If $\{a,b,c\}$ is a basis for $\Bbb R^3$, is $\{a+b,b-c,c-a\}$ is also a basis for $\Bbb R^3?$

Question

If $\{a,b,c\}$ is a basis for $\Bbb R^3$, is $\{a+b,b-c,c-a\}$ is also a basis for $\Bbb R^3?$

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What I thought was to look at the standard basis: $$a = (1,0,0)$$ $$b = (0,1,0)$$ $$c = (0,0,1)$$

For this basis we get: $$a+b = (1,1,0)$$ $$b-c = (0,1,-1)$$ $$c-a = (-1,0,1)$$

To check if the new vectors are linear independent, put them as rows in a matrix and rank, you get: $$\begin{bmatrix}1&1&0\\0&1&-1\\0&0&2\end{bmatrix}$$

Namely, the vectors are linear independent.

Is it true to conclude from this basis, the standard basis, for all the bases in $\Bbb R^3$ regarding the question above?

Answer 1

Yes it is enough to check that the matrix $$ C = \begin{bmatrix}1&1&0\\0&1&-1\\-1&0&1\end{bmatrix} $$ has linearly independent columns/rows. But the reason why this works even though you are just checking a special case is as follows:

To get the new set of vectors $\{a + b, b - c, c - a\}$ from your original basis $\{a, b, c\}$, you can write the basis vectors $a, b, c$ as $1 \times 3$ row vectors and stack them on top of each other to get a $3 \times 3$ matrix: $$ \begin{bmatrix}a\\b\\c\end{bmatrix} $$ It looks like a column vector but each row is whole vector so you have a matrix but matrix-matrix multiplication is nice so that you can pretend like it is a column vector when you multiply $C$ to it: $$ C \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1&1&0\\0&1&-1\\-1&0&1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}a + b\\b - c\\c - a\end{bmatrix} $$ Well you get your new basis stacked on top of one another as a matrix. Now you know that the matrix $\begin{bmatrix}a\\b\\c\end{bmatrix}$ has linearly independent rows because the rows form a basis. Also you have already checked that $C$ has linearly independent rows. Thus their product $\begin{bmatrix}a + b\\b - c\\c - a\end{bmatrix}$ must also have linearly independent rows i.e. your new set of vectors is a basis.

Answer 2

I don't entirely understand what you mean in your last paragraph, and your matrix really ought to be $$ \begin{bmatrix}1&1&0\\0&1&-1\\-1&0&1\end{bmatrix} $$ However, it seems you have actually taken this matrix, then applied a couple of row operations to get the upper triangular matrix that you have.

And yes, that is the standard way of checking that a set of vectors is a basis: Take your set of vectors, expressed in any convenient basis (in this case, the basis consisting of $(a,b,c)$ is very convenient), as the rows (or more conventionally columns) of a matrix and check that the determinant is non-zero. In this case the determinant of your upper triangular matrix is clearly $2$, which is non-zero.

The standard basis is irrelevant in this particular problem. In the basis given by $(a,b,c)$, the vectors $a,b$ and $c$ are indeed given by the coordinates you have listed, which means that $a+b, b-c$ and $c-a$ are also given by the coordinates you have found, and we can use the matrix method.