MHB Q01 are linearly independent vectors, so are....

karush
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Let A be invertible. Show that, if $\textbf{$v_i,v_2,v_j$}$ are linearly independent vectors, so are \textbf{$Av_1,Av_2,Av_3$}

https://drive.google.com/file/d/1OuHxfUdACbpK4E5aca2oBzdaxGR0IYKv/view?usp=sharing
p57.png


ok I think this is the the definition we need for this practice exam question,
However I tried to insert using a link but not successful
I thot if we use a link the image would always be there unless we delete its source

as to the question... not real sure of the answer since one $c_n$ may equal 0 and another may not

Anyway Mahalo...
 
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Yes, that is the definition of "linearly independent" you need. Now, what are you trying to prove?

You have "Show that, if $v_1$, $v_2$, $v_3$ are linearly independent, then so are $Av_1$, $Av_2$, $Av_3$" but what is "A"? If it is a general linear transformation this is not true. If A is an INVERTIBLE linear transformation then it is true and can be shown by applying $A^{-1}$ to both sides of$x_1Av_1+ x_2Av_2+ x_3Av_3= 0$.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

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