# [SOLVED]R^3

#### karush

##### Well-known member
View attachment 9046

ok I thot if we get all zeros in 4th column of an augmented matrix that this would be a dependent linear set of vectors
but this example says independent

So I did this

View attachment 9047

#### HallsofIvy

##### Well-known member
MHB Math Helper
You say "
I thot if we get all zeros in 4th column of an augmented matrix that this would be a dependent linear set of vectors"

You "thot" wrong. First, we didn't "get" all zeros in the 4th column, they were set to 0 initially because these are "homogeneous" equations. The question is whether or not the solution to the equation is all zeros. If, as here, the solution is the "trivial" solution, with all values 0, then the vectors are independent.

For your work, yes, the vectors $$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$, $$\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}$$ and $$\begin{bmatrix}1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$ are linearly independent and, since there are three of them, form a basis for $$R^3$$. Personally, I wouldn't have used "matrices" to prove they are independent. From $$c_1\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}+ c_2 \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}+ c_3\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ I derive the three equations $$c_1+ c_2+ c_3= 0$$, $$c_1+ c_2= 0$$, and $$c_1= 0$$. The third equation immediately tells us that $$c_1= 0$$. Setting $$c_1= 0$$ in the second equation, it becomes $$0+ c_2= c_2= 0$$ so $$c_2= 0$$. Setting both $$c_1$$ and $$c_2$$ equal to 0 in the first equation, it becomes $$0+ 0+ c_3= 0$$ so $$c_3= 0$$. All three coefficients must be 0 so these vectors are independent and so form a basis for $$R^3$$.

Since they are a basis we must be able to write $$\begin{bmatrix}3 \\ 4 \\ 5 \end{bmatrix}$$ as a linear combination of them. That is, we can find numbers, $$c_1$$, $$c_2$$, and $$c_3$$ such that $$c_1\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}+ c_2\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}+ c_3\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}3 \\ 4 \\ 5 \end{bmatrix}$$.

That gives the three equations $$c_1+ c_2+ c_3= 3$$ $$c_1+ c_2= 4$$, and $$c_1= 5$$. Since $$c_1= 5$$, the second equation becomes $$5+ c_2= 4$$ so $$c_2= -1$$. Then the first equation becomes $$5- 1+ c_3= 4+ c_3= 3$$ so $$c_3= -1$$.

Yes, the vector written $$\begin{bmatrix}3 \\ 4 \\ 5\end{bmatrix}$$ in the "standard basis" will be written $$\begin{bmatrix}5 \\ -1 \\ -1 \end{bmatrix}$$ in this matrix.

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#### karush

##### Well-known member
ok really appreciate all the explanation.

I notice on the DE there is a lot a views