$ is a basis of $\mathbb{R}^m$ .Then, $\{Au_{r+1},\ldots,Au_m\}$ is a basis of $\textrm{Im}A$ . Try to prove it (it is not difficult) and ask if you have any doubt.

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